Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra More Ques

I. Choose the Correct Answer.

 

Question 1. The HCF of \( x^2 - y^2 \); \( x^3 - y^3 \) ... \( x^n - y^n \) where \( n \in N \) is
(1) x - y
(2) x + y
(4) do not intersect
Answer: (1) x - y
In simple words: When finding the HCF of algebraic expressions like \( x^n - y^n \) for any natural number \(n\), the common factor that divides all terms is always \( x-y \). This is a fundamental property of these types of expressions.

🎯 Exam Tip: When finding the HCF of expressions like \(x^n - y^n\), always look for the common factor that divides all terms, which is \(x-y\).

 

Question 2. Which of the following is correct.
(i) Every polynomial has finite number of multiples
(ii) LCM of two polynomials of degree "2" may be a constant
(iii) HCF of 2 polynomials may be a constant
(iv) Degree of HCF of two polynomials is always less than degree of L.C.M.
(1) (i) and (iii)
(2) (iii) and (iv)
(3) (iii) only
(4) (iv) only
Answer: (3) (iii) only
In simple words: When dealing with polynomials, it is true that the Highest Common Factor (HCF) of two polynomials can sometimes be a constant number, like 1 or 5. The other statements are not always correct.

🎯 Exam Tip: Remember that the HCF of polynomials can be a constant if the polynomials share only numerical common factors.

 

Question 3. The HCF of \( x^2 - 2xy + y^2 \) and \( x^4 - y^4 \) is ............
(1) 1
(2) x + y
(3) x - y
(4) x² – y²
Answer: (3) x - y
In simple words: First, we factorize the given expressions. \( x^2 - 2xy + y^2 \) is \((x-y)^2\). And \( x^4 - y^4 \) factors into \((x-y)(x+y)(x^2+y^2)\). The HCF is the largest common factor of both, which is \((x-y)\). This is the common part that divides both expressions.

🎯 Exam Tip: Always factorize polynomials completely to find their HCF or LCM. Remember common algebraic identities like difference of squares and perfect squares.

 

Question 4. The L.C.M. of \( a^k \), \( a^{k+3} \), \( a^{k+5} \) where \( K \in N \) is ............
(3) \( a^{k+6} \)
(4) \( a^{k+9} \)
Answer: (1) \( a^{k+5} \)
In simple words: To find the Least Common Multiple (LCM) of terms that have the same base but different powers, we simply pick the term with the biggest power. Here, \( k+5 \) is the biggest power among \( k \), \( k+3 \), and \( k+5 \). So, the LCM is \( a^{k+5} \).

🎯 Exam Tip: For terms with the same base, the LCM is the term with the highest power, and the HCF is the term with the lowest power.

 

Question 5. The LCM of \( (x + 1)^2 (x – 3) \) and \( (x^2 – 9) (x + 1) \) is
(1) \( (x + 1)^3 (x^2 – 9) \)
(2) \( (x + 1)^2 (x^2 – 9) \)
(3) \( (x + 1)^2 (x − 3) \)
(4) \( (x – 9) (x + 1) \)
Answer: (2) \( (x + 1)^2(x^2 – 9) \)
In simple words: First, we factorize the second expression: \( (x^2 - 9)(x + 1) \) becomes \( (x - 3)(x + 3)(x + 1) \). The first expression is \( (x + 1)^2 (x - 3) \). To find the LCM, we take the highest power of each unique factor. So, the LCM is \( (x+1)^2 (x-3)(x+3) \), which can also be written as \( (x+1)^2 (x^2-9) \).

🎯 Exam Tip: When finding the LCM of polynomials, ensure all expressions are fully factored into their prime polynomial components first.

 

Question 6. If \( \frac{a^{3}}{a-b} \) is added with \( \frac{b^{3}}{b-a} \) then the new expression is
(1) \( a^2 – ab + b^2 \)
(2) \( a^2 + ab + b^2 \)
(3) \( a^3 + b^3 \)
(4) \( a^3 – b^3 \)
Answer: (2) \( a^2 + ab + b^2 \)
In simple words: To add the fractions, we change \( \frac{b^3}{b-a} \) to \( -\frac{b^3}{a-b} \). Then we combine them to \( \frac{a^3 - b^3}{a-b} \). Using the formula \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \), we can cancel \( (a-b) \) from top and bottom, leaving \( a^2+ab+b^2 \).

🎯 Exam Tip: Always look for common denominators or ways to make denominators common, especially by factoring out a negative sign, when adding or subtracting rational expressions.

 

Question 7. The solution set of \( x + \frac { 1 }{ x } = \frac { 5 }{ 2 } \) is .............
(1) \( 2,\frac { 1 }{ 2 } \)
(2) \( 2,-\frac { 1 }{ 2 } \)
(3) \( -2, - \frac { 1 }{ 2 } \)
(4) \( -2, \frac { 7 }{ 2 } \)
Answer: (1) \( 2,\frac { 1 }{ 2 } \)
In simple words: To solve the equation, first multiply everything by \( 2x \) to remove the fractions, which gives \( 2x^2 + 2 = 5x \). Rearranging it as \( 2x^2 - 5x + 2 = 0 \) and factoring, we get \( (2x-1)(x-2) = 0 \). This means \( x = \frac{1}{2} \) or \( x = 2 \).

🎯 Exam Tip: When solving rational equations, always eliminate denominators first to convert them into standard polynomial forms, and remember to check for extraneous solutions if the original equation involved variables in the denominator.

 

Question 8. On dividing \( \frac{x^{2}-25}{x+3} \) by \( \frac{x+5}{x^{2}-9} \) is equal to ............
(1) \( (x - 5)(x + 3) \)
(2) \( (x + 5) (x – 3) \)
(3) \( (x – 5)(x – 3) \)
(4) \( (x + 5)(x + 3) \)
Answer: (3) \( (x - 5)(x – 3) \)
In simple words: To divide, we first factorize \( x^2-25 \) into \((x-5)(x+5)\) and \( x^2-9 \) into \((x-3)(x+3)\). Then, we multiply the first fraction by the second fraction's inverse. After canceling common terms like \((x+5)\) and \((x+3)\), we are left with \( (x-5)(x-3) \).

🎯 Exam Tip: Remember that dividing by a fraction is the same as multiplying by its reciprocal. Always factorize all polynomials before simplifying rational expressions.

 

Question 9. The square root of \( (x + 11)^2 – 44x \) is
(1) \( |(x – \)
(2)
(3) \( − \)
(4) \( |x − 11| \)
Answer: (4) \( |x − 11| \)
In simple words: First, expand \( (x + 11)^2 \) to \( x^2 + 22x + 121 \). Then, combine with \( -44x \) to get \( x^2 - 22x + 121 \). This expression is a perfect square, \((x - 11)^2\). So, its square root is \( |x-11| \).

🎯 Exam Tip: When finding the square root of an algebraic expression that simplifies to a perfect square, always use the absolute value notation (e.g., \( \sqrt{A^2} = |A| \)) to ensure the result is positive.

 

Question 10. If \( \alpha, \beta \) are the zeros of the polynomial \( p(x) = 4x^2 + 3x + 7 \) then \( \frac{1}{\alpha} + \frac{1}{\beta} \) is equal to
(1) \( \frac { 7 }{ 3 } \)
(3) \( \frac { 3 }{ 7 } \)
(4) \( -\frac { 3 }{ 7 } \)
Answer: (4) \( -\frac { 3 }{ 7 } \)
In simple words: For the polynomial, the sum of roots \( \alpha + \beta = -\frac{3}{4} \) and the product of roots \( \alpha \beta = \frac{7}{4} \). The expression \( \frac{1}{\alpha} + \frac{1}{\beta} \) can be written as \( \frac{\alpha + \beta}{\alpha \beta} \). Substituting the values, we get \( \frac{-\frac{3}{4}}{\frac{7}{4}} \), which simplifies to \( -\frac{3}{7} \).

🎯 Exam Tip: Always remember the formulas for the sum of roots (\( -\frac{b}{a} \)) and product of roots (\( \frac{c}{a} \)) for a quadratic equation \( ax^2 + bx + c = 0 \). They are fundamental for solving problems involving roots.

 

Question 11. The value of \( \sqrt{20+\sqrt{20+\sqrt{20+...}}} \) is ............
(1) -5
(2) 5
(3) 4
(4) -3
Answer: (2) 5
In simple words: Let the whole expression be \(x\). Then \( x = \sqrt{20+x} \). Squaring both sides gives \( x^2 = 20+x \), which simplifies to \( x^2 - x - 20 = 0 \). Factoring this gives \((x-5)(x+4)=0\). Since square roots must be positive, \(x=5\) is the correct answer.

🎯 Exam Tip: When dealing with infinite nested square roots like this, set the entire expression equal to a variable (e.g., \(x\)), then use the repeating nature to form a simple equation and solve it. Remember to discard negative solutions if the context requires a positive root.

 

Question 12. If \( \alpha \) and \( \beta \) are the roots of the equation \( ay^2 + bx + c = 0 \) then \( (\alpha + \beta)^2 \) is ............
(1) \( \frac{-b^{2}}{a^{2}} \)
(2) \( \frac{-c^{2}}{a^{2}} \)
(3) \( \frac{-b^{2}}{a^{2}} \)
(4) \( \frac { bc }{ a } \)
Answer: (3) \( \frac{-b^{2}}{a^{2}} \)
In simple words: For a quadratic equation like \( ay^2 + bx + c = 0 \), the sum of its roots, \( \alpha + \beta \), is found using the formula \( -\frac{b}{a} \). So, \( (\alpha + \beta)^2 \) means squaring this entire sum. It's important to know these basic formulas for roots.

🎯 Exam Tip: Always recall the basic formulas for sum and product of roots for a quadratic equation. Make sure you correctly apply the negative sign and squaring operations.

 

Question 13. The roots of the equation \( x^2 – 8x + 12 = 0 \) are
(1) real and equal
(2) real and rational
(4) unreal
Answer: (2) real and rational
In simple words: To find the nature of the roots, we calculate the discriminant, \( \Delta = b^2 - 4ac \). For \( x^2 - 8x + 12 = 0 \), \( \Delta = (-8)^2 - 4(1)(12) = 64 - 48 = 16 \). Since \( \Delta = 16 \) is a positive perfect square, the roots are real and rational.

🎯 Exam Tip: The discriminant (\( \Delta = b^2 - 4ac \)) tells you the nature of a quadratic equation's roots: \( \Delta > 0 \) means real and distinct roots, \( \Delta = 0 \) means real and equal roots, and \( \Delta < 0 \) means no real roots (unreal/complex).

 

Question 14. If one root of the equation is the reciprocal of the other root in \( ax^2 + bx + c = 0 \) then ............
(1) a = c
(2) a = b
(3) b = c
(4) c = 0
Answer: (1) a = c
In simple words: If one root is the reciprocal of the other (say, \( \alpha \) and \( \frac{1}{\alpha} \)), then their product is \( \alpha \times \frac{1}{\alpha} = 1 \). Since the product of roots is also \( \frac{c}{a} \), we have \( 1 = \frac{c}{a} \), which means \( a = c \).

🎯 Exam Tip: When roots have special relationships (like being reciprocals or negatives of each other), use the sum and product of roots formulas to find relationships between the coefficients.

 

Question 15. If \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + 2x + 8 = 0 \) then the value of \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \) is
(1) \( \frac { 1 }{ 2 } \)
(2) 6
(3) \( \frac { 3 }{ 2 } \)
(4) \( -\frac { 3 }{ 2 } \)
Answer: (4) \( -\frac { 3 }{ 2 } \)
In simple words: For the given equation, the sum of roots \( \alpha + \beta = -2 \) and the product of roots \( \alpha \beta = 8 \). The expression \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \) simplifies to \( \frac{\alpha^2 + \beta^2}{\alpha \beta} \). Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \), we get \( (-2)^2 - 2(8) = 4 - 16 = -12 \). So the final answer is \( \frac{-12}{8} = -\frac{3}{2} \).

🎯 Exam Tip: When asked for expressions involving sums of squares of roots (like \( \alpha^2 + \beta^2 \)), always relate them back to the sum and product of roots using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \).

 

Question 16. If \( \begin{pmatrix} x+y & x-y \\ 7 & 6 \end{pmatrix} = \begin{pmatrix} 10 & 2 \\ 7 & z \end{pmatrix} \) then x, y, z are
(2) 6, 6, 4
(3) 6, 4, 6
(4) 4, 4, 6
Answer: (3) 6, 4, 6
In simple words: When two matrices are equal, their matching elements must be equal. From this, we get \( x+y=10 \), \( x-y=2 \), and \( z=6 \). Solving the first two equations simultaneously gives \( x=6 \) and \( y=4 \). So, \( x=6, y=4, z=6 \).

🎯 Exam Tip: When two matrices are stated to be equal, always equate their corresponding elements to form a system of equations, which can then be solved for the unknown variables.

 

Question 17. If \( \begin{pmatrix} -1 & -2 & 4 \end{pmatrix} \begin{pmatrix} 2 \\ a \\ -3 \end{pmatrix} = -10 \) then the value of "a" is ............
(1) 2
(2) -4
(3) 4
(4) -2
Answer: (4) -2
In simple words: To solve for \(a\), we perform the matrix multiplication: \((-1)(2) + (-2)(a) + (4)(-3) = -10\). This simplifies to \( -2 - 2a - 12 = -10 \). Combining terms, \( -14 - 2a = -10 \). Adding 14 to both sides gives \( -2a = 4 \), so \( a = -2 \).

🎯 Exam Tip: Be careful with signs and order of operations when multiplying matrices. Remember that the number of columns in the first matrix must equal the number of rows in the second matrix for multiplication to be defined.

 

Question 18. The matrix A given by \( (a_{ij})_{2 \times 2} \) if \( a_{ij} = i - j \) is .............
(1) \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \)
(2) \( \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \)
(3) \( \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \)
(4) \( \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix} \)
Answer: (2) \( \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \)
In simple words: We calculate each element using the rule \( a_{ij} = i - j \). For example, \( a_{11} = 1-1=0 \), \( a_{12} = 1-2=-1 \), \( a_{21} = 2-1=1 \), and \( a_{22} = 2-2=0 \). Placing these elements in order gives the matrix \( \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \).

🎯 Exam Tip: When constructing a matrix from a given rule for its elements (\(a_{ij}\)), systematically calculate each element by substituting the row index \(i\) and column index \(j\) into the rule.

 

Question 19. If A is of order \( 4 \times 3 \) and B is of order \( 3 \times 4 \) then the order of BA is ............
(1) \( 3 \times 4 \)
(2) \( 4 \times 4 \)
(3) \( 3 \times 3 \)
(4) \( 4 \times 1 \)
Answer: (3) \( 3 \times 3 \)
In simple words: If A is \( 4 \times 3 \) and B is \( 3 \times 4 \), then for the product BA, we look at B's rows (3) and A's columns (3). Since B is \( 3 \times 4 \) and A is \( 4 \times 3 \), the number of columns in B (4) matches the number of rows in A (4). The resulting matrix BA will have the number of rows of B (3) and the number of columns of A (3), making its order \( 3 \times 3 \).

🎯 Exam Tip: The order of a product matrix \(MN\) is (rows of M) x (columns of N). The multiplication is only defined if (columns of M) equals (rows of N).

 

Question 20. If \( \begin{pmatrix} 4 & 3 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ -2 \\ x \end{pmatrix} = 6 \) then "x" is ............
(1) 1
(2) 2
(3) 3
(4) 4
Answer: (4) 4
In simple words: We multiply the row matrix by the column matrix: \((4 \times 1) + (3 \times -2) + (2 \times x)\). This sums to \( 4 - 6 + 2x \). Setting this equal to 6, we get \( -2 + 2x = 6 \). Solving for \(x\), we find \( 2x = 8 \), so \( x=4 \).

🎯 Exam Tip: Be methodical when performing matrix multiplication, especially with multiple terms. Ensure each element's product is correctly added to form the resulting scalar or matrix element.

II. Answer the Following.

 

Question 1. Solve \( x + y = 7; y + z = 4; z + x = 1 \)
Answer: We have a system of three linear equations: \( x + y = 7 \), \( y + z = 4 \), and \( z + x = 1 \). Adding all three equations together yields \( (x+y) + (y+z) + (z+x) = 7+4+1 \), which simplifies to \( 2x + 2y + 2z = 12 \). Dividing by 2, we get \( x+y+z = 6 \). Now, we can substitute each original equation back into this new sum. For example, using \( x+y=7 \) in \( x+y+z=6 \) gives \( 7+z=6 \), so \( z=-1 \). Similarly, using \( y+z=4 \) gives \( 4+x=6 \), so \( x=2 \). And using \( z+x=1 \) gives \( 1+y=6 \), so \( y=5 \). Therefore, the solution is \( x=2, y=5, z=-1 \). This method of adding all equations is very efficient for symmetric systems.
In simple words: Add all three equations together to get \( 2(x+y+z) = 12 \), so \( x+y+z = 6 \). Then substitute each original equation back into this sum to find the individual values of \(x\), \(y\), and \(z\). The solution is \( x=2, y=5, z=-1 \).

🎯 Exam Tip: For systems of equations where variables appear symmetrically (e.g., \(x+y\), \(y+z\), \(z+x\)), summing all equations can often quickly lead to the sum of all variables, which simplifies finding individual values.

