OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Exercise 10 (C)

Question 1. Without solving, find the nature of the roots of the following equations :
(i) \( 3x^2 - 7x + 5 = 0 \)
(ii) \( 4x^2 + 4x + 1 = 0 \)
(iii) \( 3x^2 + 7x + 2 = 0 \)
(iv) \( x^2 + px - q^2 = 0 \)
Answer:
(i) Given quadratic equation is \( 3x^2 - 7x + 5 = 0 \).
We compare this with the standard form \( ax^2 + bx + c = 0 \).
Here, \( a = 3 \), \( b = -7 \), and \( c = 5 \).
The discriminant D is calculated as \( b^2 - 4ac \).
\( D = (-7)^2 - 4 \times 3 \times 5 \)
\( D = 49 - 60 \)
\( D = -11 \)
Since \( D < 0 \), the roots of this quadratic equation are imaginary. Imaginary roots mean there are no real number solutions for the equation.
(ii) Given quadratic equation is \( 4x^2 + 4x + 1 = 0 \).
Comparing with \( ax^2 + bx + c = 0 \), we have \( a = 4 \), \( b = 4 \), and \( c = 1 \).
The discriminant D is \( b^2 - 4ac \).
\( D = (4)^2 - 4 \times 4 \times 1 \)
\( D = 16 - 16 \)
\( D = 0 \)
Since \( D = 0 \), the roots of this quadratic equation are real and equal. This means the quadratic has exactly one unique real solution.
(iii) Given quadratic equation is \( 3x^2 + 7x + 2 = 0 \).
Comparing with \( ax^2 + bx + c = 0 \), we have \( a = 3 \), \( b = 7 \), and \( c = 2 \).
The discriminant D is \( b^2 - 4ac \).
\( D = (7)^2 - 4 \times 3 \times 2 \)
\( D = 49 - 24 \)
\( D = 25 \)
Since \( D > 0 \), the roots of this quadratic equation are real, distinct, and rational. Rational roots can be expressed as a fraction of two integers.
(iv) Given quadratic equation is \( x^2 + px - q^2 = 0 \).
Comparing with \( ax^2 + bx + c = 0 \), we have \( a = 1 \), \( b = p \), and \( c = -q^2 \).
The discriminant D is \( b^2 - 4ac \).
\( D = (p)^2 - 4 \times 1 \times (-q^2) \)
\( D = p^2 + 4q^2 \)
For the roots to be real, the discriminant \( D \) must be greater than or equal to 0. Since \( p^2 \) is always greater than or equal to 0, and \( 4q^2 \) is also always greater than or equal to 0, their sum \( p^2 + 4q^2 \) will always be greater than or equal to 0.
Therefore, the roots are always real. They will be unequal (distinct) if either \( p \neq 0 \) or \( q \neq 0 \) (or both). The roots will be real and equal only if \( p = 0 \) and \( q = 0 \) at the same time, which makes D = 0.
In simple words: To find the nature of roots, we calculate the discriminant, D. If D is negative, roots are imaginary. If D is zero, roots are real and equal. If D is positive, roots are real and different.

🎯 Exam Tip: Remember the three conditions for the discriminant (D = \( b^2 - 4ac \)): D < 0 for imaginary roots, D = 0 for real and equal roots, and D > 0 for real and distinct roots. This is a fundamental concept in quadratic equations.

Question 2. If the equation \( (1 + m^2) x^2 + 2mcx + c^2 - a^2 = 0 \) has equal roots, show that \( c^2 = a^2 (1 + m^2) \).
Answer: Given quadratic equation is \( (1 + m^2) x^2 + 2mcx + c^2 - a^2 = 0 \).
We compare this with the general quadratic equation \( Ax^2 + Bx + C = 0 \).
Here, \( A = 1 + m^2 \), \( B = 2mc \), and \( C = c^2 - a^2 \).
For the equation to have equal roots, its discriminant must be zero.
So, \( B^2 - 4AC = 0 \)
\( \implies (2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0 \)
\( \implies 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \)
Now, divide the entire equation by 4:
\( \implies m^2c^2 - (c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \)
\( \implies m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0 \)
The \( m^2c^2 \) terms cancel out:
\( \implies -c^2 + a^2 + m^2a^2 = 0 \)
\( \implies a^2 + m^2a^2 = c^2 \)
\( \implies a^2(1 + m^2) = c^2 \)
Thus, it is shown that \( c^2 = a^2(1 + m^2) \). This relationship is important for conditions where a line is tangent to a circle.
In simple words: When a quadratic equation has equal roots, its special number called the discriminant must be zero. If you put the given values into the discriminant formula and simplify, you will get the answer they asked for.

🎯 Exam Tip: Ensure careful expansion of brackets and correct sign changes when simplifying algebraic expressions. Errors in these steps are common when dealing with discriminants.

Question 3. Find the value of m so that the roots of the equation \( (4 - m)x^2 + (2m + 4) x + (8m + 1) = 0 \) may be equal.
Answer: Given quadratic equation is \( (4 - m)x^2 + (2m + 4)x + (8m + 1) = 0 \).
We compare this with the general quadratic equation \( ax^2 + bx + c = 0 \).
Here, \( a = 4 - m \), \( b = 2m + 4 \), and \( c = 8m + 1 \).
For the roots to be equal, the discriminant must be zero.
So, \( b^2 - 4ac = 0 \)
\( \implies (2m + 4)^2 - 4(4 - m)(8m + 1) = 0 \)
First, expand the terms:
\( (2m + 4)^2 = 4m^2 + 16m + 16 \)
\( 4(4 - m)(8m + 1) = 4(32m + 4 - 8m^2 - m) = 4(31m + 4 - 8m^2) = 124m + 16 - 32m^2 \)
Now substitute these back into the discriminant equation:
\( \implies (4m^2 + 16m + 16) - (124m + 16 - 32m^2) = 0 \)
\( \implies 4m^2 + 16m + 16 - 124m - 16 + 32m^2 = 0 \)
Combine like terms:
\( \implies (4m^2 + 32m^2) + (16m - 124m) + (16 - 16) = 0 \)
\( \implies 36m^2 - 108m = 0 \)
Factor out \( 36m \):
\( \implies 36m(m - 3) = 0 \)
This gives two possible values for \( m \):
\( \implies 36m = 0 \) or \( m - 3 = 0 \)
\( \implies m = 0 \) or \( m = 3 \).
Both values of m lead to the quadratic equation having equal roots, which means the parabola touches the x-axis at exactly one point.
In simple words: For the roots of a quadratic equation to be equal, its discriminant (the \( b^2 - 4ac \) part) must be zero. We set this part of the given equation to zero, solve it, and find the possible values for 'm'.