 

Question 2. Find the HCF of \( 25x^4y^7; 35x^3y^8; 45x^3y^3 \)
Answer: To find the Highest Common Factor (HCF) of \( 25x^4y^7 \), \( 35x^3y^8 \), and \( 45x^3y^3 \), we first break down each term into its prime factors.
\( 25x^4y^7 = 5^2 \cdot x^4 \cdot y^7 \)
\( 35x^3y^8 = 5 \cdot 7 \cdot x^3 \cdot y^8 \)
\( 45x^3y^3 = 3^2 \cdot 5 \cdot x^3 \cdot y^3 \) Now, we identify the common factors and take the lowest power for each. For the numerical part, the common factor is 5. For the variable \(x\), the lowest power is \(x^3\). For the variable \(y\), the lowest power is \(y^3\). Therefore, the HCF is \( 5x^3y^3 \). This method of finding common factors is very systematic.
In simple words: Find the prime factors for the numbers (25, 35, 45) and the lowest power for each common variable (\(x\) and \(y\)). The HCF is \( 5x^3y^3 \).

🎯 Exam Tip: To find the HCF of monomials, find the HCF of the coefficients and the lowest power of each common variable. For LCM, find the LCM of coefficients and the highest power of each variable present.

 

Question 3. Find the values of k for which the following equation has equal roots. \( (k – 12)x^2 + 2(k – 12)x + 2 = 0 \)
Answer: For the quadratic equation \( (k-12)x^2 + 2(k-12)x + 2 = 0 \) to have equal roots, its discriminant must be zero. The discriminant is given by \( \Delta = b^2 - 4ac \). In this equation, \( a = k-12 \), \( b = 2(k-12) \), and \( c = 2 \). Plugging these into the discriminant formula, we get \( (2(k-12))^2 - 4(k-12)(2) = 0 \). This simplifies to \( 4(k-12)^2 - 8(k-12) = 0 \). We can factor out \( 4(k-12) \), which leaves us with \( 4(k-12)((k-12) - 2) = 0 \). Further simplification yields \( 4(k-12)(k-14) = 0 \). From this, we find two possible values for \(k\): \( k-12=0 \implies k=12 \) or \( k-14=0 \implies k=14 \). These are the values for which the roots are equal.
In simple words: For equal roots, the discriminant (\(b^2-4ac\)) must be zero. For this equation, \( (2(k-12))^2 - 4(k-12)(2) = 0 \). Solving this gives \( 4(k-12)(k-14) = 0 \), so \( k=12 \) or \( k=14 \).

🎯 Exam Tip: When solving for parameters that lead to specific root conditions (like equal roots), always set the discriminant to zero. Double-check if any derived parameter values make the leading coefficient zero, as this would change the equation from quadratic to linear.

 

Question 4. Find the LCM of \( x^3 + y^3 \); \( x^3 – y^3 \); \( x^4 + x^2y^2 + y^4 \)
Answer: To find the Least Common Multiple (LCM) of \( x^3 + y^3 \), \( x^3 - y^3 \), and \( x^4 + x^2y^2 + y^4 \), we first need to factorize each expression completely.
\( x^3 + y^3 = (x+y)(x^2-xy+y^2) \)
\( x^3 - y^3 = (x-y)(x^2+xy+y^2) \)
\( x^4 + x^2y^2 + y^4 = (x^2+y^2)^2 - (xy)^2 = (x^2-xy+y^2)(x^2+xy+y^2) \) The LCM is the product of all unique factors, each raised to its highest power. In this case, it is \( (x+y)(x-y)(x^2-xy+y^2)(x^2+xy+y^2) \). This product simplifies to \( (x^3+y^3)(x^3-y^3) \), which is also \( (x^3)^2 - (y^3)^2 = x^6 - y^6 \). Factoring algebraic sums and differences of cubes is essential here.
In simple words: Factorize all three expressions. \( x^3+y^3 = (x+y)(x^2-xy+y^2) \), \( x^3-y^3 = (x-y)(x^2+xy+y^2) \), and \( x^4+x^2y^2+y^4 = (x^2-xy+y^2)(x^2+xy+y^2) \). The LCM is the product of all unique factors, which simplifies to \( (x^3+y^3)(x^3-y^3) = x^6-y^6 \).

🎯 Exam Tip: Always start by completely factoring all given polynomial expressions. The LCM includes every unique factor present in any of the polynomials, raised to its highest power found in any single polynomial.

 

Question 5. The sum of two numbers is 15. If the sum of their reciprocals is \( \frac{3}{10} \), find the numbers.
Answer: Let the two unknown numbers be \( \alpha \) and \( \beta \). Their sum is 15, so \( \alpha + \beta = 15 \). The sum of their reciprocals is \( \frac{3}{10} \), meaning \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{10} \). We can rewrite the sum of reciprocals as \( \frac{\beta + \alpha}{\alpha \beta} = \frac{3}{10} \). Substituting \( \alpha + \beta = 15 \) into this equation gives \( \frac{15}{\alpha \beta} = \frac{3}{10} \). Solving for \( \alpha \beta \), we find \( 3 \alpha \beta = 150 \), so \( \alpha \beta = 50 \). Now we have the sum (\(15\)) and product (\(50\)) of the two numbers. We can form a quadratic equation whose roots are these numbers: \( x^2 - (\alpha + \beta)x + \alpha \beta = 0 \). This becomes \( x^2 - 15x + 50 = 0 \). Factoring this quadratic, we get \((x-10)(x-5) = 0\), so the numbers are 10 and 5. This method uses the relationship between roots and coefficients.
In simple words: Let the numbers be \( \alpha \) and \( \beta \). We have \( \alpha + \beta = 15 \) and \( \frac{\alpha+\beta}{\alpha \beta} = \frac{3}{10} \). Substituting the sum, we find \( \alpha \beta = 50 \). Using these values to form a quadratic equation \( x^2 - 15x + 50 = 0 \), we solve for \(x\) to get the numbers 10 and 5.

🎯 Exam Tip: When given the sum and sum of reciprocals of two numbers, convert the reciprocal sum into terms of the original sum and product. Then, form a quadratic equation whose roots are the numbers to easily find them.

 

Question 6. For What value of k, the G.C.D. of \( [x^2 + x − (2k + 2)] \) and \( 2x^2 + kx – 12 \) is \( (x + 4) \)?
Answer: We are given two polynomials, \( p(x) = x^2 + x - (2k + 2) \) and \( g(x) = 2x^2 + kx - 12 \). Their Greatest Common Divisor (GCD) is stated to be \( (x+4) \). This means that \( (x+4) \) is a factor of both polynomials. According to the Factor Theorem, if \( (x+4) \) is a factor, then evaluating the polynomial at \( x = -4 \) should result in zero. We can use either polynomial. Let's use \( g(x) \).
Substitute \( x = -4 \) into \( g(x) \): \( 2(-4)^2 + k(-4) - 12 = 0 \)
This simplifies to \( 2(16) - 4k - 12 = 0 \), which is \( 32 - 4k - 12 = 0 \).
Further simplifying, \( 20 - 4k = 0 \).
Solving for \( k \): \( 4k = 20 \implies k = 5 \). We can check this value with \( p(x) \) as well. If \( k=5 \), then \( p(-4) = (-4)^2 + (-4) - (2(5)+2) = 16 - 4 - (10+2) = 12 - 12 = 0 \), confirming that \( k=5 \) is correct. This method uses the property that factors imply zeros.
In simple words: Since \( (x+4) \) is the GCD, it must be a factor of both polynomials. This means if we substitute \( x=-4 \) into either polynomial, the result must be zero. Using \( 2x^2 + kx - 12 = 0 \), we substitute \( x=-4 \) to get \( 2(16) - 4k - 12 = 0 \), which simplifies to \( 20 - 4k = 0 \), so \( k=5 \).

🎯 Exam Tip: If a polynomial has a factor \( (x-a) \), then \( a \) is a root of the polynomial, meaning \( P(a) = 0 \). This is a powerful tool (Factor Theorem) for finding unknown coefficients when factors are given.

 

Question 7. Simplify \( \frac{x^{2}+x-6}{x^{2}+4x+3} \)
Answer: To simplify the rational expression \( \frac{x^2+x-6}{x^2+4x+3} \), we need to factorize both the numerator and the denominator.
For the numerator, \( x^2+x-6 \), we look for two numbers that multiply to -6 and add to 1. These are 3 and -2. So, \( x^2+x-6 = (x+3)(x-2) \).
For the denominator, \( x^2+4x+3 \), we look for two numbers that multiply to 3 and add to 4. These are 3 and 1. So, \( x^2+4x+3 = (x+3)(x+1) \).
Now, the expression becomes \( \frac{(x+3)(x-2)}{(x+3)(x+1)} \). We can cancel out the common factor \( (x+3) \) from both the numerator and the denominator. The simplified expression is \( \frac{x-2}{x+1} \). Factoring is a crucial step in simplifying algebraic fractions.
In simple words: Factorize the numerator \( x^2+x-6 \) into \( (x+3)(x-2) \). Factorize the denominator \( x^2+4x+3 \) into \( (x+3)(x+1) \). Then cancel the common factor \( (x+3) \), leaving \( \frac{x-2}{x+1} \).

🎯 Exam Tip: Always factorize polynomials in the numerator and denominator completely before attempting to cancel common factors. This prevents errors from incorrect simplification.

 

Question 8. Multiply \( \frac{a^{3}b^{2}}{a-1} \) by \( \frac{a^{2}-1}{a^{2}b^{3}} \)
Answer: To multiply the two algebraic fractions \( \frac{a^3b^2}{a-1} \) and \( \frac{a^2-1}{a^2b^3} \), we first factorize any expressions that can be simplified. The term \( a^2-1 \) is a difference of squares, so it factors into \((a-1)(a+1)\).
Now, the multiplication becomes \( \frac{a^3b^2}{a-1} \times \frac{(a-1)(a+1)}{a^2b^3} \).
We can cancel out the common factor \((a-1)\) from the numerator and denominator. We also simplify the powers of \(a\) and \(b\). \(a^3\) divided by \(a^2\) leaves \(a\), and \(b^2\) divided by \(b^3\) leaves \(b\) in the denominator.
So, the simplified product is \( \frac{a(a+1)}{b} \). Factoring and canceling common terms makes multiplying fractions easier.
In simple words: Factorize \( a^2-1 \) into \((a-1)(a+1)\). Then, multiply the fractions and cancel common terms like \( (a-1) \), \( a^2 \), and \( b^2 \). The result is \( \frac{a(a+1)}{b} \).

🎯 Exam Tip: Before multiplying algebraic fractions, factorize all numerators and denominators. This allows you to cancel common factors across the fractions before multiplying, simplifying the process.

 

Question 9. If \( P = \frac{x^{2}-36}{x^{2}-49} \) and \( Q = \frac {x+6}{x+7 } \) find the value of \( \frac { P }{ Q } \)
Answer: We are given two algebraic expressions, \( P = \frac{x^2-36}{x^2-49} \) and \( Q = \frac{x+6}{x+7} \), and asked to find the value of \( \frac{P}{Q} \).
First, we factorize \(P\). The numerator \( x^2-36 \) is a difference of squares, so it becomes \((x-6)(x+6)\). The denominator \( x^2-49 \) is also a difference of squares, becoming \((x-7)(x+7)\).
So, \( P = \frac{(x-6)(x+6)}{(x-7)(x+7)} \).
Now, to find \( \frac{P}{Q} \), we divide \(P\) by \(Q\), which means multiplying \(P\) by the reciprocal of \(Q\):
\( \frac{P}{Q} = \frac{(x-6)(x+6)}{(x-7)(x+7)} \times \frac{x+7}{x+6} \).
We can cancel the common factors \((x+6)\) and \((x+7)\) from the numerator and denominator. The simplified result is \( \frac{x-6}{x-7} \). Recognizing differences of squares is very helpful in these problems.
In simple words: Factorize \( P \) to \( \frac{(x-6)(x+6)}{(x-7)(x+7)} \). To find \( \frac{P}{Q} \), multiply \( P \) by the reciprocal of \( Q \), which is \( \frac{x+7}{x+6} \). After canceling common terms, the simplified result is \( \frac{x-6}{x-7} \).

🎯 Exam Tip: When dividing algebraic fractions, always factorize all expressions completely, then multiply the first fraction by the reciprocal of the second, and cancel common factors.

 

Question 10. Simplify \( \frac { x }{ x+y } – \frac { y }{ x-y } \)
Answer: To simplify the expression \( \frac{x}{x+y} - \frac{y}{x-y} \), we need to find a common denominator for the two fractions. The common denominator will be the product of their individual denominators, which is \( (x+y)(x-y) \).
Now, we rewrite each fraction with this common denominator:
\( \frac{x(x-y)}{(x+y)(x-y)} - \frac{y(x+y)}{(x-y)(x+y)} \).
Next, expand the numerators: \( (x^2-xy) - (xy+y^2) \).
Combining these over the common denominator \( (x^2-y^2) \), we get \( \frac{x^2-xy - xy-y^2}{x^2-y^2} \).
Finally, combine like terms in the numerator: \( \frac{x^2-2xy-y^2}{x^2-y^2} \). This process of finding common denominators is fundamental for adding or subtracting fractions.
In simple words: Find a common denominator \( (x+y)(x-y) \). Rewrite both fractions with this denominator, then combine the numerators. The result is \( \frac{x^2-2xy-y^2}{x^2-y^2} \).

🎯 Exam Tip: When adding or subtracting algebraic fractions, always find a common denominator by multiplying the individual denominators. Be very careful with distributing negative signs, especially when subtracting.

 

Question 11. Find the square root of \( (x + 11)^2 – 44x \)
Answer: To find the square root of the expression \( (x + 11)^2 - 44x \), we begin by expanding the squared term: \( (x + 11)^2 = x^2 + 22x + 121 \).
Substituting this back into the expression, we have \( x^2 + 22x + 121 - 44x \).
Combining the \(x\) terms, we get \( x^2 - 22x + 121 \). This trinomial is a perfect square, specifically \((x-11)^2\).
Therefore, the square root of the original expression is \( \sqrt{(x-11)^2} \). The square root of a squared term is its absolute value, so the final answer is \( |x-11| \). Recognizing perfect squares helps simplify these problems quickly.
In simple words: Expand \( (x+11)^2 \) to \( x^2+22x+121 \). Then combine with \( -44x \) to get \( x^2-22x+121 \), which is \((x-11)^2\). The square root is \( |x-11| \).

🎯 Exam Tip: When asked for the square root of an algebraic expression, first simplify the expression to see if it becomes a perfect square. If it does, remember to use the absolute value notation for the final result.

 

Question 12. Find the square root of \( x^4 + \frac{1}{x^{4}} + 2 \)
Answer: To find the square root of \( x^4 + \frac{1}{x^4} + 2 \), we can recognize this expression as a perfect square trinomial. It matches the pattern \( A^2 + B^2 + 2AB = (A+B)^2 \).
Here, let \( A = x^2 \) and \( B = \frac{1}{x^2} \).
Then \( A^2 = (x^2)^2 = x^4 \), \( B^2 = \left(\frac{1}{x^2}\right)^2 = \frac{1}{x^4} \), and \( 2AB = 2(x^2)\left(\frac{1}{x^2}\right) = 2 \).
So, the expression can be written as \( \left(x^2 + \frac{1}{x^2}\right)^2 \).
Taking the square root, we get \( \sqrt{\left(x^2 + \frac{1}{x^2}\right)^2} \). Since \(x^2 + \frac{1}{x^2}\) will always be positive for real values of \(x\) (as squares are positive), the result is \( x^2 + \frac{1}{x^2} \). This is a common algebraic identity to remember.
In simple words: Recognize the expression as a perfect square of \( \left(x^2 + \frac{1}{x^2}\right)^2 \). Therefore, the square root is simply \( x^2 + \frac{1}{x^2} \).

🎯 Exam Tip: Be familiar with perfect square identities, such as \( (A+B)^2 = A^2 + 2AB + B^2 \). Recognizing these patterns can greatly simplify expressions involving powers and reciprocals.

 

Question 13. Solve the equation \( 2x – 1 – \frac { 2 }{ x-2 } = 3 \)
Answer: To solve the equation \( 2x - 1 - \frac{2}{x-2} = 3 \), we first eliminate the fraction by multiplying the entire equation by the denominator, which is \( (x-2) \). This gives us:
\( (2x-1)(x-2) - 2 = 3(x-2) \)
Expand the terms: \( 2x^2 - 4x - x + 2 - 2 = 3x - 6 \)
Simplify: \( 2x^2 - 5x = 3x - 6 \)
Move all terms to one side to form a quadratic equation: \( 2x^2 - 8x + 6 = 0 \)
Divide the entire equation by 2 to simplify: \( x^2 - 4x + 3 = 0 \)
Factor this quadratic equation: \( (x-3)(x-1) = 0 \)
This gives two solutions: \( x=3 \) or \( x=1 \). Since neither of these values makes the original denominator zero, both are valid solutions. The solution set is \( \{1, 3\} \). Clearing denominators is a key strategy.
In simple words: Multiply the whole equation by \( (x-2) \) to remove the fraction. This results in \( 2x^2 - 8x + 6 = 0 \), which simplifies to \( x^2 - 4x + 3 = 0 \). Factoring gives \( (x-3)(x-1)=0 \), so the solutions are \( x=1 \) and \( x=3 \).

🎯 Exam Tip: When solving rational equations, always multiply by the LCD to eliminate denominators. After solving the resulting polynomial equation, check if any of the solutions make the original denominators zero; such solutions are extraneous and must be excluded.

 

Question 14. Find the roots of \( \sqrt { 2 } x^2 + 7x + 5\sqrt { 2 } = 0 \)
Answer: To find the roots of the quadratic equation \( \sqrt{2} x^2 + 7x + 5\sqrt{2} = 0 \), we can use the factorization method by splitting the middle term. We need two numbers whose product is \( ac = (\sqrt{2})(5\sqrt{2}) = 10 \) and whose sum is \( b=7 \). The numbers 2 and 5 fit this criteria.
Rewrite the equation: \( \sqrt{2} x^2 + 2x + 5x + 5\sqrt{2} = 0 \)
Group and factor: \( \sqrt{2} x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0 \)
Factor out the common binomial: \( (x + \sqrt{2})(\sqrt{2} x + 5) = 0 \)
Setting each factor to zero gives the roots:
\( x + \sqrt{2} = 0 \implies x = -\sqrt{2} \)
\( \sqrt{2} x + 5 = 0 \implies \sqrt{2} x = -5 \implies x = -\frac{5}{\sqrt{2}} \) The second root can be rationalized to \( -\frac{5\sqrt{2}}{2} \). The roots are \( -\sqrt{2} \) and \( -\frac{5\sqrt{2}}{2} \). Splitting the middle term is an effective technique for factoring quadratics.
In simple words: Split the middle term \( 7x \) into \( 2x+5x \). Factor by grouping: \( \sqrt{2}x(x+\sqrt{2}) + 5(x+\sqrt{2}) = 0 \). This gives \( (x+\sqrt{2})(\sqrt{2}x+5)=0 \). The roots are \( x=-\sqrt{2} \) and \( x=-\frac{5}{\sqrt{2}} \) (or \( -\frac{5\sqrt{2}}{2} \)).

🎯 Exam Tip: When solving quadratic equations with irrational coefficients, try splitting the middle term. Look for two numbers that multiply to \(ac\) and add to \(b\), even if \(a\) or \(c\) involve radicals.

 

Question 15. Solve \( \sqrt {x+5 } = 2x + 3 \) using formula method.
Answer:
Given the equation: \( \sqrt {x+5 } = 2x + 3 \)
To solve this, we first square both sides of the equation.
\( (\sqrt { x+5 })^2 = (2x + 3)^2 \)
\( x + 5 = 4x^2 + 12x + 9 \)
Now, rearrange the terms to form a standard quadratic equation (making one side equal to zero).
\( 0 = 4x^2 + 12x - x + 9 - 5 \)
\( 0 = 4x^2 + 11x + 4 \)
This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 4, b = 11, c = 4 \).
We can use the quadratic formula to find the values of x:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute the values of a, b, and c into the formula.
\( x = \frac{-11 \pm \sqrt{11^2 - 4(4)(4)}}{2(4)} \)
\( x = \frac{-11 \pm \sqrt{121 - 64}}{8} \)
\( x = \frac{-11 \pm \sqrt{57}}{8} \)
The two possible solutions for x are \( \frac{-11 + \sqrt{57}}{8} \) and \( \frac{-11 - \sqrt{57}}{8} \).
The solution set is \( \left\{ \frac{-11 + \sqrt{57}}{8}, \frac{-11 - \sqrt{57}}{8} \right\} \).
In simple words: First, remove the square root by squaring both sides. Then, rearrange the equation into the standard quadratic form. Finally, use the quadratic formula to find the two possible values for x. Remember to check if both solutions are valid in the original equation, especially when dealing with square roots.

🎯 Exam Tip: Always verify your solutions by substituting them back into the original equation, especially when dealing with square roots, to ensure no extraneous roots are included.

 

Question 16. The sum of a number and its reciprocal is \( \frac { 37 }{ 6 } \). Find the number.
Answer:
Let the required number be "x".
Its reciprocal is \( \frac { 1 }{ x } \).
According to the given information, the sum of the number and its reciprocal is \( \frac { 37 }{ 6 } \).
So, we write the equation:
\( x + \frac { 1 }{ x } = \frac { 37 }{ 6 } \)
To solve this, multiply the entire equation by \( 6x \) to eliminate the denominators.
\( 6x(x) + 6x(\frac { 1 }{ x }) = 6x(\frac { 37 }{ 6 }) \)
\( 6x^2 + 6 = 37x \)
Rearrange the terms to form a quadratic equation:
\( 6x^2 - 37x + 6 = 0 \)
Now, we can solve this quadratic equation by factoring or using the quadratic formula. Let's factor by splitting the middle term. We need two numbers that multiply to \( 6 \times 6 = 36 \) and add up to -37. These numbers are -1 and -36.
\( 6x^2 - x - 36x + 6 = 0 \)
Factor by grouping:
\( x(6x - 1) - 6(6x - 1) = 0 \)
\( (6x - 1)(x - 6) = 0 \)
Set each factor to zero to find the possible values of x:
\( 6x - 1 = 0 \implies 6x = 1 \implies x = \frac { 1 }{ 6 } \)
\( x - 6 = 0 \implies x = 6 \)
The required number is \( \frac { 1 }{ 6 } \) or \( 6 \). Both numbers satisfy the condition.
In simple words: We set up an equation where a number plus its flip (reciprocal) equals \( \frac { 37 }{ 6 } \). We turn this into a quadratic equation and solve it, finding two possible numbers that fit the description: \( \frac { 1 }{ 6 } \) and 6.

🎯 Exam Tip: When dealing with sums of numbers and their reciprocals, always clear the denominators to form a quadratic equation, which is typically easier to solve.

 

Question 17. Determine the nature of the roots of the equation \( 2x^2 + x - 1 = 0 \).
Answer:
Given the quadratic equation: \( 2x^2 + x - 1 = 0 \)
This equation is in the standard form \( ax^2 + bx + c = 0 \).
Comparing the given equation with the standard form, we have:
\( a = 2 \)
\( b = 1 \)
\( c = -1 \)
The nature of the roots of a quadratic equation is determined by its discriminant, \( \Delta \), which is calculated as \( \Delta = b^2 - 4ac \).
Let's calculate the discriminant:
\( \Delta = (1)^2 - 4(2)(-1) \)
\( \Delta = 1 - (-8) \)
\( \Delta = 1 + 8 \)
\( \Delta = 9 \)
Since the discriminant \( \Delta = 9 \) is greater than 0 (\( \Delta > 0 \)) and is a perfect square, the roots are real and unequal (distinct) and also rational. If it were not a perfect square but still positive, the roots would be real and unequal (distinct) but irrational.
In simple words: We look at a special number called the discriminant, which is \( b^2 - 4ac \). If this number is positive and a perfect square (like 9), it means the solutions (roots) to the equation are real numbers, different from each other, and can be written as simple fractions.

🎯 Exam Tip: Remember the three conditions for the discriminant \( \Delta \): if \( \Delta > 0 \), roots are real and distinct; if \( \Delta = 0 \), roots are real and equal; if \( \Delta < 0 \), roots are not real (complex).

 

Question 18. Find the value of k for which the given equation \( 9x^2 + 3kx + 4 = 0 \) has real and equal roots.
Answer:
Given the quadratic equation: \( 9x^2 + 3kx + 4 = 0 \)
This equation is in the standard form \( ax^2 + bx + c = 0 \).
Comparing the given equation with the standard form, we have:
\( a = 9 \)
\( b = 3k \)
\( c = 4 \)
For a quadratic equation to have real and equal roots, its discriminant (\( \Delta \)) must be equal to zero. The discriminant is given by the formula \( \Delta = b^2 - 4ac \).
Set the discriminant to zero:
\( b^2 - 4ac = 0 \)
Substitute the values of a, b, and c into the formula:
\( (3k)^2 - 4(9)(4) = 0 \)
\( 9k^2 - 144 = 0 \)
Now, solve for k:
\( 9k^2 = 144 \)
\( k^2 = \frac { 144 }{ 9 } \)
\( k^2 = 16 \)
Take the square root of both sides:
\( k = \sqrt { 16 } \)
\( k = \pm 4 \)
Thus, the values of k for which the equation has real and equal roots are \( +4 \) and \( -4 \). This ensures the quadratic has exactly one unique real solution.
In simple words: For an equation to have roots that are real and exactly the same, a special value called the discriminant must be zero. We use this rule to set up an equation with 'k' and solve it, finding that 'k' can be either 4 or -4.

🎯 Exam Tip: Remember that real and equal roots imply the discriminant is exactly zero, which leads to only one distinct solution for k in such problems.

 

Question 19. If one root of the equation \( 3x^2 – 10x + 3 = 0 \) is \( \frac { 1 }{ 3 } \) find the other root.
Answer:
Given the quadratic equation: \( 3x^2 – 10x + 3 = 0 \)
Let \( \alpha \) and \( \beta \) be the roots of this equation.
Comparing the given equation with the standard form \( ax^2 + bx + c = 0 \), we have \( a = 3, b = -10, c = 3 \).
We know the sum of the roots is \( \alpha + \beta = -\frac{b}{a} \) and the product of the roots is \( \alpha \beta = \frac{c}{a} \).
Sum of the roots: \( \alpha + \beta = -(\frac{-10}{3}) = \frac { 10 }{ 3 } \)
Product of the roots: \( \alpha \beta = \frac { 3 }{ 3 } = 1 \)
We are given that one of the roots is \( \frac { 1 }{ 3 } \). Let's say \( \alpha = \frac { 1 }{ 3 } \).
Now, we can use the product of the roots formula to find the other root (\( \beta \)). This method is often simpler when one root is given.
\( \alpha \beta = 1 \)
Substitute \( \alpha = \frac { 1 }{ 3 } \):
\( \frac { 1 }{ 3 } \times \beta = 1 \)
Multiply both sides by 3 to solve for \( \beta \):
\( \beta = 3 \)
Therefore, the other root of the equation is 3.
In simple words: For any quadratic equation, if you multiply its two solutions (roots), you get \( c/a \). Since we know one solution and \( c/a \), we can easily find the other solution by simple division.

🎯 Exam Tip: When one root of a quadratic equation is given, using the product of roots formula (\( \alpha \beta = \frac{c}{a} \)) is usually the most efficient way to find the other root.

 

Question 20. Form the quadratic equation whose roots are \( 3 + \sqrt { 7 } \); \( 3 – \sqrt { 7 } \).
Answer:
Given the roots of the quadratic equation are \( \alpha = 3 + \sqrt { 7 } \) and \( \beta = 3 – \sqrt { 7 } \).
A quadratic equation can be formed using the formula: \( x^2 - (\text{Sum of the roots})x + (\text{Product of the roots}) = 0 \).
First, calculate the sum of the roots:
Sum of the roots \( = \alpha + \beta = (3 + \sqrt { 7 }) + (3 – \sqrt { 7 }) \)
\( = 3 + 3 + \sqrt { 7 } – \sqrt { 7 } \)
\( = 6 \)
Next, calculate the product of the roots:
Product of the roots \( = \alpha \beta = (3 + \sqrt { 7 })(3 – \sqrt { 7 }) \)
This is in the form \( (a+b)(a-b) = a^2 - b^2 \).
\( = 3^2 - (\sqrt { 7 })^2 \)
\( = 9 - 7 \)
\( = 2 \)
Now, substitute the sum and product of the roots into the quadratic equation formula:
\( x^2 - (6)x + 2 = 0 \)
So, the required quadratic equation is \( x^2 - 6x + 2 = 0 \). This equation has these specific irrational roots.
In simple words: To build a quadratic equation from its solutions, first add the solutions together and then multiply them. After that, use the formula: \( x^2 \) minus the sum, plus the product, all equal to zero.

🎯 Exam Tip: Remember the general formula for forming a quadratic equation from its roots: \( x^2 - (\alpha + \beta)x + \alpha \beta = 0 \). This simplifies the process for conjugate roots.

 

Question 21. If \( \alpha \) and \( \beta \) are the roots of the equation \( 3x^2 – 5x + 2 = 0 \), then find the value of \( \alpha – \beta \).
Answer:
Given the quadratic equation: \( 3x^2 – 5x + 2 = 0 \)
Comparing this with the standard form \( ax^2 + bx + c = 0 \), we have \( a = 3, b = -5, c = 2 \).
We know the sum of the roots: \( \alpha + \beta = -\frac{b}{a} = -(\frac{-5}{3}) = \frac { 5 }{ 3 } \)
And the product of the roots: \( \alpha \beta = \frac{c}{a} = \frac { 2 }{ 3 } \)
We need to find the value of \( \alpha – \beta \). We know the identity \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta \). This identity helps us find the difference of roots without finding the roots themselves.
Let's calculate \( (\alpha - \beta)^2 \):
\( (\alpha - \beta)^2 = (\frac { 5 }{ 3 })^2 - 4(\frac { 2 }{ 3 }) \)
\( (\alpha - \beta)^2 = \frac { 25 }{ 9 } - \frac { 8 }{ 3 } \)
To subtract these fractions, find a common denominator, which is 9.
\( (\alpha - \beta)^2 = \frac { 25 }{ 9 } - \frac { 8 \times 3 }{ 3 \times 3 } \)
\( (\alpha - \beta)^2 = \frac { 25 }{ 9 } - \frac { 24 }{ 9 } \)
\( (\alpha - \beta)^2 = \frac { 1 }{ 9 } \)
Now, take the square root of both sides to find \( \alpha – \beta \):
\( \alpha - \beta = \sqrt { \frac { 1 }{ 9 } } \)
\( \alpha - \beta = \pm \frac { 1 }{ 3 } \)
The value of \( \alpha – \beta \) is \( \pm \frac { 1 }{ 3 } \). The sign depends on which root is designated as \( \alpha \) or \( \beta \).
In simple words: First, find the sum and product of the roots from the given equation. Then, use a special formula that relates the difference of the roots to their sum and product. After calculating, we find that the difference between the roots can be either \( \frac{1}{3} \) or \( -\frac{1}{3} \).

🎯 Exam Tip: Always remember the algebraic identity \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta \), as it directly calculates the square of the difference of roots using their sum and product, which are easily found from the quadratic coefficients.

 

Question 22. Determine the matrix A = \( (a_{ij})_{3 \times 2} \) if \( a_{ij} = 3i – 2j \).
Answer:
The matrix A is a \( 3 \times 2 \) matrix, meaning it has 3 rows and 2 columns. Its elements are denoted by \( a_{ij} \), where 'i' is the row number and 'j' is the column number.
The general form of a \( 3 \times 2 \) matrix is:
\[ A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} \]
The rule for each element is given as \( a_{ij} = 3i – 2j \). Let's calculate each element:
For the first row (i = 1):
\( a_{11} = 3(1) - 2(1) = 3 - 2 = 1 \)
\( a_{12} = 3(1) - 2(2) = 3 - 4 = -1 \)
For the second row (i = 2):
\( a_{21} = 3(2) - 2(1) = 6 - 2 = 4 \)
\( a_{22} = 3(2) - 2(2) = 6 - 4 = 2 \)
For the third row (i = 3):
\( a_{31} = 3(3) - 2(1) = 9 - 2 = 7 \)
\( a_{32} = 3(3) - 2(2) = 9 - 4 = 5 \)
Now, substitute these calculated values into the matrix form:
\[ A = \begin{bmatrix} 1 & -1 \\ 4 & 2 \\ 7 & 5 \end{bmatrix} \]
In simple words: We are given a rule to make a matrix with 3 rows and 2 columns. For each spot in the matrix, we use its row number (i) and column number (j) in the given rule (\( 3i - 2j \)) to find the number that goes in that spot. After calculating all six numbers, we put them into the matrix.

🎯 Exam Tip: Pay close attention to the order of operations and the signs when calculating matrix elements using a given formula, especially with subtraction involved.

 

Question 23. If \( A = \begin{bmatrix} 1 & 0 & 3 \\ 4 & 5 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 2 & 5 \\ 7 & 3 & 1 \end{bmatrix} \) find the value of \( B-A \).
Answer:
Given two matrices:
\[ A = \begin{bmatrix} 1 & 0 & 3 \\ 4 & 5 & 2 \end{bmatrix} \]
\[ B = \begin{bmatrix} -1 & 2 & 5 \\ 7 & 3 & 1 \end{bmatrix} \]
To find \( B - A \), we subtract the corresponding elements of matrix A from matrix B. Both matrices must have the same dimensions for subtraction to be possible, which they do (both are \( 2 \times 3 \) matrices).
\[ B - A = \begin{bmatrix} -1 & 2 & 5 \\ 7 & 3 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 3 \\ 4 & 5 & 2 \end{bmatrix} \]
Subtract each element: \( (B_{ij} - A_{ij}) \)
\[ B - A = \begin{bmatrix} (-1 - 1) & (2 - 0) & (5 - 3) \\ (7 - 4) & (3 - 5) & (1 - 2) \end{bmatrix} \]
\[ B - A = \begin{bmatrix} -2 & 2 & 2 \\ 3 & -2 & -1 \end{bmatrix} \]
So, the resulting matrix \( B - A \) is \( \begin{bmatrix} -2 & 2 & 2 \\ 3 & -2 & -1 \end{bmatrix} \). Matrix subtraction involves performing the operation on elements in the same position.
In simple words: To subtract matrices, you simply subtract the numbers in the same position from one matrix to the other. You do this for every single number, like subtracting top-left from top-left, and so on.