🎯 Exam Tip: Always remember to fully expand and simplify all terms, especially when multiplying binomials, to avoid errors in solving for the variable 'm'.

Question 4. If the roots of \( ax^2 + x + b = 0 \) are real and unequal, show that the roots of \( \frac{x^2+1}{x}=4\sqrt{ab} \) are imaginary.
Answer: Given the first quadratic equation is \( ax^2 + x + b = 0 \).
For this equation, the roots are real and unequal. This means its discriminant must be greater than 0.
So, \( D_1 = 1^2 - 4(a)(b) > 0 \)
\( \implies 1 - 4ab > 0 \) ...(1)
Now, consider the second equation: \( \frac{x^2+1}{x}=4\sqrt{ab} \).
To convert this into a standard quadratic form, multiply both sides by \( x \):
\( \implies x^2 + 1 = 4\sqrt{ab} x \)
Rearrange the terms to get a standard quadratic equation:
\( \implies x^2 - 4\sqrt{ab} x + 1 = 0 \) ...(2)
Let's find the discriminant of this second quadratic equation. Comparing with \( Ax^2 + Bx + C = 0 \), we have \( A = 1 \), \( B = -4\sqrt{ab} \), and \( C = 1 \).
The discriminant \( D_2 \) is \( B^2 - 4AC \).
\( D_2 = (-4\sqrt{ab})^2 - 4 \times 1 \times 1 \)
\( D_2 = 16ab - 4 \)
\( D_2 = 4(4ab - 1) \)
From equation (1), we know that \( 1 - 4ab > 0 \).
This implies that \( 4ab - 1 < 0 \).
Since \( 4ab - 1 \) is negative, when we multiply it by 4, \( D_2 \) will also be negative.
So, \( D_2 = 4(4ab - 1) < 0 \).
Because the discriminant \( D_2 \) is less than 0, the roots of the second equation are imaginary. This shows how conditions on one equation can tell us about the nature of roots in another related equation.
In simple words: If the first equation has real and different roots, it means a certain part of its formula (the discriminant) is positive. Using this information, we rewrite the second equation into a standard form and calculate its discriminant. We then see that this second discriminant must be negative, which means its roots are imaginary.

🎯 Exam Tip: Pay close attention to inequalities and how they transform when terms are moved or multiplied by negative numbers. A small sign error can lead to a wrong conclusion about the nature of roots.

Question 5. Find a so that the sum of the roots of the equation \( ax^2 + 2x + 3a = 0 \) may be equal to their product.
Answer: Given quadratic equation is \( ax^2 + 2x + 3a = 0 \).
For a quadratic equation \( Ax^2 + Bx + C = 0 \):
Sum of roots \( = -\frac{B}{A} \)
Product of roots \( = \frac{C}{A} \)
In our given equation, \( A = a \), \( B = 2 \), and \( C = 3a \).
So, the sum of the roots \( = -\frac{2}{a} \).
And the product of the roots \( = \frac{3a}{a} = 3 \).
The problem states that the sum of the roots is equal to their product.
Therefore, \( -\frac{2}{a} = 3 \)
To solve for \( a \), multiply both sides by \( a \):
\( \implies -2 = 3a \)
Now, divide by 3:
\( \implies a = -\frac{2}{3} \).
This value of 'a' ensures that the relationship between the roots of the equation holds true.
In simple words: For a quadratic equation, we know how to find the sum and product of its roots using 'a', 'b', and 'c'. The question says these two values are equal. So we set the sum formula equal to the product formula and then solve to find 'a'.

🎯 Exam Tip: Always correctly identify the coefficients 'a', 'b', and 'c' from the given quadratic equation before applying the sum and product of roots formulas. Be careful with signs.

Question 6. If \( \alpha, \beta \) are the roots of the equation \( x^2 + x + 1 = 0 \), find the value of \( \alpha^3 + \beta^3 \).
Answer: Given quadratic equation is \( x^2 + x + 1 = 0 \).
If \( \alpha \) and \( \beta \) are the roots of \( Ax^2 + Bx + C = 0 \), then:
Sum of roots \( \alpha + \beta = -\frac{B}{A} \)
Product of roots \( \alpha\beta = \frac{C}{A} \)
For the given equation, \( A = 1 \), \( B = 1 \), \( C = 1 \).
So, \( \alpha + \beta = -\frac{1}{1} = -1 \).
And, \( \alpha\beta = \frac{1}{1} = 1 \).
We need to find the value of \( \alpha^3 + \beta^3 \). We use the algebraic identity:
\( \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \)
Now, substitute the values of \( \alpha + \beta \) and \( \alpha\beta \) into the identity:
\( \alpha^3 + \beta^3 = (-1)^3 - 3(1)(-1) \)
\( \implies \alpha^3 + \beta^3 = -1 - (-3) \)
\( \implies \alpha^3 + \beta^3 = -1 + 3 \)
\( \implies \alpha^3 + \beta^3 = 2 \).
This problem demonstrates how identities are used to simplify expressions involving roots without directly solving for the roots themselves.
In simple words: First, find the sum and product of the roots from the given equation. Then, use a special algebraic formula for \( \alpha^3 + \beta^3 \) that involves these sum and product values. Put the numbers in the formula to get the final answer.

🎯 Exam Tip: Memorize key algebraic identities like \( \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \) and \( \alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) \) as they are frequently used in problems involving roots of quadratic equations.

Question 7. If \( \alpha, \beta \) are the roots of the equation \( x^2 + px + q = 0 \), find the value of
(a) \( \alpha^3\beta + \alpha\beta^3 \)
(b) \( \alpha^4 + \alpha^2\beta^2 + \beta^4 \)
Answer: Given quadratic equation is \( x^2 + px + q = 0 \).
Since \( \alpha \) and \( \beta \) are its roots, we know:
Sum of roots \( \alpha + \beta = -p \)
Product of roots \( \alpha\beta = q \)
(a) We need to find the value of \( \alpha^3\beta + \alpha\beta^3 \).
Factor out \( \alpha\beta \) from the expression:
\( \alpha^3\beta + \alpha\beta^3 = \alpha\beta(\alpha^2 + \beta^2) \)
Now, use the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\( = \alpha\beta((\alpha + \beta)^2 - 2\alpha\beta) \)
Substitute the values of \( \alpha + \beta \) and \( \alpha\beta \):
\( = q((-p)^2 - 2q) \)
\( = q(p^2 - 2q) \).
(b) We need to find the value of \( \alpha^4 + \alpha^2\beta^2 + \beta^4 \).
This expression can be rewritten by completing the square for \( \alpha^4 + 2\alpha^2\beta^2 + \beta^4 \):
\( \alpha^4 + \alpha^2\beta^2 + \beta^4 = (\alpha^4 + 2\alpha^2\beta^2 + \beta^4) - \alpha^2\beta^2 \)
\( = (\alpha^2 + \beta^2)^2 - (\alpha\beta)^2 \)
Again, use the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\( = ((\alpha + \beta)^2 - 2\alpha\beta)^2 - (\alpha\beta)^2 \)
Substitute the values of \( \alpha + \beta \) and \( \alpha\beta \):
\( = ((-p)^2 - 2q)^2 - (q)^2 \)
\( = (p^2 - 2q)^2 - q^2 \)
Now, expand \( (p^2 - 2q)^2 \):
\( = (p^2)^2 - 2(p^2)(2q) + (2q)^2 - q^2 \)
\( = p^4 - 4p^2q + 4q^2 - q^2 \)
\( = p^4 - 4p^2q + 3q^2 \).
These calculations show how to find symmetric polynomial expressions of roots in terms of coefficients of the quadratic equation.
In simple words: For both parts, we use the sum and product of the roots (which are \( -p \) and \( q \)). We rewrite the given expressions using known algebraic formulas until they only contain sum and product terms. Then we just substitute \( -p \) and \( q \) into these simplified expressions.

🎯 Exam Tip: Factoring common terms and using algebraic identities for sums and differences of squares or cubes is crucial for simplifying expressions involving powers of roots.

Question 8. If the roots of the equation \( x^2 + px + 7 = 0 \) are denoted by \( \alpha \) and \( \beta \), and \( \alpha^2 + \beta^2 = 22 \), find p.
Answer: Given quadratic equation is \( x^2 + px + 7 = 0 \).
If \( \alpha \) and \( \beta \) are the roots, then:
Sum of roots \( \alpha + \beta = -\frac{p}{1} = -p \)
Product of roots \( \alpha\beta = \frac{7}{1} = 7 \)
We are also given that \( \alpha^2 + \beta^2 = 22 \).
We know the identity: \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \).
Substitute the given value and the sum/product values into this identity:
\( 22 = (-p)^2 - 2(7) \)
\( \implies 22 = p^2 - 14 \)
To solve for \( p^2 \), add 14 to both sides:
\( \implies 22 + 14 = p^2 \)
\( \implies 36 = p^2 \)
Now, take the square root of both sides to find \( p \):
\( \implies p = \pm\sqrt{36} \)
\( \implies p = \pm 6 \).
This shows how to find unknown coefficients when a relationship between the roots is provided.
In simple words: We first write down the sum and product of roots using 'p' from the equation. Then, we use the given information about \( \alpha^2 + \beta^2 \) along with an algebraic formula to set up an equation in 'p'. Solving this equation gives us the two possible values for 'p'.

🎯 Exam Tip: Remember that taking the square root can result in both positive and negative values. Always include both possibilities unless the context of the question restricts the variable to a specific sign.

Question 9. If \( \alpha, \beta \) are the roots of the equation \( 3x^2 - 6x + 4 = 0 \), find the value of \( \left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3 \alpha \beta \)
Answer: Given quadratic equation is \( 3x^2 - 6x + 4 = 0 \).
If \( \alpha \) and \( \beta \) are the roots, then:
Sum of roots \( \alpha + \beta = -\frac{-6}{3} = 2 \)
Product of roots \( \alpha\beta = \frac{4}{3} \)
We need to find the value of the expression \( \left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3 \alpha \beta \).
Let's simplify each part of the expression:
First term: \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} \).
Using \( \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \):
\( = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} \)
Substitute the values: \( = \frac{(2)^2 - 2(4/3)}{4/3} = \frac{4 - 8/3}{4/3} = \frac{12/3 - 8/3}{4/3} = \frac{4/3}{4/3} = 1 \).
Second term: \( 2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) = 2\left(\frac{\beta + \alpha}{\alpha\beta}\right) \)
Substitute the values: \( = 2\left(\frac{2}{4/3}\right) = 2\left(\frac{2 \times 3}{4}\right) = 2\left(\frac{6}{4}\right) = 2\left(\frac{3}{2}\right) = 3 \).
Third term: \( 3\alpha\beta \)
Substitute the value: \( = 3\left(\frac{4}{3}\right) = 4 \).
Now, add the simplified terms together:
The total value \( = 1 + 3 + 4 = 8 \).
This problem shows that even complex expressions involving roots can be simplified by expressing them in terms of the sum and product of roots.
In simple words: First, find the sum and product of the roots from the given equation. Then, break the complex expression into simpler parts. Rewrite each part using the sum and product of roots. Finally, put all the numbers together to get the answer.