🎯 Exam Tip: Remember that matrix addition and subtraction are only possible if both matrices have the exact same number of rows and columns. Perform the operation element-wise.

 

Question 24. Find if \( \begin{bmatrix} 2 & x & 3 \end{bmatrix} \begin{bmatrix} 0 \\ x \\ -1 \end{bmatrix} = 13 \).
Answer:
Given the matrix equation:
\[ \begin{bmatrix} 2 & x & 3 \end{bmatrix} \begin{bmatrix} 0 \\ x \\ -1 \end{bmatrix} = 13 \]
This is a multiplication of a \( 1 \times 3 \) matrix by a \( 3 \times 1 \) matrix, which will result in a \( 1 \times 1 \) matrix (a scalar value).
To perform the multiplication, multiply the elements of the row matrix by the corresponding elements of the column matrix and sum them up:
\( (2 \times 0) + (x \times x) + (3 \times -1) = 13 \)
\( 0 + x^2 - 3 = 13 \)
Now, solve this simple algebraic equation for x:
\( x^2 - 3 = 13 \)
Add 3 to both sides:
\( x^2 = 13 + 3 \)
\( x^2 = 16 \)
Take the square root of both sides:
\( x = \pm \sqrt{16} \)
\( x = \pm 4 \)
The value of x can be either 4 or -4. Both values satisfy the given matrix equation.
In simple words: We multiply the numbers in the first matrix by the numbers in the second matrix, matching their positions, and add them up. This sum is given as 13. We then solve the simple equation for 'x' and find that 'x' can be either 4 or -4.

🎯 Exam Tip: When multiplying a row matrix by a column matrix, remember to multiply corresponding elements and then sum the products. Always consider both positive and negative roots when solving for a squared variable.

 

Question 25. If \( A = \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \) find BA.
Answer:
Given matrices:
\[ A = \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} \]
\[ B = \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \]
The order of matrix A is \( 3 \times 2 \) (3 rows, 2 columns).
The order of matrix B is \( 2 \times 3 \) (2 rows, 3 columns).
To find the product BA, the number of columns in B must equal the number of rows in A. Here, columns of B = 3, rows of A = 3. So, multiplication BA is not possible with the given dimensions in the order BA. The source seems to have provided the solution for AB, not BA, or there is a misinterpretation of the input. Let's calculate AB based on the provided solution steps from the source, which indicates multiplication is performed as \( B_{row} \times A_{col} \). If the question actually meant to find the product AB, then:
Order of A is \( 3 \times 2 \). Order of B is \( 2 \times 3 \).
Number of columns in A (2) equals number of rows in B (2). So, AB is possible and the resulting matrix will be of order \( 3 \times 3 \). However, looking at the provided solution, it seems to perform \( B_{2 \times 3} \times A_{3 \times 2} \), which gives a \( 2 \times 2 \) matrix. This means the matrices might have been meant as \( A_{3 \times 2} \) and \( B_{2 \times 3} \), and the problem asks for BA, not AB as I initially thought. Let's re-examine the OCR image's solution for Question 25. The image shows calculation for BA as \( \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} \). This multiplication is invalid because the number of columns in B (3) does not equal the number of rows in A (3). So, the operation BA as written in the question cannot be performed directly. **Correction based on source solution logic for BA:** The source solution shows: \[ BA = \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} \] This is invalid. It looks like the source performed a different operation or there is a fundamental error in the source material's problem setup or solution. If we assume the question intended for the matrices to be: \( A = \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \) and the operation is \( BA \), this is not possible. Let's assume the question meant to find \( A \times B \). \[ A \times B = \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \] This would result in a \( 3 \times 3 \) matrix. \( AB = \begin{bmatrix} (1 \cdot 3 + 0 \cdot 4) & (1 \cdot 1 + 0 \cdot 6) & (1 \cdot (-2) + 0 \cdot 0) \\ (3 \cdot 3 + (-1) \cdot 4) & (3 \cdot 1 + (-1) \cdot 6) & (3 \cdot (-2) + (-1) \cdot 0) \\ (2 \cdot 3 + 4 \cdot 4) & (2 \cdot 1 + 4 \cdot 6) & (2 \cdot (-2) + 4 \cdot 0) \end{bmatrix} \) \( AB = \begin{bmatrix} (3+0) & (1+0) & (-2+0) \\ (9-4) & (3-6) & (-6+0) \\ (6+16) & (2+24) & (-4+0) \end{bmatrix} \) \( AB = \begin{bmatrix} 3 & 1 & -2 \\ 5 & -3 & -6 \\ 22 & 26 & -4 \end{bmatrix} \) This result for AB does not match the provided solution of \( \begin{bmatrix} 2 & -9 \\ 22 & -6 \end{bmatrix} \) which is a \( 2 \times 2 \) matrix. The source solution for Question 25 shows: \[ BA = \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} (3 \cdot 1 + 1 \cdot 3 + (-2) \cdot 2) & (3 \cdot 0 + 1 \cdot (-1) + (-2) \cdot 4) \\ (4 \cdot 1 + 6 \cdot 3 + 0 \cdot 2) & (4 \cdot 0 + 6 \cdot (-1) + 0 \cdot 4) \end{bmatrix} \] \[ BA = \begin{bmatrix} (3+3-4) & (0-1-8) \\ (4+18+0) & (0-6+0) \end{bmatrix} \] \[ BA = \begin{bmatrix} 2 & -9 \\ 22 & -6 \end{bmatrix} \] This calculation is actually correct if B is \( 2 \times 3 \) and A is \( 3 \times 2 \). The problem statement in the PDF has them switched in definition: \( A = \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} \) is \( 3 \times 2 \) \( B = \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \) is \( 2 \times 3 \) The product BA is possible and its order would be \( 2 \times 2 \). The source's calculation for BA is indeed valid. My earlier confusion was based on visually misinterpreting the matrix dimensions or the initial problem definition in the source. So, the question is "find BA" for \( A = \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \). Let's re-write the answer correctly. Given matrices:
\[ A = \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} \]
\[ B = \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \]
The order of matrix B is \( 2 \times 3 \) (2 rows, 3 columns).
The order of matrix A is \( 3 \times 2 \) (3 rows, 2 columns).
Since the number of columns in B (3) is equal to the number of rows in A (3), the product BA is possible. The resulting matrix BA will have an order of \( 2 \times 2 \) (rows of B by columns of A).
To calculate BA:
\[ BA = \begin{bmatrix} 3 & 1 & -2 \\ 4 & 6 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ 2 & 4 \end{bmatrix} \]
For element (1,1) of BA: \( (3 \times 1) + (1 \times 3) + (-2 \times 2) = 3 + 3 - 4 = 2 \)
For element (1,2) of BA: \( (3 \times 0) + (1 \times -1) + (-2 \times 4) = 0 - 1 - 8 = -9 \)
For element (2,1) of BA: \( (4 \times 1) + (6 \times 3) + (0 \times 2) = 4 + 18 + 0 = 22 \)
For element (2,2) of BA: \( (4 \times 0) + (6 \times -1) + (0 \times 4) = 0 - 6 + 0 = -6 \)
So, the product matrix BA is:
\[ BA = \begin{bmatrix} 2 & -9 \\ 22 & -6 \end{bmatrix} \]
In simple words: To find the product BA, we multiply the rows of matrix B by the columns of matrix A. We multiply the numbers in each row of B by the corresponding numbers in each column of A and then add them up. This gives us each new number for the resulting matrix.

🎯 Exam Tip: Always check if matrix multiplication is possible by comparing the inner dimensions (columns of the first matrix with rows of the second). The outer dimensions give the size of the resulting matrix.

 

Question 26. Find the unknowns a, b, c, d, x, y in the given matrix equation.
\[ \begin{bmatrix} d+1 & 10+a \\ 3b-2 & a-4 \end{bmatrix} = \begin{bmatrix} 2 & 2a+1 \\ b-5 & 4c \end{bmatrix} \]
Answer:
Given the matrix equation:
\[ \begin{bmatrix} d+1 & 10+a \\ 3b-2 & a-4 \end{bmatrix} = \begin{bmatrix} 2 & 2a+1 \\ b-5 & 4c \end{bmatrix} \]
For two matrices to be equal, their corresponding elements must be equal. We equate the elements at each position to form a system of equations.
1. Equating elements at position (1,1):
\( d + 1 = 2 \)
\( d = 2 - 1 \)
\( d = 1 \)
2. Equating elements at position (1,2):
\( 10 + a = 2a + 1 \)
\( 10 - 1 = 2a - a \)
\( 9 = a \)
3. Equating elements at position (2,1):
\( 3b - 2 = b - 5 \)
\( 3b - b = -5 + 2 \)
\( 2b = -3 \)
\( b = \frac{-3}{2} \)
4. Equating elements at position (2,2):
\( a - 4 = 4c \)
Substitute the value of \( a = 9 \) (found in step 2) into this equation:
\( 9 - 4 = 4c \)
\( 5 = 4c \)
\( c = \frac{5}{4} \)
The values of the unknowns are \( a = 9, b = -\frac{3}{2}, c = \frac{5}{4} \), and \( d = 1 \). This method relies on the fundamental property of matrix equality.
In simple words: When two matrices are said to be equal, it means every number in the same spot in both matrices is exactly the same. So, we make small equations for each matching spot and solve them one by one to find the values of the hidden letters.

🎯 Exam Tip: Systematically equate corresponding elements, starting with variables that are easier to isolate. Solve one variable at a time and substitute its value into other equations if needed.

 

Question 27. Prove that \( \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} \) and \( \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix} \) are inverse to each other through multiplication.
Answer:
Let matrix A be \( \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} \) and matrix B be \( \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix} \).
Two matrices A and B are inverses of each other if their product, in both orders (AB and BA), results in the identity matrix (I). For \( 2 \times 2 \) matrices, the identity matrix is \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
First, let's calculate the product AB:
\[ AB = \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix} \]
\[ AB = \begin{bmatrix} (3 \times 2 + 5 \times -1) & (3 \times -5 + 5 \times 3) \\ (1 \times 2 + 2 \times -1) & (1 \times -5 + 2 \times 3) \end{bmatrix} \]
\[ AB = \begin{bmatrix} (6 - 5) & (-15 + 15) \\ (2 - 2) & (-5 + 6) \end{bmatrix} \]
\[ AB = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
Since AB equals the identity matrix, let's now calculate BA to confirm.
\[ BA = \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix} \times \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} \]
\[ BA = \begin{bmatrix} (2 \times 3 + (-5) \times 1) & (2 \times 5 + (-5) \times 2) \\ (-1 \times 3 + 3 \times 1) & (-1 \times 5 + 3 \times 2) \end{bmatrix} \]
\[ BA = \begin{bmatrix} (6 - 5) & (10 - 10) \\ (-3 + 3) & (-5 + 6) \end{bmatrix} \]
\[ BA = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
Since both AB and BA result in the identity matrix, it is proven that the given matrices are inverses of each other. This is the definition of an inverse matrix.
In simple words: To prove two matrices are inverses, you just multiply them together in both possible orders. If both multiplications give you the special "identity matrix" (which has ones on the diagonal and zeros elsewhere), then they are indeed inverses of each other.

🎯 Exam Tip: For matrices to be inverses, their product in *both* orders (AB and BA) must yield the identity matrix. If only one order works, they are not inverses.

III. Answer the Following Questions.

 

Question 1. Solve \( x – \frac { y }{ 5 } = 6 \); \( y − \frac { z }{ 7 } = 8 \); \( z − \frac { x }{ 2 } = 10 \).
Answer:
Given the system of linear equations:
1. \( x – \frac { y }{ 5 } = 6 \)
2. \( y − \frac { z }{ 7 } = 8 \)
3. \( z − \frac { x }{ 2 } = 10 \)

First, simplify each equation by clearing the fractions:
From (1), multiply by 5:
\( 5x - y = 30 \) ....(A)

From (2), multiply by 7:
\( 7y - z = 56 \) ....(B)

From (3), multiply by 2:
\( 2z - x = 20 \) ....(C)

Now we have a cleaner system:
(A) \( 5x - y = 30 \)
(B) \( 7y - z = 56 \)
(C) \( -x + 2z = 20 \)

Let's try to eliminate one variable. We can multiply (A) by 7 to eliminate y when combined with (B).
Multiply (A) by 7: \( 35x - 7y = 210 \) ....(D)
Add (D) and (B):
\( (35x - 7y) + (7y - z) = 210 + 56 \)
\( 35x - z = 266 \) ....(E)

Now we have a system with only x and z:
(C) \( -x + 2z = 20 \)
(E) \( 35x - z = 266 \)

Multiply (E) by 2 to eliminate z:
\( 2(35x - z) = 2(266) \)
\( 70x - 2z = 532 \) ....(F)

Add (C) and (F):
\( (-x + 2z) + (70x - 2z) = 20 + 532 \)
\( 69x = 552 \)
\( x = \frac{552}{69} \)
\( x = 8 \)

Now that we have x, substitute \( x = 8 \) into equation (A) to find y:
\( 5(8) - y = 30 \)
\( 40 - y = 30 \)
\( -y = 30 - 40 \)
\( -y = -10 \)
\( y = 10 \)

Finally, substitute \( y = 10 \) into equation (B) to find z (or x into C):
Using (B):
\( 7(10) - z = 56 \)
\( 70 - z = 56 \)
\( -z = 56 - 70 \)
\( -z = -14 \)
\( z = 14 \)

Let's check with (C) using \( x = 8 \) and \( z = 14 \):
\( -8 + 2(14) = -8 + 28 = 20 \). This is correct.
The values are \( x = 8, y = 10 \), and \( z = 14 \). Solving systems of equations often involves methodical elimination or substitution.
In simple words: We have three equations with three unknowns. First, we get rid of all fractions. Then, we combine the equations to remove one variable at a time, until we find the value of one variable. Once we know one, we put it back into the equations to find the others.

🎯 Exam Tip: When solving systems of equations with fractions, always start by multiplying each equation by its least common multiple of denominators to eliminate fractions and simplify calculations.

 

Question 2. Solve for x,y and z using the given 3 equations \( \frac { 2 }{ y } – \frac { 4 }{ z } + \frac { 3 }{ x } = 3 \); \( \frac { 5 }{ x } – \frac { 4 }{ y } – \frac { 8 }{ z } = 8 \); \( \frac { 6 }{ y } + \frac { 6 }{ z } +\frac { 1 }{ x } = 2 \).
Answer:
Given the system of equations:
1. \( \frac { 3 }{ x } + \frac { 2 }{ y } – \frac { 4 }{ z } = 3 \)
2. \( \frac { 5 }{ x } – \frac { 4 }{ y } – \frac { 8 }{ z } = 8 \)
3. \( \frac { 1 }{ x } + \frac { 6 }{ y } + \frac { 6 }{ z } = 2 \)

These equations involve reciprocals of x, y, and z. To simplify, let's make a substitution:
Let \( a = \frac { 1 }{ x } \), \( b = \frac { 1 }{ y } \), and \( c = \frac { 1 }{ z } \).
Substitute these into the given equations:
1. \( 3a + 2b - 4c = 3 \) ....(1)
2. \( 5a - 4b - 8c = 8 \) ....(2)
3. \( a + 6b + 6c = 2 \) ....(3)

Now we have a standard system of three linear equations in a, b, and c.

Step 1: Eliminate 'b' from (1) and (2).
Multiply equation (1) by 2: \( 2(3a + 2b - 4c) = 2(3) \)
\( 6a + 4b - 8c = 6 \) ....(4)
Add equation (4) and (2):
\( (6a + 4b - 8c) + (5a - 4b - 8c) = 6 + 8 \)
\( 11a - 16c = 14 \) ....(5)

Step 2: Eliminate 'b' from (1) and (3).
Multiply equation (1) by 3: \( 3(3a + 2b - 4c) = 3(3) \)
\( 9a + 6b - 12c = 9 \) ....(6)
Subtract equation (3) from (6):
\( (9a + 6b - 12c) - (a + 6b + 6c) = 9 - 2 \)
\( 9a - a + 6b - 6b - 12c - 6c = 7 \)
\( 8a - 18c = 7 \) ....(7)

Now we have a system of two equations with 'a' and 'c':
(5) \( 11a - 16c = 14 \)
(7) \( 8a - 18c = 7 \)

Step 3: Eliminate 'a' from (5) and (7).
To eliminate 'a', multiply (5) by 8 and (7) by 11. (LCM of 11, 8 is 88).
Multiply (5) by 8: \( 8(11a - 16c) = 8(14) \)
\( 88a - 128c = 112 \) ....(8)
Multiply (7) by 11: \( 11(8a - 18c) = 11(7) \)
\( 88a - 198c = 77 \) ....(9)
Subtract equation (9) from (8):
\( (88a - 128c) - (88a - 198c) = 112 - 77 \)
\( 88a - 88a - 128c + 198c = 35 \)
\( 70c = 35 \)
\( c = \frac { 35 }{ 70 } \)
\( c = \frac { 1 }{ 2 } \)

Step 4: Substitute 'c' back to find 'a'.
Substitute \( c = \frac { 1 }{ 2 } \) into equation (7):
\( 8a - 18(\frac { 1 }{ 2 }) = 7 \)
\( 8a - 9 = 7 \)
\( 8a = 7 + 9 \)
\( 8a = 16 \)
\( a = \frac { 16 }{ 8 } \)
\( a = 2 \)

Step 5: Substitute 'a' and 'c' back to find 'b'.
Substitute \( a = 2 \) and \( c = \frac { 1 }{ 2 } \) into equation (1):
\( 3(2) + 2b - 4(\frac { 1 }{ 2 }) = 3 \)
\( 6 + 2b - 2 = 3 \)
\( 4 + 2b = 3 \)
\( 2b = 3 - 4 \)
\( 2b = -1 \)
\( b = -\frac { 1 }{ 2 } \)

Now that we have a, b, and c, we can find x, y, and z using our initial substitutions:
\( a = \frac { 1 }{ x } \implies 2 = \frac { 1 }{ x } \implies x = \frac { 1 }{ 2 } \)
\( b = \frac { 1 }{ y } \implies -\frac { 1 }{ 2 } = \frac { 1 }{ y } \implies y = -2 \)
\( c = \frac { 1 }{ z } \implies \frac { 1 }{ 2 } = \frac { 1 }{ z } \implies z = 2 \)

The values are \( x = \frac { 1 }{ 2 }, y = -2 \), and \( z = 2 \). This method of substitution simplifies the complex reciprocal equations into a solvable linear system.
In simple words: This problem involves equations with \( 1/x, 1/y, 1/z \). To make it easier, we pretend these are new variables (a, b, c). Then, we solve for a, b, and c using standard methods like adding and subtracting equations. Finally, we convert these back to find the original x, y, and z.

🎯 Exam Tip: For equations with reciprocals of variables, always introduce new variables for the reciprocals (e.g., \( a = \frac{1}{x} \)) to transform the system into a more manageable linear form.

 

Question 3. 100 pencils are to be kept inside three types of boxes A, B and C. If 5 boxes of type A, 3 boxes of type B, 2 boxes of type C are used 6 pencils are left out. If 3 boxes of type A, 5 boxes of type B, 2 boxes of type C are used 2 pencils are left out. If 2 boxes of type A, 4 boxes of type B and 4 boxes of type C are used, there is a space for 4 pencils. Find the number of pencils that each box can hold.
Answer:
Let 'x' be the number of pencils a box of type A can hold.
Let 'y' be the number of pencils a box of type B can hold.
Let 'z' be the number of pencils a box of type C can hold.
The total number of pencils available is 100.

From the first condition:
5 boxes of type A, 3 boxes of type B, and 2 boxes of type C are used, and 6 pencils are left out. This means the total capacity of these boxes is \( 100 - 6 = 94 \) pencils.
So, \( 5x + 3y + 2z = 94 \) ....(1)

From the second condition:
3 boxes of type A, 5 boxes of type B, and 2 boxes of type C are used, and 2 pencils are left out. This means the total capacity is \( 100 - 2 = 98 \) pencils.
So, \( 3x + 5y + 2z = 98 \) ....(2)

From the third condition:
2 boxes of type A, 4 boxes of type B, and 4 boxes of type C are used, and there is space for 4 pencils. This means the total capacity is \( 100 + 4 = 104 \) pencils.
So, \( 2x + 4y + 4z = 104 \) ....(3)

We now have a system of three linear equations:
(1) \( 5x + 3y + 2z = 94 \)
(2) \( 3x + 5y + 2z = 98 \)
(3) \( 2x + 4y + 4z = 104 \)

Step 1: Simplify equation (3) by dividing by 2:
\( x + 2y + 2z = 52 \) ....(4)

Step 2: Eliminate 'z' from equations (1) and (4).
Subtract equation (4) from equation (1):
\( (5x + 3y + 2z) - (x + 2y + 2z) = 94 - 52 \)
\( 4x + y = 42 \) ....(5)

Step 3: Eliminate 'z' from equations (2) and (4).
Multiply equation (4) by 2: \( 2(x + 2y + 2z) = 2(52) \)
\( 2x + 4y + 4z = 104 \) ....(6)
This is the same as equation (3). So, let's use (1) and (2) to eliminate 'z'. Subtract equation (1) from (2):
\( (3x + 5y + 2z) - (5x + 3y + 2z) = 98 - 94 \)
\( -2x + 2y = 4 \)
Divide by 2:
\( -x + y = 2 \)
Rearrange this to \( y - x = 2 \) or \( y = x + 2 \) ....(7)

Now we have a system of two equations with 'x' and 'y':
(5) \( 4x + y = 42 \)
(7) \( y - x = 2 \)

Substitute \( y = x + 2 \) from (7) into (5):
\( 4x + (x + 2) = 42 \)
\( 5x + 2 = 42 \)
\( 5x = 42 - 2 \)
\( 5x = 40 \)
\( x = \frac{40}{5} \)
\( x = 8 \)

Now that we have x, substitute \( x = 8 \) into equation (7) to find y:
\( y = 8 + 2 \)
\( y = 10 \)

Finally, substitute \( x = 8 \) and \( y = 10 \) into equation (4) to find z:
\( 8 + 2(10) + 2z = 52 \)
\( 8 + 20 + 2z = 52 \)
\( 28 + 2z = 52 \)
\( 2z = 52 - 28 \)
\( 2z = 24 \)
\( z = \frac{24}{2} \)
\( z = 12 \)

So, a box of type A can hold 8 pencils, a box of type B can hold 10 pencils, and a box of type C can hold 12 pencils. This problem demonstrates setting up and solving a real-world system of linear equations.
In simple words: We create three math problems (equations) based on the number of pencils, boxes, and leftovers. We then solve these equations step-by-step to find out how many pencils each kind of box (A, B, C) can hold.

🎯 Exam Tip: Carefully translate word problems into mathematical equations, paying attention to phrases like "left out" (subtract from total) and "space for" (add to total) to correctly form the constants.

 

Question 4. What 2 masons earn in a day is earned by 3 male workers in a day. The daily wages of 15 female workers is Rs.30 more than the total daily wages of 5 masons and 3 male workers. If one mason, one male worker and 2 female workers are engaged for a day, the builder has to pay Rs.160 as wages. Find the daily wages of a mason, a male worker and a female worker.
Answer:
Let 'x' be the daily wage of a mason.
Let 'y' be the daily wage of a male worker.
Let 'z' be the daily wage of a female worker.

From the first condition:
2 masons' daily earnings = 3 male workers' daily earnings
\( 2x = 3y \)
Rearrange this to form an equation:
\( 2x - 3y = 0 \) ....(1)

From the second condition:
Daily wages of 15 female workers = Rs.30 + (daily wages of 5 masons + daily wages of 3 male workers)
\( 15z = 30 + (5x + 3y) \)
Rearrange this:
\( -5x - 3y + 15z = 30 \) ....(2)

From the third condition:
If 1 mason, 1 male worker, and 2 female workers are paid for a day, the total wage is Rs.160.
\( x + y + 2z = 160 \) ....(3)

We now have a system of three linear equations:
(1) \( 2x - 3y = 0 \)
(2) \( -5x - 3y + 15z = 30 \)
(3) \( x + y + 2z = 160 \)

Step 1: Eliminate 'y' from (1) and (2).
Since 'y' terms have the same coefficient with the same sign, subtract (2) from (1).
\( (2x - 3y) - (-5x - 3y + 15z) = 0 - 30 \)
\( 2x + 5x - 3y + 3y - 15z = -30 \)
\( 7x - 15z = -30 \) ....(4)

Step 2: Eliminate 'y' from (1) and (3).
Multiply (3) by 3: \( 3(x + y + 2z) = 3(160) \)
\( 3x + 3y + 6z = 480 \) ....(5)
Add (1) and (5):
\( (2x - 3y) + (3x + 3y + 6z) = 0 + 480 \)
\( 5x + 6z = 480 \) ....(6)

Now we have a system of two equations with 'x' and 'z':
(4) \( 7x - 15z = -30 \)
(6) \( 5x + 6z = 480 \)

Step 3: Eliminate 'z' from (4) and (6).
Multiply (4) by 2: \( 2(7x - 15z) = 2(-30) \)
\( 14x - 30z = -60 \) ....(7)
Multiply (6) by 5: \( 5(5x + 6z) = 5(480) \)
\( 25x + 30z = 2400 \) ....(8)
Add (7) and (8):
\( (14x - 30z) + (25x + 30z) = -60 + 2400 \)
\( 39x = 2340 \)
\( x = \frac{2340}{39} \)
\( x = 60 \)

Step 4: Substitute 'x' back to find 'y'.
Substitute \( x = 60 \) into equation (1):
\( 2(60) - 3y = 0 \)
\( 120 - 3y = 0 \)
\( -3y = -120 \)
\( y = \frac{-120}{-3} \)
\( y = 40 \)

Step 5: Substitute 'x' and 'y' back to find 'z'.
Substitute \( x = 60 \) and \( y = 40 \) into equation (3):
\( 60 + 40 + 2z = 160 \)
\( 100 + 2z = 160 \)
\( 2z = 160 - 100 \)
\( 2z = 60 \)
\( z = \frac{60}{2} \)
\( z = 30 \)

The daily wages are: Mason = Rs.60, Male worker = Rs.40, Female worker = Rs.30. This detailed step-by-step solution ensures accuracy for such complex word problems.
In simple words: We set up three algebraic equations from the three given facts about wages. We then solve this system of equations by carefully removing one variable at a time until we find the wage for masons, male workers, and female workers.

🎯 Exam Tip: When dealing with multiple conditions in word problems, clearly define variables and translate each condition into a separate equation. Organize your elimination steps to avoid errors.

 

Question 5. Find the G.C.D. of \( x^3 - 10x^2 + 31x - 30 \) and \( 2x^3 - 8x^2 + 2x + 12 \)
Answer:
Let \( p(x) = x^3 - 10x^2 + 31x - 30 \)
Let \( g(x) = 2x^3 - 8x^2 + 2x + 12 \)
First, we can factor out 2 from \( g(x) \):
\( g(x) = 2(x^3 - 4x^2 + x + 6) \)
Now, we perform polynomial long division of \( p(x) \) by \( (x^3 - 4x^2 + x + 6) \).
\[ \require{enclose} \begin{array}{r} \quad 1 \\ x^3-4x^2+x+6 \enclose{longdiv}{x^3-10x^2+31x-30} \\ \underline{x^3-4x^2+\;x+\;6} \\ -6x^2+30x-36 \\ \underline{-6x^2+30x-36} \\ 0 \end{array} \] Now we divide \( (x^3 - 4x^2 + x + 6) \) by \( (-6x^2 + 30x - 36) \).
First, factor out -6 from the remainder: \( -6(x^2 - 5x + 6) \). We use \( (x^2 - 5x + 6) \) as the new divisor.
\[ \require{enclose} \begin{array}{r} \quad x+1 \\ x^2-5x+6 \enclose{longdiv}{x^3-4x^2+\;x+\;6} \\ \underline{x^3-5x^2+\;6x} \\ x^2-5x+6 \\ \underline{x^2-5x+6} \\ 0 \end{array} \] Since the remainder is 0, the G.C.D. is the last divisor, which is \( x^2 - 5x + 6 \).
In simple words: We find the Greatest Common Divisor by using polynomial long division. We simplify one polynomial by factoring out a number, then divide the first polynomial by the simplified part of the second. We continue dividing the previous divisor by the remainder until we get zero. The last divisor is our G.C.D.

🎯 Exam Tip: Always check if any constant can be factored out from the polynomials first to simplify the division process. Remember that the G.C.D. is the last non-zero divisor.

 

Question 6. The G.C.D of \( x^4 + 3x^3 + 5x^2 + 26x + 56 \) and \( x^4 + 2x^3 - 4x^2 - x + 28 \) is \( x^2 + 5x + 7 \). Find their L.C.M.
Answer:
Let \( p(x) = x^4 + 3x^3 + 5x^2 + 26x + 56 \)
Let \( g(x) = x^4 + 2x^3 - 4x^2 - x + 28 \)
Given G.C.D. \( = x^2 + 5x + 7 \).
We know that for two polynomials \( p(x) \) and \( g(x) \), \( \text{LCM} \times \text{GCD} = p(x) \times g(x) \).
So, \( \text{LCM} = \frac{p(x) \times g(x)}{\text{GCD}} \).

First, divide \( p(x) \) by the G.C.D. \( (x^2 + 5x + 7) \):
\[ \require{enclose} \begin{array}{r} x^2-2x+8 \\ x^2+5x+7 \enclose{longdiv}{x^4+3x^3+\;5x^2+26x+56} \\ \underline{x^4+5x^3+\;7x^2} \\ -2x^3-2x^2+26x \\ \underline{-2x^3-10x^2-14x} \\ 8x^2+40x+56 \\ \underline{8x^2+40x+56} \\ 0 \end{array} \] So, \( p(x) = (x^2 + 5x + 7)(x^2 - 2x + 8) \).

Now, we can find L.C.M. as \( \text{LCM} = \frac{(x^2 + 5x + 7)(x^2 - 2x + 8) \times g(x)}{(x^2 + 5x + 7)} \)
\( \text{LCM} = (x^2 - 2x + 8) \times g(x) \)
\( \text{LCM} = (x^2 - 2x + 8)(x^4 + 2x^3 - 4x^2 - x + 28) \)
We also need to divide \( g(x) \) by \( (x^2 + 5x + 7) \).
\[ \require{enclose} \begin{array}{r} x^2-3x+4 \\ x^2+5x+7 \enclose{longdiv}{x^4+2x^3-4x^2-x+28} \\ \underline{x^4+5x^3+7x^2} \\ -3x^3-11x^2-x \\ \underline{-3x^3-15x^2-21x} \\ 4x^2+20x+28 \\ \underline{4x^2+20x+28} \\ 0 \end{array} \] So, \( g(x) = (x^2 + 5x + 7)(x^2 - 3x + 4) \).

Therefore, \( \text{LCM} = \frac{p(x) \times g(x)}{\text{GCD}} = \frac{(x^2 + 5x + 7)(x^2 - 2x + 8) \times (x^2 + 5x + 7)(x^2 - 3x + 4)}{x^2 + 5x + 7} \)
\( \text{LCM} = (x^2 - 2x + 8)(x^2 + 5x + 7)(x^2 - 3x + 4) \)
This is the least common multiple of the two given polynomials.
In simple words: The LCM of two polynomials is found using the formula: \( \text{LCM} = \frac{\text{Polynomial 1} \times \text{Polynomial 2}}{\text{GCD}} \). We used long division to factor the first polynomial by the given GCD. Then we used the formula to get the final LCM.

🎯 Exam Tip: When given the GCD, you can simplify the process by dividing one of the polynomials by the GCD first. This helps in factoring and then applying the LCM formula efficiently.

 

Question 7. Find the values of "a" and "b" given that \( p(x) = (x^2 + 3x + 2)(x^2 - 4x + a) \); \( g(x) = (x^2 - 6x + 9) \times (x^2 + 4x + b) \) and their G.C.D is \( (x + 2)(x - 3) \)
Answer:
Given \( p(x) = (x^2 + 3x + 2)(x^2 - 4x + a) \)
Factor the first part of \( p(x) \): \( x^2 + 3x + 2 = (x + 1)(x + 2) \)
So, \( p(x) = (x + 1)(x + 2)(x^2 - 4x + a) \)

Given \( g(x) = (x^2 - 6x + 9)(x^2 + 4x + b) \)
Factor the first part of \( g(x) \): \( x^2 - 6x + 9 = (x - 3)^2 = (x - 3)(x - 3) \)
So, \( g(x) = (x - 3)(x - 3)(x^2 + 4x + b) \)

The G.C.D. is given as \( (x + 2)(x - 3) \).

Since \( (x + 2) \) is a factor of \( p(x) \), this is already present.
Since \( (x - 3) \) is a factor of \( p(x) \), it must be a factor of \( (x^2 - 4x + a) \).
If \( (x - 3) \) is a factor, then substituting \( x = 3 \) into \( (x^2 - 4x + a) \) should give 0.
\( (3)^2 - 4(3) + a = 0 \)
\( 9 - 12 + a = 0 \)
\( -3 + a = 0 \)
\( \implies a = 3 \)

Since \( (x - 3) \) is a factor of \( g(x) \), this is already present.
Since \( (x + 2) \) is a factor of \( g(x) \), it must be a factor of \( (x^2 + 4x + b) \).
If \( (x + 2) \) is a factor, then substituting \( x = -2 \) into \( (x^2 + 4x + b) \) should give 0.
\( (-2)^2 + 4(-2) + b = 0 \)
\( 4 - 8 + b = 0 \)
\( -4 + b = 0 \)
\( \implies b = 4 \)
Thus, the values are \( a = 3 \) and \( b = 4 \).
In simple words: We factor the known parts of both polynomials \( p(x) \) and \( g(x) \). Since the G.C.D. factors must be present in both polynomials, we use the remaining factors of the G.C.D. and set the other parts of the polynomials to zero when substituting the root, which helps us find the unknown values 'a' and 'b'.

🎯 Exam Tip: Remember the Factor Theorem: if \( (x-k) \) is a factor of a polynomial, then substituting \( x=k \) into the polynomial will result in zero. Use this to find unknown coefficients.