🎯 Exam Tip: When evaluating complex expressions, simplify each part separately first. This reduces the chance of making calculation errors and makes the entire process clearer.

Question 10. If \( \alpha, \beta \) are the roots of \( ax^2 + bx + c = 0 \), find the values of
(i) \( \left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2 \)
(ii) \( \frac{\alpha^3}{\beta}+\frac{\beta^3}{\alpha} \)
Answer: Given quadratic equation is \( ax^2 + bx + c = 0 \).
If \( \alpha \) and \( \beta \) are the roots, then:
Sum of roots \( \alpha + \beta = -\frac{b}{a} \)
Product of roots \( \alpha\beta = \frac{c}{a} \)
(i) We need to find the value of \( \left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2 \).
First, combine the fractions inside the parenthesis:
\( \left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2 = \left(\frac{\alpha^2 - \beta^2}{\alpha\beta}\right)^2 \)
\( = \frac{(\alpha^2 - \beta^2)^2}{(\alpha\beta)^2} \)
Use the identity \( \alpha^2 - \beta^2 = (\alpha - \beta)(\alpha + \beta) \):
\( = \frac{((\alpha - \beta)(\alpha + \beta))^2}{(\alpha\beta)^2} = \frac{(\alpha - \beta)^2 (\alpha + \beta)^2}{(\alpha\beta)^2} \)
Use the identity \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \):
\( = \frac{((\alpha + \beta)^2 - 4\alpha\beta)(\alpha + \beta)^2}{(\alpha\beta)^2} \)
Now, substitute \( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \):
\( = \frac{\left(\left(-\frac{b}{a}\right)^2 - 4\frac{c}{a}\right)\left(-\frac{b}{a}\right)^2}{\left(\frac{c}{a}\right)^2} \)
\( = \frac{\left(\frac{b^2}{a^2} - \frac{4c}{a}\right)\frac{b^2}{a^2}}{\frac{c^2}{a^2}} \)
\( = \frac{\left(\frac{b^2 - 4ac}{a^2}\right)\frac{b^2}{a^2}}{\frac{c^2}{a^2}} \)
Multiply the fractions in the numerator and simplify by cancelling \( \frac{1}{a^2} \):
\( = \frac{\frac{b^2(b^2 - 4ac)}{a^4}}{\frac{c^2}{a^2}} \)
\( = \frac{b^2(b^2 - 4ac)}{a^4} \times \frac{a^2}{c^2} \)
\( = \frac{b^2(b^2 - 4ac)}{a^2c^2} \).
(ii) We need to find the value of \( \frac{\alpha^3}{\beta}+\frac{\beta^3}{\alpha} \).
Combine the fractions:
\( \frac{\alpha^3}{\beta}+\frac{\beta^3}{\alpha} = \frac{\alpha^4 + \beta^4}{\alpha\beta} \)
Now, we use the identity for \( \alpha^4 + \beta^4 \):
\( \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 \)
And \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \).
So, \( \alpha^4 + \beta^4 = ((\alpha + \beta)^2 - 2\alpha\beta)^2 - 2(\alpha\beta)^2 \)
Substitute this back into the expression:
\( = \frac{((\alpha + \beta)^2 - 2\alpha\beta)^2 - 2(\alpha\beta)^2}{\alpha\beta} \)
Now, substitute \( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \):
\( = \frac{\left(\left(-\frac{b}{a}\right)^2 - 2\frac{c}{a}\right)^2 - 2\left(\frac{c}{a}\right)^2}{\frac{c}{a}} \)
\( = \frac{\left(\frac{b^2}{a^2} - \frac{2c}{a}\right)^2 - \frac{2c^2}{a^2}}{\frac{c}{a}} \)
\( = \frac{\left(\frac{b^2 - 2ac}{a^2}\right)^2 - \frac{2c^2}{a^2}}{\frac{c}{a}} \)
\( = \frac{\frac{(b^2 - 2ac)^2}{a^4} - \frac{2c^2a^2}{a^4}}{\frac{c}{a}} \)
\( = \frac{(b^2 - 2ac)^2 - 2a^2c^2}{a^4} \times \frac{a}{c} \)
\( = \frac{(b^2 - 2ac)^2 - 2a^2c^2}{a^3c} \)
Expanding the numerator:
\( = \frac{b^4 - 4ab^2c + 4a^2c^2 - 2a^2c^2}{a^3c} \)
\( = \frac{b^4 - 4ab^2c + 2a^2c^2}{a^3c} \).
This demonstrates finding values of complex symmetric expressions of roots using only the coefficients of the quadratic equation. Mastering these algebraic manipulations is key to solving such problems.
In simple words: For both parts, we use the sum of roots (\( -b/a \)) and product of roots (\( c/a \)). We rewrite the expressions using common algebraic formulas, making sure they only contain the sum and product of roots. Then, we replace those sums and products with \( -b/a \) and \( c/a \) and simplify the fractions to get the final answer.

🎯 Exam Tip: Be meticulous with algebraic simplification, especially when dealing with squares of fractions and common denominators. A single error in a step can lead to a completely different final expression.