 

Question 8. Find the other polynomial \( g(x) \), given that LCM, HCF and \( p(x) \) as \( (x - 1)(x - 2)(x^2 - 3x + 3) \); \( x - 1 \) and \( x^3 - 4x^2 + 6x - 3 \) respectively.
Answer:
Given:
\( \text{L.C.M.} = (x - 1)(x - 2)(x^2 - 3x + 3) \)
\( \text{H.C.F.} = (x - 1) \)
\( p(x) = x^3 - 4x^2 + 6x - 3 \)

We know the relationship: \( p(x) \times g(x) = \text{L.C.M.} \times \text{H.C.F.} \).
So, \( g(x) = \frac{\text{L.C.M.} \times \text{H.C.F.}}{p(x)} \).

Let's divide \( p(x) \) by \( (x - 1) \). We can use synthetic division or long division.
Using long division for \( x^3 - 4x^2 + 6x - 3 \) divided by \( x - 1 \):
\[ \require{enclose} \begin{array}{r} x^2-3x+3 \\ x-1 \enclose{longdiv}{x^3-4x^2+6x-3} \\ \underline{x^3-x^2} \\ -3x^2+6x \\ \underline{-3x^2+3x} \\ 3x-3 \\ \underline{3x-3} \\ 0 \end{array} \] So, \( p(x) = (x - 1)(x^2 - 3x + 3) \).

Now substitute this into the formula for \( g(x) \):
\( g(x) = \frac{(x - 1)(x - 2)(x^2 - 3x + 3) \times (x - 1)}{(x - 1)(x^2 - 3x + 3)} \)
Cancel out the common terms \( (x - 1) \) and \( (x^2 - 3x + 3) \):
\( g(x) = (x - 2)(x - 1) \)
\( g(x) = x^2 - x - 2x + 2 \)
\( g(x) = x^2 - 3x + 2 \)
The other polynomial is \( x^2 - 3x + 2 \).
In simple words: We use the formula that states the product of two polynomials is equal to the product of their LCM and HCF. First, we factor \( p(x) \) using the given HCF. Then, we substitute all the known values into the formula and simplify to find the unknown polynomial \( g(x) \).

🎯 Exam Tip: Always remember the fundamental relation: \( \text{LCM} \times \text{HCF} = p(x) \times g(x) \). This formula is key to solving problems where three of the four quantities are given. Factoring polynomials accurately is also crucial.

 

Question 9. Divide \( \frac{2x^2+x-3}{(x-1)^2} \) by \( \frac{2x^2+5x+3}{x^2-1} \)
Answer:
To divide one fraction by another, we multiply the first fraction by the reciprocal of the second.
\( \frac{2x^2+x-3}{(x-1)^2} \div \frac{2x^2+5x+3}{x^2-1} = \frac{2x^2+x-3}{(x-1)^2} \times \frac{x^2-1}{2x^2+5x+3} \)

First, factor each polynomial:
1. \( 2x^2 + x - 3 \): We look for factors that multiply to \( 2 \times -3 = -6 \) and add to 1. These are 3 and -2.
\( 2x^2 + 3x - 2x - 3 = x(2x + 3) - 1(2x + 3) = (2x + 3)(x - 1) \)

2. \( (x - 1)^2 \): This is already factored.

3. \( x^2 - 1 \): This is a difference of squares.
\( x^2 - 1 = (x + 1)(x - 1) \)

4. \( 2x^2 + 5x + 3 \): We look for factors that multiply to \( 2 \times 3 = 6 \) and add to 5. These are 2 and 3.
\( 2x^2 + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1) \)

Now substitute the factored forms back into the expression:
\( \frac{(2x+3)(x-1)}{(x-1)(x-1)} \times \frac{(x+1)(x-1)}{(2x+3)(x+1)} \)

Cancel out common factors from the numerator and denominator:
\( \frac{\cancel{(2x+3)}\cancel{(x-1)}}{\cancel{(x-1)}(x-1)} \times \frac{\cancel{(x+1)}\cancel{(x-1)}}{\cancel{(2x+3)}\cancel{(x+1)}} \)
After cancellation, we are left with \( 1 \).
The simplified expression is \( 1 \).
In simple words: To divide fractions, we flip the second fraction and multiply. We then break down each top and bottom part into its smallest multiplications (factors). After that, we cross out any matching parts from the top and bottom. What's left over is the answer. In this case, all parts canceled out, leaving 1.

🎯 Exam Tip: Always factor all polynomials completely before attempting to cancel terms. This helps avoid errors and ensures you simplify the expression to its lowest terms.

 

Question 10. Simplify \( \frac{x-3}{x^2-x-6} + \frac{2x-1}{2x^2+5x-3} - \frac{2x+5}{x^2+5x+6} \)
Answer:
First, factor each denominator:
1. \( x^2 - x - 6 \): Factors that multiply to -6 and add to -1 are -3 and 2.
\( x^2 - x - 6 = (x - 3)(x + 2) \)

2. \( 2x^2 + 5x - 3 \): Factors that multiply to \( 2 \times -3 = -6 \) and add to 5 are 6 and -1.
\( 2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x + 3)(2x - 1) \)

3. \( x^2 + 5x + 6 \): Factors that multiply to 6 and add to 5 are 2 and 3.
\( x^2 + 5x + 6 = (x + 2)(x + 3) \)

Now substitute the factored denominators into the expression:
\( \frac{x-3}{(x-3)(x+2)} + \frac{2x-1}{(2x+3)(2x-1)} - \frac{2x+5}{(x+2)(x+3)} \)

Cancel common factors in each term:
\( \frac{\cancel{(x-3)}}{\cancel{(x-3)}(x+2)} + \frac{\cancel{(2x-1)}}{(2x+3)\cancel{(2x-1)}} - \frac{2x+5}{(x+2)(x+3)} \)
\( = \frac{1}{x+2} + \frac{1}{2x+3} - \frac{2x+5}{(x+2)(x+3)} \)

Now find a common denominator, which is \( (x+2)(2x+3)(x+3) \).
However, we should notice that the last term's denominator is \( (x+2)(x+3) \). The LCM of \( (x+2) \), \( (2x+3) \), and \( (x+2)(x+3) \) is \( (x+2)(x+3)(2x+3) \).

\( = \frac{1(x+3)(2x+3)}{(x+2)(x+3)(2x+3)} + \frac{1(x+2)(x+3)}{(2x+3)(x+2)(x+3)} - \frac{(2x+5)(2x+3)}{(x+2)(x+3)(2x+3)} \)
This common denominator seems overly complex. Let's re-evaluate the terms. The third term has \( (x+2)(x+3) \) in the denominator, so it's a bit different.

Let's use the LCM of the simplified denominators: \( (x+2) \), \( (2x+3) \), and \( (x+2)(x+3) \).
The Least Common Multiple (LCM) of these three denominators is \( (x+2)(x+3)(2x+3) \).

\( = \frac{1 \cdot (x+3)(2x+3) + 1 \cdot (x+2)(x+3) - (2x+5)(2x+3)}{(x+2)(x+3)(2x+3)} \)

Expand the numerators:
\( (x+3)(2x+3) = 2x^2 + 3x + 6x + 9 = 2x^2 + 9x + 9 \)
\( (x+2)(x+3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6 \)
\( (2x+5)(2x+3) = 4x^2 + 6x + 10x + 15 = 4x^2 + 16x + 15 \)

Now substitute these back into the numerator:
Numerator \( = (2x^2 + 9x + 9) + (x^2 + 5x + 6) - (4x^2 + 16x + 15) \)
Numerator \( = 2x^2 + 9x + 9 + x^2 + 5x + 6 - 4x^2 - 16x - 15 \)
Group like terms:
Numerator \( = (2x^2 + x^2 - 4x^2) + (9x + 5x - 16x) + (9 + 6 - 15) \)
Numerator \( = -x^2 - 2x + 0 \)
Numerator \( = -x^2 - 2x \)

So the expression is \( \frac{-x^2 - 2x}{(x+2)(x+3)(2x+3)} \)
Factor out \( -x \) from the numerator: \( \frac{-x(x+2)}{(x+2)(x+3)(2x+3)} \)
Cancel \( (x+2) \) from numerator and denominator:
\( = \frac{-x}{(x+3)(2x+3)} \)
This is the simplified form of the expression.
In simple words: First, we break down each bottom part (denominator) of the fractions into its smallest multiplication parts (factors). We then cancel any matching parts on the top and bottom of each individual fraction. Next, we find the overall common bottom part for all fractions. We rewrite each fraction with this new common bottom part, making sure to adjust the top part correctly. Finally, we combine the top parts and simplify the result if possible.

🎯 Exam Tip: Always completely factor all numerators and denominators first. Then, cancel common terms within each fraction before attempting to find a common denominator for addition/subtraction. This minimizes complexity and reduces calculation errors.

 

Question 11. Find the square root of \( (6x^2 + 5x - 6)(6x^2 - x - 2)(4x^2 + 8x + 3) \)
Answer:
To find the square root, we first need to factor each quadratic polynomial.

1. Factor \( 6x^2 + 5x - 6 \):
We look for two numbers that multiply to \( 6 \times -6 = -36 \) and add to 5. These numbers are 9 and -4.
\( 6x^2 + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = (2x + 3)(3x - 2) \)

2. Factor \( 6x^2 - x - 2 \):
We look for two numbers that multiply to \( 6 \times -2 = -12 \) and add to -1. These numbers are -4 and 3.
\( 6x^2 - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (3x - 2)(2x + 1) \)

3. Factor \( 4x^2 + 8x + 3 \):
We look for two numbers that multiply to \( 4 \times 3 = 12 \) and add to 8. These numbers are 6 and 2.
\( 4x^2 + 6x + 2x + 3 = 2x(2x + 3) + 1(2x + 3) = (2x + 3)(2x + 1) \)

Now, multiply these factored forms together:
\( (2x + 3)(3x - 2) \times (3x - 2)(2x + 1) \times (2x + 3)(2x + 1) \)

Rearrange and group identical factors:
\( = (2x + 3)(2x + 3) \times (3x - 2)(3x - 2) \times (2x + 1)(2x + 1) \)
\( = (2x + 3)^2 (3x - 2)^2 (2x + 1)^2 \)

To find the square root, we take the square root of this product:
\( \sqrt{(2x + 3)^2 (3x - 2)^2 (2x + 1)^2} \)
\( = |(2x + 3)(3x - 2)(2x + 1)| \)
The absolute value is important because a square root always yields a non-negative value.
In simple words: To find the square root of this big multiplication, we first break down each of the three number groups (polynomials) into their simpler multiplication parts (factors). Then, we gather all the similar factors together. Since we are looking for a square root, we group them into pairs. The final answer will be the multiplication of one of each pair, surrounded by absolute value signs to make sure the result is positive.

🎯 Exam Tip: When finding the square root of a product of polynomials, factor each polynomial completely first. Then, group the identical factors together. The square root of \( A^2 B^2 C^2 \) is \( |ABC| \), so remember to include the absolute value for the final answer.

 

Question 12. Find the square root of \( x^4 + \frac{1}{x^4} + 2 \)
Answer:
We want to find the square root of \( x^4 + \frac{1}{x^4} + 2 \).
Notice that this expression resembles the expansion of \( (a + b)^2 = a^2 + 2ab + b^2 \).
Let \( a = x^2 \) and \( b = \frac{1}{x^2} \).
Then \( a^2 = (x^2)^2 = x^4 \).
And \( b^2 = \left(\frac{1}{x^2}\right)^2 = \frac{1}{x^4} \).
And \( 2ab = 2 \times x^2 \times \frac{1}{x^2} = 2 \times 1 = 2 \).

So, \( x^4 + \frac{1}{x^4} + 2 = (x^2)^2 + \left(\frac{1}{x^2}\right)^2 + 2x^2\left(\frac{1}{x^2}\right) \).
\( = \left(x^2 + \frac{1}{x^2}\right)^2 \).

Now, take the square root of this expression:
\( \sqrt{x^4 + \frac{1}{x^4} + 2} = \sqrt{\left(x^2 + \frac{1}{x^2}\right)^2} \)
\( = \left|x^2 + \frac{1}{x^2}\right| \)
Since \( x^2 \) is always non-negative, and \( \frac{1}{x^2} \) is also non-negative (assuming \( x \ne 0 \)), their sum \( x^2 + \frac{1}{x^2} \) is always positive. So, the absolute value sign can be removed.
\( = x^2 + \frac{1}{x^2} \)
In simple words: We need to find the square root of the given expression. We can see that the expression looks like a perfect square formula, \( (a+b)^2 \). By replacing 'a' with \( x^2 \) and 'b' with \( \frac{1}{x^2} \), we can rewrite the expression as \( \left(x^2 + \frac{1}{x^2}\right)^2 \). The square root of this will simply be \( x^2 + \frac{1}{x^2} \).

🎯 Exam Tip: Recognize perfect square patterns like \( a^2 + 2ab + b^2 \) or \( a^2 - 2ab + b^2 \). This saves significant time and simplifies complex-looking expressions quickly. Also, remember that \( \sqrt{y^2} = |y| \).

 

Question 13. If \( m - nx + 28x^2 + 12x^3 + 9x^4 \) is a perfect square, then find the values of m and n.
Answer:
First, arrange the polynomial in descending powers of \( x \):
\( 9x^4 + 12x^3 + 28x^2 - nx + m \)
We use the long division method to find the square root of the polynomial and identify the coefficients \( m \) and \( n \).
\[ \require{enclose} \begin{array}{r} 3x^2 + 2x + 4 \\ \enclose{longdiv}{9x^4 + 12x^3 + 28x^2 - nx + m} \\ \underline{9x^4} \\ 6x^2+2x \enclose{longdiv}{\;12x^3 + 28x^2} \\ \underline{\;12x^3 + \;4x^2} \\ 6x^2+4x+4 \enclose{longdiv}{\;24x^2 - nx + m} \\ \underline{\;24x^2 + 16x + 16} \\ (-n-16)x + (m-16) \end{array} \] For the given polynomial to be a perfect square, the remainder must be 0.
Therefore, \( (-n - 16)x + (m - 16) = 0 \).
This implies that the coefficients of \( x \) and the constant term in the remainder must both be zero.
So, \( -n - 16 = 0 \implies -n = 16 \implies n = -16 \)
And \( m - 16 = 0 \implies m = 16 \)
The values are \( m = 16 \) and \( n = -16 \).
In simple words: We want to find the values of 'm' and 'n' so that the given polynomial is a perfect square. We use a method similar to long division for numbers, but with polynomials. We divide the polynomial to find its square root. For it to be a perfect square, there should be no remainder. We set the remainder to zero, which helps us solve for 'm' and 'n'.

🎯 Exam Tip: When a polynomial is a perfect square, its remainder in polynomial long division for square roots must be zero. Equating the coefficients of the remainder to zero allows you to solve for unknown variables like 'm' and 'n'.

 

Question 14. If \( b + \frac{a}{x} + \frac{13}{x^2} - \frac{6}{x^3} + \frac{1}{x^4} \) is a perfect square, find the values of “a” and “b”
Answer:
First, arrange the polynomial in descending powers of \( x \):
\( \frac{1}{x^4} - \frac{6}{x^3} + \frac{13}{x^2} + \frac{a}{x} + b \)
We use the long division method to find the square root of the polynomial and identify the coefficients \( a \) and \( b \).
\[ \require{enclose} \begin{array}{r} \frac{1}{x^2} - \frac{3}{x} + 2 \\ \enclose{longdiv}{\frac{1}{x^4} - \frac{6}{x^3} + \frac{13}{x^2} + \frac{a}{x} + b} \\ \underline{\frac{1}{x^4}} \\ \frac{2}{x^2}-\frac{3}{x} \enclose{longdiv}{\;-\frac{6}{x^3} + \frac{13}{x^2}} \\ \underline{\;-\frac{6}{x^3} + \frac{9}{x^2}} \\ \frac{2}{x^2}-\frac{6}{x}+2 \enclose{longdiv}{\;\frac{4}{x^2} + \frac{a}{x} + b} \\ \underline{\;\frac{4}{x^2} - \frac{12}{x} + 4} \\ \left(\frac{a}{x} - (-\frac{12}{x})\right) + (b - 4) \\ = \left(\frac{a+12}{x}\right) + (b-4) \end{array} \] For the given polynomial to be a perfect square, the remainder must be 0.
Therefore, \( \left(\frac{a+12}{x}\right) + (b-4) = 0 \).
This implies that the coefficients of \( \frac{1}{x} \) and the constant term in the remainder must both be zero.
So, \( \frac{a+12}{x} = 0 \implies a+12 = 0 \implies a = -12 \)
And \( b - 4 = 0 \implies b = 4 \)
The values are \( a = -12 \) and \( b = 4 \).
In simple words: To find the unknown values 'a' and 'b' in this polynomial so it becomes a perfect square, we first write the polynomial from the highest power of x to the lowest. Then we use a special long division method, similar to finding the square root of a number. For the polynomial to be a perfect square, the remainder of this division must be zero. By setting the remainder to zero, we can find the exact values of 'a' and 'b'.

🎯 Exam Tip: When working with polynomials containing inverse powers of \( x \), arrange them in descending powers of \( x \) (e.g., \( x^{-4}, x^{-3}, \dots \)) to apply the long division for square roots correctly. Remember that the remainder must be zero for a perfect square.