Question 11. If the sum of the roots of the equation \( x^2 - px + q = 0 \) be m times their difference, prove that \( p^2 (m^2 - 1) = 4m^2q \).
Answer: Given quadratic equation is \( x^2 - px + q = 0 \).
Let \( \alpha \) and \( \beta \) be the roots of this equation.
Sum of roots \( \alpha + \beta = -\frac{-p}{1} = p \)
Product of roots \( \alpha\beta = \frac{q}{1} = q \)
We are given that the sum of the roots is \( m \) times their difference:
\( \alpha + \beta = m|\alpha - \beta| \)
To eliminate the absolute value and simplify, square both sides of the equation:
\( \implies (\alpha + \beta)^2 = m^2(\alpha - \beta)^2 \)
We know the identity: \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \).
Substitute this identity into the equation:
\( \implies (\alpha + \beta)^2 = m^2((\alpha + \beta)^2 - 4\alpha\beta) \)
Now, substitute the values of \( \alpha + \beta \) and \( \alpha\beta \) (which are \( p \) and \( q \)):
\( \implies p^2 = m^2(p^2 - 4q) \)
Distribute \( m^2 \) on the right side:
\( \implies p^2 = m^2p^2 - 4m^2q \)
Rearrange the terms to get \( p^2 \) terms on one side:
\( \implies p^2 - m^2p^2 = -4m^2q \)
Factor out \( p^2 \) from the left side:
\( \implies p^2(1 - m^2) = -4m^2q \)
Multiply both sides by -1 to match the desired form:
\( \implies p^2(-(1 - m^2)) = -(-4m^2q) \)
\( \implies p^2(m^2 - 1) = 4m^2q \).
Thus, the proof is complete. This problem connects the roots' sum and difference to a constant multiplier, 'm'.
In simple words: First, write down the sum and product of the roots using 'p' and 'q'. Then, use the given condition that the sum is 'm' times the difference. Square both sides to remove the absolute value, and use a formula that links the difference of roots to the sum and product. Substitute the sum and product values into this equation and simplify to get the final proof.

🎯 Exam Tip: When dealing with differences of roots or conditions involving absolute values, squaring both sides of the equation is often a useful first step to eliminate the absolute value and simplify the expression.

Question 12. If one root of the equation \( x^2 + ax + 8 = 0 \) is 4 while the equation \( x^2 + ax + b = 0 \) has equal roots, find b.
Answer: We are given two quadratic equations and conditions on their roots.
First equation: \( x^2 + ax + 8 = 0 \).
We know that 4 is one of its roots. If a value is a root of an equation, substituting it into the equation must satisfy it.
So, substitute \( x = 4 \) into the first equation:
\( (4)^2 + a(4) + 8 = 0 \)
\( \implies 16 + 4a + 8 = 0 \)
\( \implies 4a + 24 = 0 \)
To find \( a \), subtract 24 from both sides and then divide by 4:
\( \implies 4a = -24 \)
\( \implies a = -6 \).
Now, consider the second equation: \( x^2 + ax + b = 0 \).
This equation has equal roots. For a quadratic equation to have equal roots, its discriminant must be zero.
The discriminant is \( B^2 - 4AC \). For this equation, \( A = 1 \), \( B = a \), \( C = b \).
So, \( a^2 - 4(1)(b) = 0 \)
\( \implies a^2 - 4b = 0 \)
We found the value of \( a \) to be -6 from the first part of the problem. Substitute this value of \( a \) into the discriminant equation:
\( \implies (-6)^2 - 4b = 0 \)
\( \implies 36 - 4b = 0 \)
To find \( b \), add \( 4b \) to both sides:
\( \implies 36 = 4b \)
Now, divide by 4:
\( \implies b = \frac{36}{4} \)
\( \implies b = 9 \).
This problem shows how conditions on roots can be used to determine unknown coefficients in quadratic equations.
In simple words: First, use the fact that \( x = 4 \) is a root of the first equation to find the value of 'a'. Then, use this 'a' in the second equation. Since the second equation has equal roots, its discriminant (the \( b^2 - 4ac \) part) must be zero. Set it to zero and solve for 'b'.

🎯 Exam Tip: Remember that if a value is a root of an equation, it must satisfy the equation when substituted. Also, apply the discriminant condition correctly for equal roots (D=0) and distinct roots (D>0).

Question 13. Find the value of a for which one root of the quadratic equation \( (a^2 - 5a + 3) x^2 + (3a - 1) x + 2 = 0 \) is twice as large as the other.
Answer: Given quadratic equation is \( (a^2 - 5a + 3) x^2 + (3a - 1) x + 2 = 0 \).
Let the roots of the equation be \( \alpha \) and \( 2\alpha \), since one root is twice as large as the other.
For a quadratic equation \( Ax^2 + Bx + C = 0 \):
Sum of roots \( = -\frac{B}{A} \)
Product of roots \( = \frac{C}{A} \)
In our given equation, \( A = a^2 - 5a + 3 \), \( B = 3a - 1 \), and \( C = 2 \).
Sum of roots: \( \alpha + 2\alpha = 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3} \) ...(1)
Product of roots: \( \alpha \times 2\alpha = 2\alpha^2 = \frac{2}{a^2 - 5a + 3} \) ...(2)
From equation (2), we can simplify by dividing by 2:
\( \alpha^2 = \frac{1}{a^2 - 5a + 3} \)
Now, square equation (1):
\( (3\alpha)^2 = \left(-\frac{3a - 1}{a^2 - 5a + 3}\right)^2 \)
\( \implies 9\alpha^2 = \frac{(3a - 1)^2}{(a^2 - 5a + 3)^2} \)
Substitute the expression for \( \alpha^2 \) from (2) into this squared equation:
\( \implies 9\left(\frac{1}{a^2 - 5a + 3}\right) = \frac{(3a - 1)^2}{(a^2 - 5a + 3)^2} \)
Assuming \( a^2 - 5a + 3 \neq 0 \), we can multiply both sides by \( (a^2 - 5a + 3)^2 \):
\( \implies 9(a^2 - 5a + 3) = (3a - 1)^2 \)
Expand both sides:
\( \implies 9a^2 - 45a + 27 = 9a^2 - 6a + 1 \)
Subtract \( 9a^2 \) from both sides:
\( \implies -45a + 27 = -6a + 1 \)
Add \( 45a \) to both sides:
\( \implies 27 = 39a + 1 \)
Subtract 1 from both sides:
\( \implies 26 = 39a \)
Divide by 39 to find \( a \):
\( \implies a = \frac{26}{39} \)
\( \implies a = \frac{2}{3} \).
This value of 'a' ensures the roots have the specified relationship. If \( a^2 - 5a + 3 = 0 \), the equation would not be quadratic, which is a condition to keep in mind.
In simple words: Let the roots be \( \alpha \) and \( 2\alpha \). Write down the sum and product of these roots using the formulas for quadratic equations. You will get two equations. Solve these two equations by replacing \( \alpha^2 \) from one into the other. Simplify the resulting equation to find the value of 'a'.