 

Question 15. Solve \( \frac{1}{x+1} + \frac{4}{3x+6} = \frac{2}{3} \)
Answer:
First, simplify the second term's denominator:
\( 3x + 6 = 3(x + 2) \)
So the equation becomes: \( \frac{1}{x+1} + \frac{4}{3(x+2)} = \frac{2}{3} \)

Find the Least Common Multiple (LCM) of the denominators: \( 3(x+1)(x+2) \).
Multiply every term by the LCM to clear the denominators:
\( 3(x+1)(x+2) \left(\frac{1}{x+1}\right) + 3(x+1)(x+2) \left(\frac{4}{3(x+2)}\right) = 3(x+1)(x+2) \left(\frac{2}{3}\right) \)

Simplify each term:
\( 3(x+2) + 4(x+1) = 2(x+1)(x+2) \)

Expand both sides of the equation:
Left side: \( 3x + 6 + 4x + 4 = 7x + 10 \)
Right side: \( 2(x^2 + 2x + x + 2) = 2(x^2 + 3x + 2) = 2x^2 + 6x + 4 \)

Now, set the expanded sides equal to each other:
\( 7x + 10 = 2x^2 + 6x + 4 \)

Move all terms to one side to form a quadratic equation:
\( 0 = 2x^2 + 6x - 7x + 4 - 10 \)
\( 0 = 2x^2 - x - 6 \)

Solve the quadratic equation \( 2x^2 - x - 6 = 0 \) by factoring:
We look for two numbers that multiply to \( 2 \times -6 = -12 \) and add to -1. These numbers are -4 and 3.
\( 2x^2 - 4x + 3x - 6 = 0 \)
\( 2x(x - 2) + 3(x - 2) = 0 \)
\( (x - 2)(2x + 3) = 0 \)

Set each factor to zero to find the solutions:
\( x - 2 = 0 \implies x = 2 \)
\( 2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2} \)
The solution set is \( \left\{2, -\frac{3}{2}\right\} \).
In simple words: First, we make the denominators simple. Then, we find the smallest common number that all denominators can divide into (LCM). We multiply every part of the equation by this LCM to get rid of the fractions. This gives us a simpler equation, usually a quadratic one. We then solve this quadratic equation to find the values of \( x \).

🎯 Exam Tip: Always check your solutions by plugging them back into the original equation to ensure they don't make any denominator zero, as such values are extraneous and must be discarded.

 

Question 16. A two-digit number is such that the product of the digits is 14. When 45 is added to the number, the digits interchange their places. Find the number (solve by completing square method)
Answer:
Let the ten's digit be \( x \).
Let the unit digit be \( y \).
The product of the digits is 14, so \( xy = 14 \).
This means the unit digit \( y = \frac{14}{x} \).

The original two-digit number can be written as \( 10x + y \).
Substituting \( y = \frac{14}{x} \), the number is \( 10x + \frac{14}{x} \).

When the digits interchange their places, the new number is \( 10y + x \).
Substituting \( y = \frac{14}{x} \), the new number is \( 10\left(\frac{14}{x}\right) + x = \frac{140}{x} + x \).

According to the problem, when 45 is added to the original number, the digits interchange places:
\( \left(10x + \frac{14}{x}\right) + 45 = \frac{140}{x} + x \)

Multiply the entire equation by \( x \) to eliminate the fractions (assuming \( x \ne 0 \)):
\( 10x^2 + 14 + 45x = 140 + x^2 \)

Rearrange the terms to form a quadratic equation:
\( 10x^2 - x^2 + 45x + 14 - 140 = 0 \)
\( 9x^2 + 45x - 126 = 0 \)

Divide the entire equation by 9 to simplify:
\( x^2 + 5x - 14 = 0 \)

Now, solve this quadratic equation by the completing the square method.
\( x^2 + 5x = 14 \)
To complete the square, add \( \left(\frac{5}{2}\right)^2 \) to both sides:
\( x^2 + 5x + \left(\frac{5}{2}\right)^2 = 14 + \left(\frac{5}{2}\right)^2 \)
\( \left(x + \frac{5}{2}\right)^2 = 14 + \frac{25}{4} \)
\( \left(x + \frac{5}{2}\right)^2 = \frac{56}{4} + \frac{25}{4} \)
\( \left(x + \frac{5}{2}\right)^2 = \frac{81}{4} \)

Take the square root of both sides:
\( x + \frac{5}{2} = \pm\sqrt{\frac{81}{4}} \)
\( x + \frac{5}{2} = \pm\frac{9}{2} \)

Solve for \( x \):
Case 1: \( x + \frac{5}{2} = \frac{9}{2} \)
\( x = \frac{9}{2} - \frac{5}{2} = \frac{4}{2} = 2 \)

Case 2: \( x + \frac{5}{2} = -\frac{9}{2} \)
\( x = -\frac{9}{2} - \frac{5}{2} = -\frac{14}{2} = -7 \)

Since \( x \) represents a digit of a number, it must be a positive single digit. So \( x = 2 \).
Now find \( y \) using \( y = \frac{14}{x} \):
\( y = \frac{14}{2} = 7 \)

The original two-digit number is \( 10x + y = 10(2) + 7 = 20 + 7 = 27 \).
Check: Product of digits \( 2 \times 7 = 14 \) (Correct).
If 45 is added to 27: \( 27 + 45 = 72 \). The digits are interchanged (Correct).
In simple words: We set up the problem using 'x' for the tens digit. We express the units digit and the number itself using 'x'. Then, we write an equation based on the condition that adding 45 swaps the digits. We simplify this equation into a quadratic form. We then use the "completing the square" method to solve for 'x'. Since 'x' must be a positive digit, we find the correct 'x' and then 'y', which gives us the final two-digit number.

🎯 Exam Tip: For digit-based word problems, represent the number using variables for digits (e.g., \( 10x+y \)). Clearly translate the conditions into algebraic equations. When solving quadratic equations, be sure to consider only valid solutions (e.g., positive digits for physical quantities).

 

Question 17. A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of walk is 120 sqm assuming the width of walk be 'x' form the equation then solve it by formula method.
Answer:
Dimensions of the garden:
Length \( L = 16 \) m
Width \( W = 10 \) m
Area of the garden \( = L \times W = 16 \times 10 = 160 \) sq.m

Let the uniform width of the concrete walk be \( x \) meters.
When the walk surrounds the garden, the new dimensions of the garden including the walk will be:
New Length \( = 16 + 2x \) m (since \( x \) is added to both ends of the length)
New Width \( = 10 + 2x \) m (since \( x \) is added to both ends of the width)

Area of the garden with the walk \( = (16 + 2x)(10 + 2x) \)
Expand this expression:
\( = 16(10) + 16(2x) + 2x(10) + 2x(2x) \)
\( = 160 + 32x + 20x + 4x^2 \)
\( = 4x^2 + 52x + 160 \) sq.m

The area of the walk is the difference between the area of the garden with the walk and the area of the garden.
Area of walk \( = (\text{Area of garden with walk}) - (\text{Area of garden}) \)
Given Area of walk \( = 120 \) sq.m
So, \( 120 = (4x^2 + 52x + 160) - 160 \)
\( 120 = 4x^2 + 52x \)

Rearrange the equation to the standard quadratic form \( ax^2 + bx + c = 0 \):
\( 4x^2 + 52x - 120 = 0 \)

Divide the entire equation by 4 to simplify it:
\( x^2 + 13x - 30 = 0 \)

Now, solve this quadratic equation using the formula method:
For \( ax^2 + bx + c = 0 \), the solutions are \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Here, \( a = 1, b = 13, c = -30 \).

Substitute these values into the formula:
\( x = \frac{-(13) \pm \sqrt{(13)^2 - 4(1)(-30)}}{2(1)} \)
\( x = \frac{-13 \pm \sqrt{169 + 120}}{2} \)
\( x = \frac{-13 \pm \sqrt{289}}{2} \)
\( x = \frac{-13 \pm 17}{2} \)

We have two possible solutions for \( x \):
Case 1: \( x = \frac{-13 + 17}{2} = \frac{4}{2} = 2 \)
Case 2: \( x = \frac{-13 - 17}{2} = \frac{-30}{2} = -15 \)

Since the width of the walk cannot be negative, we discard \( x = -15 \).
Therefore, the uniform width of the walk is \( x = 2 \) meters.
In simple words: First, we find the area of the garden. Then, we imagine the garden with the path around it, which makes it bigger. We find the area of this bigger shape. The area of the path itself is the difference between these two areas. We set up an equation with this information, which turns into a quadratic equation. We solve this equation using the special formula, and since width cannot be negative, we choose the positive answer. So, the path is 2 meters wide.

🎯 Exam Tip: For geometry word problems, drawing a simple diagram can help visualize the dimensions. Always define variables clearly and ensure you use the correct formula for areas. Remember to check for realistic solutions, as negative lengths or widths are usually not valid.

 

Question 18. If \( \alpha \) and \( \beta \) are the roots of the equation \( 3x^2 - 5x + 2 = 0 \) find the value of
(i) \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \)
(ii) \( \alpha - \beta \)
(iii) \( \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha} \)
Answer:
Given the quadratic equation: \( 3x^2 - 5x + 2 = 0 \)
Comparing this with the standard quadratic equation \( ax^2 + bx + c = 0 \), we have:
\( a = 3, b = -5, c = 2 \)

For a quadratic equation, the sum of the roots \( (\alpha + \beta) \) and the product of the roots \( (\alpha\beta) \) are given by:
\( \alpha + \beta = -\frac{b}{a} = -\frac{(-5)}{3} = \frac{5}{3} \)
\( \alpha\beta = \frac{c}{a} = \frac{2}{3} \)

(i) Find the value of \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \):
To add these fractions, find a common denominator:
\( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} \)
We know that \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \).
Substitute this into the expression:
\( = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} \)

Now, substitute the values of \( (\alpha + \beta) \) and \( \alpha\beta \):
\( = \frac{\left(\frac{5}{3}\right)^2 - 2\left(\frac{2}{3}\right)}{\frac{2}{3}} \)
\( = \frac{\frac{25}{9} - \frac{4}{3}}{\frac{2}{3}} \)
To subtract in the numerator, find a common denominator (9):
\( = \frac{\frac{25}{9} - \frac{12}{9}}{\frac{2}{3}} = \frac{\frac{13}{9}}{\frac{2}{3}} \)
To divide fractions, multiply by the reciprocal:
\( = \frac{13}{9} \times \frac{3}{2} = \frac{13}{3 \times 2} = \frac{13}{6} \)

(ii) Find the value of \( \alpha - \beta \):
We know that \( (\alpha - \beta)^2 = \alpha^2 + \beta^2 - 2\alpha\beta \).
Also, \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \).
So, \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 2\alpha\beta - 2\alpha\beta \)
\( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \)

Substitute the values of \( (\alpha + \beta) \) and \( \alpha\beta \):
\( (\alpha - \beta)^2 = \left(\frac{5}{3}\right)^2 - 4\left(\frac{2}{3}\right) \)
\( = \frac{25}{9} - \frac{8}{3} \)
To subtract, find a common denominator (9):
\( = \frac{25}{9} - \frac{24}{9} = \frac{1}{9} \)
Now, take the square root of both sides:
\( \alpha - \beta = \pm\sqrt{\frac{1}{9}} = \pm\frac{1}{3} \)

(iii) Find the value of \( \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha} \):
To add these fractions, find a common denominator:
\( \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta} \)
We know the identity \( \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \).
Substitute this into the expression:
\( = \frac{(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)}{\alpha\beta} \)

Now, substitute the values of \( (\alpha + \beta) \) and \( \alpha\beta \):
\( = \frac{\left(\frac{5}{3}\right)^3 - 3\left(\frac{2}{3}\right)\left(\frac{5}{3}\right)}{\frac{2}{3}} \)
\( = \frac{\frac{125}{27} - 3 \times \frac{10}{9}}{\frac{2}{3}} \)
\( = \frac{\frac{125}{27} - \frac{30}{9}}{\frac{2}{3}} \)
To subtract in the numerator, find a common denominator (27):
\( = \frac{\frac{125}{27} - \frac{90}{27}}{\frac{2}{3}} = \frac{\frac{35}{27}}{\frac{2}{3}} \)
To divide fractions, multiply by the reciprocal:
\( = \frac{35}{27} \times \frac{3}{2} = \frac{35}{9 \times 2} = \frac{35}{18} \)
In simple words: For a given quadratic equation, we first find the sum and product of its roots using special formulas. Then, for each sub-question, we rewrite the expression in terms of the sum and product of the roots. Finally, we plug in the calculated sum and product values to get the numerical answer. Remember that \( \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \) and \( \alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) \).

🎯 Exam Tip: Master the relationships between roots and coefficients for a quadratic equation: \( \alpha + \beta = -b/a \) and \( \alpha\beta = c/a \). Also, memorize common algebraic identities for \( \alpha^2+\beta^2 \), \( (\alpha-\beta)^2 \), and \( \alpha^3+\beta^3 \) in terms of \( \alpha+\beta \) and \( \alpha\beta \).

 

Question 18. If \( \alpha \) and \( \beta \) are the roots of the equation \( 3x^2 – 5x + 2 = 0 \), then find the value of
(iii) \( \frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha} \)
Answer: We need to find \( \frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha} \). First, we find the values for \( \alpha + \beta \) and \( \alpha\beta \) from the equation \( 3x^2 – 5x + 2 = 0 \).
Comparing with \( ax^2 + bx + c = 0 \), we have \( a = 3, b = -5, c = 2 \).
Sum of the roots: \( \alpha + \beta = \frac{-b}{a} = \frac{-(-5)}{3} = \frac{5}{3} \)
Product of the roots: \( \alpha\beta = \frac{c}{a} = \frac{2}{3} \)
Now, let's simplify the expression:
\( \frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha} = \frac{\alpha^{3}+\beta^{3}}{\alpha\beta} \)
We know the identity \( a^3+b^3 = (a+b)^3-3ab(a + b) \).
So, \( \alpha^{3}+\beta^{3} = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) \)
Substitute these into the expression:
\( \frac{\alpha^{3}+\beta^{3}}{\alpha\beta} = \frac{(\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)}{\alpha\beta} \)
Now, plug in the values of \( \alpha + \beta \) and \( \alpha\beta \):
\( = \frac{(\frac{5}{3})^3 - 3(\frac{2}{3})(\frac{5}{3})}{\frac{2}{3}} \)
\( = \frac{\frac{125}{27} - \frac{30}{9}}{\frac{2}{3}} \)
To subtract in the numerator, find a common denominator (27):
\( = \frac{\frac{125}{27} - \frac{30 \times 3}{9 \times 3}}{\frac{2}{3}} = \frac{\frac{125}{27} - \frac{90}{27}}{\frac{2}{3}} \)
\( = \frac{\frac{125 - 90}{27}}{\frac{2}{3}} = \frac{\frac{35}{27}}{\frac{2}{3}} \)
To divide fractions, multiply by the reciprocal of the denominator:
\( = \frac{35}{27} \times \frac{3}{2} \)
Now, simplify by canceling common factors (3 in 27):
\( = \frac{35}{9 \times 2} = \frac{35}{18} \)
In simple words: First, find the sum and product of the roots from the given equation. Then, use a special formula to expand \( \alpha^3 + \beta^3 \) and substitute the values you found. Finally, do the math to get the answer.

🎯 Exam Tip: Remember key algebraic identities like \( a^3+b^3 = (a+b)^3-3ab(a+b) \) and \( a^2+b^2 = (a+b)^2-2ab \) as they are essential for simplifying expressions involving roots.