🎯 Exam Tip: When setting up equations based on relationships between roots, it is common to have two variables (the root and the coefficient). Solving this system of equations (often by substitution or elimination) is key to finding the unknown coefficient.

Question 14. If \( \alpha, \beta \) are the roots of the equation \( ax^2 - bx + b = 0 \), prove that \( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{b}{a}}=0 \).
Answer: Given quadratic equation is \( ax^2 - bx + b = 0 \).
If \( \alpha \) and \( \beta \) are the roots, then:
Sum of roots \( \alpha + \beta = -\frac{-b}{a} = \frac{b}{a} \)
Product of roots \( \alpha\beta = \frac{b}{a} \)
We need to prove that \( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{b}{a}}=0 \).
Let's consider the first two terms of the expression: \( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}} \).
To add these square roots, we can find a common denominator:
\( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}} = \frac{\sqrt{\alpha}}{\sqrt{\beta}} + \frac{\sqrt{\beta}}{\sqrt{\alpha}} \)
\( = \frac{\sqrt{\alpha}\sqrt{\alpha} + \sqrt{\beta}\sqrt{\beta}}{\sqrt{\alpha}\sqrt{\beta}} \)
\( = \frac{\alpha + \beta}{\sqrt{\alpha\beta}} \)
Now, substitute the values of \( \alpha + \beta \) and \( \alpha\beta \) from the given quadratic equation:
\( = \frac{\frac{b}{a}}{\sqrt{\frac{b}{a}}} \)
This expression simplifies to \( \sqrt{\frac{b}{a}} \). (Because \( \frac{X}{\sqrt{X}} = \sqrt{X} \), where \( X = \frac{b}{a} \)).
So, we have shown that \( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}} = \sqrt{\frac{b}{a}} \).
Now, substitute this back into the original expression we need to prove:
\( \sqrt{\frac{b}{a}} - \sqrt{\frac{b}{a}} = 0 \).
Thus, it is proven that \( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{b}{a}}=0 \). This identity cleverly links the coefficients to the square roots of the ratios of the roots.
In simple words: First, write down the sum and product of the roots using the given equation. Then, take the first two parts of the expression you need to prove. Simplify them by finding a common denominator for the square roots. You'll find that these two parts simplify to \( \sqrt{b/a} \). So, the whole expression becomes \( \sqrt{b/a} - \sqrt{b/a} \), which is 0.

🎯 Exam Tip: When proving identities involving roots, always start by expressing the sum and product of roots in terms of the coefficients. Algebraic manipulation of expressions to incorporate these sum/product terms is a key skill.

Question 15. If \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + x - 7 = 0 \), form the equation whose roots are \( \alpha^2 \) and \( \beta^2 \).
Answer: Given quadratic equation is \( x^2 + x - 7 = 0 \).
Let \( \alpha \) and \( \beta \) be its roots.
Sum of roots \( \alpha + \beta = -\frac{1}{1} = -1 \)
Product of roots \( \alpha\beta = \frac{-7}{1} = -7 \)
We need to form a new quadratic equation whose roots are \( \alpha^2 \) and \( \beta^2 \).
For a new quadratic equation \( X^2 - S'X + P' = 0 \), where \( S' \) is the sum of new roots and \( P' \) is the product of new roots.
New sum of roots \( S' = \alpha^2 + \beta^2 \).
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\( S' = (-1)^2 - 2(-7) \)
\( S' = 1 + 14 \)
\( S' = 15 \).
New product of roots \( P' = \alpha^2\beta^2 \).
\( P' = (\alpha\beta)^2 \)
\( P' = (-7)^2 \)
\( P' = 49 \).
Now, substitute \( S' \) and \( P' \) into the formula for the new quadratic equation:
\( x^2 - S'x + P' = 0 \)
\( x^2 - 15x + 49 = 0 \).
This is the required quadratic equation. Finding new equations based on transformations of roots is a common application of sum and product of roots.
In simple words: First, find the sum and product of roots from the given equation. Then, calculate the sum and product of the *new* roots (\( \alpha^2 \) and \( \beta^2 \)) using these initial values and algebraic formulas. Finally, use the new sum and product to write the new quadratic equation in the form \( x^2 - (\text{new sum})x + (\text{new product}) = 0 \).

🎯 Exam Tip: Clearly distinguish between the original sum/product of roots and the new sum/product of roots for the transformed equation. Organize your steps to avoid confusion, especially when roots are squares or other functions of the original roots.

Question 16. If \( \alpha \) and \( \beta \) are the roots of the equation \( 2x^2 + 3x + 2 = 0 \), find the equation whose roots are \( \alpha + 1 \) and \( \beta + 1 \).
Answer: Given quadratic equation is \( 2x^2 + 3x + 2 = 0 \).
Let \( \alpha \) and \( \beta \) be its roots.
Sum of roots \( \alpha + \beta = -\frac{3}{2} \)
Product of roots \( \alpha\beta = \frac{2}{2} = 1 \)
We need to form a new quadratic equation whose roots are \( \alpha + 1 \) and \( \beta + 1 \).
For a new quadratic equation \( X^2 - S'X + P' = 0 \):
New sum of roots \( S' = (\alpha + 1) + (\beta + 1) \)
\( S' = \alpha + \beta + 2 \)
Substitute the value of \( \alpha + \beta \):
\( S' = -\frac{3}{2} + 2 \)
\( S' = -\frac{3}{2} + \frac{4}{2} \)
\( S' = \frac{1}{2} \).
New product of roots \( P' = (\alpha + 1)(\beta + 1) \)
Expand the product:
\( P' = \alpha\beta + \alpha + \beta + 1 \)
Substitute the values of \( \alpha\beta \) and \( \alpha + \beta \):
\( P' = 1 + \left(-\frac{3}{2}\right) + 1 \)
\( P' = 2 - \frac{3}{2} \)
\( P' = \frac{4}{2} - \frac{3}{2} \)
\( P' = \frac{1}{2} \).
Now, substitute \( S' \) and \( P' \) into the formula for the new quadratic equation:
\( x^2 - S'x + P' = 0 \)
\( x^2 - \frac{1}{2}x + \frac{1}{2} = 0 \)
To get integer coefficients, multiply the entire equation by 2:
\( 2x^2 - x + 1 = 0 \).
This shows how to find a new quadratic equation when the roots are shifted by an additive constant.
In simple words: First, find the sum and product of the original roots. Then, for the new roots (\( \alpha+1 \) and \( \beta+1 \)), calculate their sum and product. You can do this by using the original sum and product values. Finally, use these new sum and product values to write down the new quadratic equation.

🎯 Exam Tip: Be careful when expanding products of transformed roots. Ensure you include all cross-terms (like \( \alpha \) and \( \beta \) in \( (\alpha+1)(\beta+1) \)) before substituting the sum and product of original roots.

Question 17. Find the equation whose roots are \( \frac{\alpha}{\beta} \) and \( \frac{\beta}{\alpha} \), where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + 2x + 3 = 0 \).
Answer: Given quadratic equation is \( x^2 + 2x + 3 = 0 \).
Let \( \alpha \) and \( \beta \) be its roots.
Sum of roots \( \alpha + \beta = -\frac{2}{1} = -2 \)
Product of roots \( \alpha\beta = \frac{3}{1} = 3 \)
We need to form a new quadratic equation whose roots are \( \frac{\alpha}{\beta} \) and \( \frac{\beta}{\alpha} \).
For a new quadratic equation \( X^2 - S'X + P' = 0 \):
New sum of roots \( S' = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \)
To combine these fractions, find a common denominator:
\( S' = \frac{\alpha^2 + \beta^2}{\alpha\beta} \)
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\( S' = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} \)
Substitute the values of \( \alpha + \beta \) and \( \alpha\beta \):
\( S' = \frac{(-2)^2 - 2(3)}{3} \)
\( S' = \frac{4 - 6}{3} \)
\( S' = \frac{-2}{3} \).
New product of roots \( P' = \left(\frac{\alpha}{\beta}\right) \times \left(\frac{\beta}{\alpha}\right) \)
\( P' = 1 \).
Now, substitute \( S' \) and \( P' \) into the formula for the new quadratic equation:
\( x^2 - S'x + P' = 0 \)
\( x^2 - \left(-\frac{2}{3}\right)x + 1 = 0 \)
\( x^2 + \frac{2}{3}x + 1 = 0 \)
To get integer coefficients, multiply the entire equation by 3:
\( 3x^2 + 2x + 3 = 0 \).
This illustrates how to derive new equations when roots are ratios of the original roots.
In simple words: First, find the sum and product of the original roots. Then, calculate the sum and product of the new roots (\( \alpha/\beta \) and \( \beta/\alpha \)). Use the original sum and product values in these calculations. Finally, use the new sum and product to form the new quadratic equation.

🎯 Exam Tip: The product of reciprocal roots \( (\alpha/\beta) \times (\beta/\alpha) \) will always be 1, which can be a quick check or simplification. Remember to handle fractions carefully when finding the sum of new roots.

Question 18. If \( \alpha \) and \( \beta \) are the roots of the equation \( 2x^2 - 3x + 1 = 0 \), form the equation whose roots are \( \frac{\alpha}{2\beta+3} \) and \( \frac{\beta}{2\alpha+3} \).
Answer: Given quadratic equation is \( 2x^2 - 3x + 1 = 0 \).
Let \( \alpha \) and \( \beta \) be its roots.
Sum of roots \( \alpha + \beta = -\frac{-3}{2} = \frac{3}{2} \)
Product of roots \( \alpha\beta = \frac{1}{2} \)
We need to form a new quadratic equation whose roots are \( r_1 = \frac{\alpha}{2\beta+3} \) and \( r_2 = \frac{\beta}{2\alpha+3} \).
For a new quadratic equation \( X^2 - S'X + P' = 0 \):
New sum of roots \( S' = r_1 + r_2 = \frac{\alpha}{2\beta+3} + \frac{\beta}{2\alpha+3} \)
To add these fractions, find a common denominator:
\( S' = \frac{\alpha(2\alpha+3) + \beta(2\beta+3)}{(2\beta+3)(2\alpha+3)} \)
Expand the numerator and denominator:
Numerator: \( 2\alpha^2 + 3\alpha + 2\beta^2 + 3\beta = 2(\alpha^2 + \beta^2) + 3(\alpha + \beta) \)
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
Numerator: \( 2((\alpha + \beta)^2 - 2\alpha\beta) + 3(\alpha + \beta) \)
Denominator: \( 4\alpha\beta + 6\alpha + 6\beta + 9 = 4\alpha\beta + 6(\alpha + \beta) + 9 \)
Now, substitute the values of \( \alpha + \beta = \frac{3}{2} \) and \( \alpha\beta = \frac{1}{2} \).
Calculate the Numerator (N):
\( N = 2\left(\left(\frac{3}{2}\right)^2 - 2\left(\frac{1}{2}\right)\right) + 3\left(\frac{3}{2}\right) \)
\( N = 2\left(\frac{9}{4} - 1\right) + \frac{9}{2} \)
\( N = 2\left(\frac{9}{4} - \frac{4}{4}\right) + \frac{9}{2} \)
\( N = 2\left(\frac{5}{4}\right) + \frac{9}{2} \)
\( N = \frac{5}{2} + \frac{9}{2} = \frac{14}{2} = 7 \).
Calculate the Denominator (D):
\( D = 4\left(\frac{1}{2}\right) + 6\left(\frac{3}{2}\right) + 9 \)
\( D = 2 + 9 + 9 = 20 \).
So, the new sum \( S' = \frac{N}{D} = \frac{7}{20} \).
New product of roots \( P' = r_1 \times r_2 = \left(\frac{\alpha}{2\beta+3}\right) \times \left(\frac{\beta}{2\alpha+3}\right) \)
\( P' = \frac{\alpha\beta}{(2\beta+3)(2\alpha+3)} \)
We already calculated the denominator for \( S' \), which is 20.
\( P' = \frac{\alpha\beta}{4\alpha\beta + 6(\alpha + \beta) + 9} \)
Substitute the values:
\( P' = \frac{1/2}{20} \)
\( P' = \frac{1}{2 \times 20} = \frac{1}{40} \).
Now, substitute \( S' \) and \( P' \) into the formula for the new quadratic equation:
\( x^2 - S'x + P' = 0 \)
\( x^2 - \frac{7}{20}x + \frac{1}{40} = 0 \)
To get integer coefficients, multiply the entire equation by 40:
\( 40x^2 - 14x + 1 = 0 \).
This problem involves more complex root transformations, requiring careful algebraic expansion and substitution. It tests your ability to handle multiple terms in the numerator and denominator.
In simple words: First, find the sum and product of the original roots. Then, for the new, more complex roots, calculate their sum and product. You'll need to expand and simplify carefully, using the original sum and product values in the calculations. Finally, use these new sum and product values to write the new quadratic equation.