 

Question 19. If \( \alpha \) and \( \beta \) are the roots of the equation \( 3x^2 – 6x + 1 = 0 \), form the equation whose roots are
(i) \( \alpha^2\beta; \beta^2\alpha \)
(ii) \( 2\alpha + \beta; 2\beta + \alpha \)
Answer: We are given the quadratic equation \( 3x^2 – 6x + 1 = 0 \).
Comparing with \( Ax^2 + Bx + C = 0 \), we have \( A=3, B=-6, C=1 \).
Sum of the roots: \( \alpha + \beta = \frac{-B}{A} = \frac{-(-6)}{3} = \frac{6}{3} = 2 \)
Product of the roots: \( \alpha\beta = \frac{C}{A} = \frac{1}{3} \)

(i) The new roots are \( \alpha^2\beta \) and \( \beta^2\alpha \).
First, find the sum of these new roots:
Sum \( = \alpha^2\beta + \beta^2\alpha = \alpha\beta(\alpha + \beta) \)
Substitute the values of \( \alpha\beta \) and \( \alpha + \beta \):
Sum \( = (\frac{1}{3})(2) = \frac{2}{3} \)
Next, find the product of these new roots:
Product \( = (\alpha^2\beta)(\beta^2\alpha) = \alpha^3\beta^3 = (\alpha\beta)^3 \)
Substitute the value of \( \alpha\beta \):
Product \( = (\frac{1}{3})^3 = \frac{1}{27} \)
The quadratic equation is given by \( x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0 \).
So, \( x^2 - \frac{2}{3}x + \frac{1}{27} = 0 \)
To remove fractions, multiply the entire equation by 27:
\( 27x^2 - 27(\frac{2}{3})x + 27(\frac{1}{27}) = 0 \)
\( 27x^2 - 18x + 1 = 0 \)

(ii) The new roots are \( 2\alpha + \beta \) and \( 2\beta + \alpha \).
First, find the sum of these new roots:
Sum \( = (2\alpha + \beta) + (2\beta + \alpha) = 3\alpha + 3\beta = 3(\alpha + \beta) \)
Substitute the value of \( \alpha + \beta \):
Sum \( = 3(2) = 6 \)
Next, find the product of these new roots:
Product \( = (2\alpha + \beta)(2\beta + \alpha) \)
Expand the product:
\( = 4\alpha\beta + 2\alpha^2 + 2\beta^2 + \alpha\beta \)
Combine like terms:
\( = 5\alpha\beta + 2\alpha^2 + 2\beta^2 = 5\alpha\beta + 2(\alpha^2 + \beta^2) \)
Use the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\( = 5\alpha\beta + 2((\alpha + \beta)^2 - 2\alpha\beta) \)
Substitute the values of \( \alpha + \beta \) and \( \alpha\beta \):
\( = 5(\frac{1}{3}) + 2((2)^2 - 2(\frac{1}{3})) \)
\( = \frac{5}{3} + 2(4 - \frac{2}{3}) \)
\( = \frac{5}{3} + 2(\frac{12-2}{3}) = \frac{5}{3} + 2(\frac{10}{3}) \)
\( = \frac{5}{3} + \frac{20}{3} = \frac{25}{3} \)
The quadratic equation is given by \( x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0 \).
So, \( x^2 - 6x + \frac{25}{3} = 0 \)
To remove fractions, multiply the entire equation by 3:
\( 3x^2 - 18x + 25 = 0 \)
In simple words: For each part, first find the sum and product of the new roots using the values of \( \alpha + \beta \) and \( \alpha\beta \) from the original equation. Then, use the standard formula to write the new quadratic equation. Remember to clear any fractions by multiplying the entire equation by the common denominator.

🎯 Exam Tip: When forming a new quadratic equation, always remember the formula: \( x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0 \). Be careful with algebraic expansions and fraction arithmetic.

 

Question 20. Find X and Y if \( X - Y = \begin{bmatrix} 2 & 1 \\ 4 & 3 \\ 0 & 6 \end{bmatrix} \) and \( X + Y = \begin{bmatrix} 12 & 13 \\ 6 & 5 \\ 4 & 8 \end{bmatrix} \).
Answer: Let the given equations be:
\( X - Y = \begin{bmatrix} 2 & 1 \\ 4 & 3 \\ 0 & 6 \end{bmatrix} \) .....(1)
\( X + Y = \begin{bmatrix} 12 & 13 \\ 6 & 5 \\ 4 & 8 \end{bmatrix} \) .....(2)

To find X, add equation (1) and equation (2):
\( (X - Y) + (X + Y) = \begin{bmatrix} 2 & 1 \\ 4 & 3 \\ 0 & 6 \end{bmatrix} + \begin{bmatrix} 12 & 13 \\ 6 & 5 \\ 4 & 8 \end{bmatrix} \)
\( 2X = \begin{bmatrix} 2+12 & 1+13 \\ 4+6 & 3+5 \\ 0+4 & 6+8 \end{bmatrix} \)
\( 2X = \begin{bmatrix} 14 & 14 \\ 10 & 8 \\ 4 & 14 \end{bmatrix} \)
Now, divide each element of the matrix by 2 to find X:
\( X = \frac{1}{2} \begin{bmatrix} 14 & 14 \\ 10 & 8 \\ 4 & 14 \end{bmatrix} = \begin{bmatrix} 7 & 7 \\ 5 & 4 \\ 2 & 7 \end{bmatrix} \)

To find Y, subtract equation (2) from equation (1) (or (1) from (2), then change sign):
\( (X - Y) - (X + Y) = \begin{bmatrix} 2 & 1 \\ 4 & 3 \\ 0 & 6 \end{bmatrix} - \begin{bmatrix} 12 & 13 \\ 6 & 5 \\ 4 & 8 \end{bmatrix} \)
\( -2Y = \begin{bmatrix} 2-12 & 1-13 \\ 4-6 & 3-5 \\ 0-4 & 6-8 \end{bmatrix} \)
\( -2Y = \begin{bmatrix} -10 & -12 \\ -2 & -2 \\ -4 & -2 \end{bmatrix} \)
Now, divide each element of the matrix by -2 to find Y:
\( Y = \frac{1}{-2} \begin{bmatrix} -10 & -12 \\ -2 & -2 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 1 \\ 2 & 1 \end{bmatrix} \)
In simple words: Treat these matrix equations like normal algebraic equations. Add the two equations to find \( 2X \), then divide by 2 to get X. Subtract the equations to find \( -2Y \), then divide by -2 to get Y. Each step involves adding or subtracting corresponding elements in the matrices.

🎯 Exam Tip: When adding or subtracting matrices, remember that only elements in the exact same position are combined. Always double-check your arithmetic, especially with negative numbers.

 

Question 21. Solve for x, y: \( \begin{bmatrix} x^2 \\ y^2 \end{bmatrix} + \begin{bmatrix} 6x \\ -9y \end{bmatrix} = \begin{bmatrix} -5 \\ -18 \end{bmatrix} \).
Answer: We are given the matrix equation:
\( \begin{bmatrix} x^2 \\ y^2 \end{bmatrix} + \begin{bmatrix} 6x \\ -9y \end{bmatrix} = \begin{bmatrix} -5 \\ -18 \end{bmatrix} \)
To solve this, we equate the corresponding elements of the matrices. This gives us two separate equations:
1) \( x^2 + 6x = -5 \)
2) \( y^2 - 9y = -18 \)

Let's solve the first equation for x:
\( x^2 + 6x = -5 \)
Move -5 to the left side to form a standard quadratic equation:
\( x^2 + 6x + 5 = 0 \)
Factor the quadratic equation (find two numbers that multiply to 5 and add to 6, which are 5 and 1):
\( (x + 5)(x + 1) = 0 \)
Set each factor to zero to find the possible values of x:
\( x + 5 = 0 \implies x = -5 \)
\( x + 1 = 0 \implies x = -1 \)

Now, let's solve the second equation for y:
\( y^2 - 9y = -18 \)
Move -18 to the left side to form a standard quadratic equation:
\( y^2 - 9y + 18 = 0 \)
Factor the quadratic equation (find two numbers that multiply to 18 and add to -9, which are -6 and -3):
\( (y - 6)(y - 3) = 0 \)
Set each factor to zero to find the possible values of y:
\( y - 6 = 0 \implies y = 6 \)
\( y - 3 = 0 \implies y = 3 \)

Thus, the values for x are -5 or -1, and the values for y are 6 or 3.
In simple words: When two matrices are equal, their matching parts must also be equal. This helps us turn the matrix problem into two simple equations for x and y. Then, we solve these quadratic equations by moving all terms to one side and factoring them to find the possible values for x and y.

🎯 Exam Tip: Always remember that corresponding elements in equal matrices are equal. For quadratic equations, factoring or using the quadratic formula are the most common methods to find the roots.

 

Question 22. Given the matrix \( A = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \), show that \( A^2 – 7A + 10I_3 = 0 \).
Answer: We are given the matrix \( A = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \). We need to show that \( A^2 – 7A + 10I_3 = 0 \).
First, calculate \( A^2 = A \times A \):
\( A^2 = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \times \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \)
\( = \begin{bmatrix} (3)(3)+(2)(1)+(0)(0) & (3)(2)+(2)(4)+(0)(0) & (3)(0)+(2)(0)+(0)(5) \\ (1)(3)+(4)(1)+(0)(0) & (1)(2)+(4)(4)+(0)(0) & (1)(0)+(4)(0)+(0)(5) \\ (0)(3)+(0)(1)+(5)(0) & (0)(2)+(0)(4)+(5)(0) & (0)(0)+(0)(0)+(5)(5) \end{bmatrix} \)
\( = \begin{bmatrix} 9+2+0 & 6+8+0 & 0+0+0 \\ 3+4+0 & 2+16+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+25 \end{bmatrix} \)
\( = \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} \)

Next, calculate \( 7A \):
\( 7A = 7 \times \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} = \begin{bmatrix} 7 \times 3 & 7 \times 2 & 7 \times 0 \\ 7 \times 1 & 7 \times 4 & 7 \times 0 \\ 7 \times 0 & 7 \times 0 & 7 \times 5 \end{bmatrix} = \begin{bmatrix} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{bmatrix} \)

Then, calculate \( 10I_3 \), where \( I_3 \) is the 3x3 identity matrix:
\( 10I_3 = 10 \times \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} \)

Now, substitute these into the expression \( A^2 – 7A + 10I_3 \):
\( A^2 – 7A + 10I_3 = \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} - \begin{bmatrix} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{bmatrix} + \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} \)
Perform the subtraction first:
\( = \begin{bmatrix} 11-21 & 14-14 & 0-0 \\ 7-7 & 18-28 & 0-0 \\ 0-0 & 0-0 & 25-35 \end{bmatrix} + \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} \)
\( = \begin{bmatrix} -10 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & -10 \end{bmatrix} + \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} \)
Perform the addition:
\( = \begin{bmatrix} -10+10 & 0+0 & 0+0 \\ 0+0 & -10+10 & 0+0 \\ 0+0 & 0+0 & -10+10 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
This is the zero matrix. Thus, \( A^2 – 7A + 10I_3 = 0 \) is shown.
In simple words: To prove the equation, first calculate \( A \) multiplied by itself (which is \( A^2 \)). Then, multiply \( A \) by 7 and the identity matrix \( I_3 \) by 10. Finally, combine these three resulting matrices using subtraction and addition. If the final matrix is all zeros, the equation is proven.

🎯 Exam Tip: When performing matrix operations, ensure you follow the correct order (multiplication before addition/subtraction). Remember that for multiplication, rows of the first matrix are multiplied by columns of the second.

 

Question 23. Given \( A = \begin{bmatrix} 2 & 3 & -1 \\ 4 & 1 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -2 \\ 3 & -3 \\ 2 & 6 \end{bmatrix} \), show that \( (AB)^T = B^T A^T \).
Answer: We are given the matrices \( A = \begin{bmatrix} 2 & 3 & -1 \\ 4 & 1 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -2 \\ 3 & -3 \\ 2 & 6 \end{bmatrix} \). We need to show that \( (AB)^T = B^T A^T \).

First, calculate \( AB \):
\( AB = \begin{bmatrix} 2 & 3 & -1 \\ 4 & 1 & 5 \end{bmatrix} \times \begin{bmatrix} 1 & -2 \\ 3 & -3 \\ 2 & 6 \end{bmatrix} \)
\( = \begin{bmatrix} (2)(1)+(3)(3)+(-1)(2) & (2)(-2)+(3)(-3)+(-1)(6) \\ (4)(1)+(1)(3)+(5)(2) & (4)(-2)+(1)(-3)+(5)(6) \end{bmatrix} \)
\( = \begin{bmatrix} 2+9-2 & -4-9-6 \\ 4+3+10 & -8-3+30 \end{bmatrix} \)
\( = \begin{bmatrix} 9 & -19 \\ 17 & 19 \end{bmatrix} \)

Next, find the transpose of \( AB \), denoted as \( (AB)^T \). This involves swapping rows and columns:
\( (AB)^T = \begin{bmatrix} 9 & 17 \\ -19 & 19 \end{bmatrix} \) .....(1)

Now, find the transpose of B, \( B^T \), and the transpose of A, \( A^T \):
\( B^T = \begin{bmatrix} 1 & 3 & 2 \\ -2 & -3 & 6 \end{bmatrix} \)
\( A^T = \begin{bmatrix} 2 & 4 \\ 3 & 1 \\ -1 & 5 \end{bmatrix} \)

Then, calculate \( B^T A^T \):
\( B^T A^T = \begin{bmatrix} 1 & 3 & 2 \\ -2 & -3 & 6 \end{bmatrix} \times \begin{bmatrix} 2 & 4 \\ 3 & 1 \\ -1 & 5 \end{bmatrix} \)
\( = \begin{bmatrix} (1)(2)+(3)(3)+(2)(-1) & (1)(4)+(3)(1)+(2)(5) \\ (-2)(2)+(-3)(3)+(6)(-1) & (-2)(4)+(-3)(1)+(6)(5) \end{bmatrix} \)
\( = \begin{bmatrix} 2+9-2 & 4+3+10 \\ -4-9-6 & -8-3+30 \end{bmatrix} \)
\( = \begin{bmatrix} 9 & 17 \\ -19 & 19 \end{bmatrix} \) .....(2)

From (1) and (2), we can see that \( (AB)^T = B^T A^T \). This verifies the property.
In simple words: First, multiply matrices A and B, then flip the result (transpose it). This is the left side. For the right side, flip B and A separately (transpose them), then multiply these flipped matrices. If both sides give the same matrix, the property is proven.

🎯 Exam Tip: Remember the property \( (AB)^T = B^T A^T \) for matrix transposes. Matrix multiplication is not commutative, so the order \( B^T A^T \) is crucial; \( A^T B^T \) would generally give a different result.

 

Question 24. Draw the graph of \( y = x^2 \) and hence solve \( x^2 – 4x – 5 = 0 \).
Answer: We need to draw the graph of \( y = x^2 \) and use it to solve the equation \( x^2 – 4x – 5 = 0 \).

**Step 1: Create a table of values for \( y = x^2 \).**
Let's choose x values from -4 to 5.

Plot these points: (-4, 16), (-3, 9), (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9), (4, 16), (5, 25). Join these points to form a smooth curve (a parabola).

**Step 2: Rewrite the equation to be solved.**
We have \( y = x^2 \) and the equation \( x^2 – 4x – 5 = 0 \).
Substitute \( y \) for \( x^2 \) in the second equation: \( y – 4x – 5 = 0 \).
Rearrange this to get an equation for a straight line: \( y = 4x + 5 \).

**Step 3: Create a table of values for the line \( y = 4x + 5 \).**
Use the same x values as for the parabola.

Plot these points: (-3, -7), (-1, 1), (0, 5), (2, 13), (3, 17), (5, 25). Join these points to form a straight line.

**Step 4: Find the intersection points.**
The graph shows that the straight line \( y = 4x + 5 \) intersects the parabola \( y = x^2 \) at two points. By looking at the graph, these intersection points are (-1, 1) and (5, 25).
The x-coordinates of these intersection points are the solutions to the equation \( x^2 – 4x – 5 = 0 \).
Thus, the solutions are \( x = -1 \) and \( x = 5 \).
The solution set is \( \{ -1, 5 \} \).
In simple words: First, draw the graph of \( y = x^2 \). Then, change the equation you need to solve into a straight line equation, like \( y = 4x + 5 \). Draw this line on the same graph. The x-values where the line and the curve cross are the answers to the equation.

🎯 Exam Tip: Always clearly label both curves on your graph. The x-coordinates of the intersection points are the roots of the quadratic equation you are solving, not the y-coordinates.

 

Question 25. Draw the graph of \( y = 2x^2 + x – 6 \) and hence solve \( 2x^2 + x – 10 = 0 \).
Answer: We need to draw the graph of \( y = 2x^2 + x – 6 \) and use it to solve the equation \( 2x^2 + x – 10 = 0 \).

**Step 1: Create a table of values for \( y = 2x^2 + x – 6 \).**
Let's choose x values from -4 to 4.

Plot these points: (-4, 22), (-3, 9), (-2, 0), (-1, -5), (0, -6), (1, -3), (2, 4), (3, 15), (4, 30). Join these points to form a smooth curve (a parabola).

**Step 2: Rewrite the equation to be solved.**
We have \( y = 2x^2 + x – 6 \) and the equation \( 2x^2 + x – 10 = 0 \).
Subtract the equation \( 2x^2 + x – 10 = 0 \) from the graph equation \( y = 2x^2 + x – 6 \):
\( y - (2x^2 + x – 10) = (2x^2 + x – 6) - (2x^2 + x – 10) \)
\( y - 2x^2 - x + 10 = 2x^2 + x - 6 - 2x^2 - x + 10 \)
\( y - (2x^2 + x - 6) = 4 \)
So, \( y = 4 \). This is the equation of a straight line parallel to the x-axis.

**Step 3: Draw the line \( y = 4 \).**
Draw a horizontal straight line at \( y = 4 \) on the same graph.

**Step 4: Find the intersection points.**
The graph shows that the straight line \( y = 4 \) intersects the parabola \( y = 2x^2 + x – 6 \) at two points. By looking at the graph, these intersection points are approximately (-2.5, 4) and (2, 4).
The x-coordinates of these intersection points are the solutions to the equation \( 2x^2 + x – 10 = 0 \).
Thus, the solutions are \( x = -2.5 \) and \( x = 2 \).
The solution set is \( \{ -2.5, 2 \} \).
In simple words: First, draw the main curve \( y = 2x^2 + x – 6 \). To solve \( 2x^2 + x – 10 = 0 \), subtract this equation from the first one to find a simple straight line (here, \( y=4 \)). Draw this line on your graph. The x-values where the line and curve meet are the answers.

🎯 Exam Tip: When using graphs to solve equations, the method of subtracting the equation from the graph equation is very useful. Ensure your scale is consistent and clearly marked on both axes.