🎯 Exam Tip: For complex fractions or expressions, calculate the numerator and denominator separately using the sum and product of original roots. This modular approach helps manage complexity and reduces errors.

Question 20. Given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 = x + 7 \). (i) Prove that (a) \( \frac { 1 }{ \alpha } = \frac{\alpha-1}{7} \) and (b) \( \alpha^3 = 8\alpha + 7 \). (ii) Find the numerical value of \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \).
Answer:
(i) (a) To prove \( \frac { 1 }{ \alpha } = \frac{\alpha-1}{7} \):
The given quadratic equation is \( x^2 = x + 7 \). This can be rewritten as \( x^2 - x - 7 = 0 \).
Since \( \alpha \) is a root of this equation, it must satisfy the equation:
\( \alpha^2 = \alpha + 7 \)
Now, we can rearrange this equation:
\( \alpha^2 - \alpha = 7 \)
Next, factor out \( \alpha \) from the left side:
\( \alpha(\alpha - 1) = 7 \)
Finally, divide both sides by \( 7\alpha \) to get the desired form:
\( \frac{ \alpha(\alpha - 1) }{ 7\alpha } = \frac{ 7 }{ 7\alpha } \)
\( \implies \frac{ \alpha - 1 }{ 7 } = \frac{ 1 }{ \alpha } \)
So, \( \frac { 1 }{ \alpha } = \frac{\alpha-1}{7} \). This proves the first part.

(b) To prove \( \alpha^3 = 8\alpha + 7 \):
We start again from the relation \( \alpha^2 = \alpha + 7 \).
To get \( \alpha^3 \), we multiply the entire equation by \( \alpha \):
\( \alpha \cdot (\alpha^2) = \alpha \cdot (\alpha + 7) \)
\( \alpha^3 = \alpha^2 + 7\alpha \)
Now, we substitute \( \alpha^2 = \alpha + 7 \) back into this equation:
\( \alpha^3 = (\alpha + 7) + 7\alpha \)
Combine the terms involving \( \alpha \):
\( \alpha^3 = 8\alpha + 7 \). This proves the second part.

(ii) To find the numerical value of \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \):
From the given equation \( x^2 - x - 7 = 0 \), we can find the sum and product of the roots.
Sum of roots: \( \alpha + \beta = -(\text{coefficient of } x) / (\text{coefficient of } x^2) = -(-1)/1 = 1 \).
Product of roots: \( \alpha\beta = (\text{constant term}) / (\text{coefficient of } x^2) = -7/1 = -7 \).
Now, let's simplify the expression \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \):
First, find a common denominator:
\( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha \cdot \alpha}{\beta \cdot \alpha}+\frac{\beta \cdot \beta}{\alpha \cdot \beta} = \frac{\alpha^2+\beta^2}{\alpha\beta} \)
We know that \( \alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \).
So, substitute this into the expression:
\( \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} \)
Now, substitute the values of \( (\alpha+\beta) \) and \( \alpha\beta \):
\( \frac{(1)^2 - 2(-7)}{-7} \)
\( = \frac{1 + 14}{-7} \)
\( = \frac{15}{-7} = -\frac{15}{7} \).
In simple words: For part (i), we use the fact that if a value is a root, it fits into the equation. We then rearrange the equation to show what was asked. For part (ii), we use simple rules to find the sum and product of the roots from the given equation. Then, we rewrite the complex fraction using these sum and product values to find the final number.

🎯 Exam Tip: Always remember that if 'r' is a root of an equation, substituting 'r' for 'x' makes the equation true. This property is key for proving relationships between roots and coefficients.

Question 21. Given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - x + 7 = 0 \), find (i) the numerical value of \( \frac{\alpha}{\beta+3}+\frac{\beta}{\alpha+3} \) ; (ii) an equation whose roots are \( \frac { \alpha }{ \beta+3 } \) and \( \frac { \beta }{ \alpha +3 } \).
Answer:
First, from the given equation \( x^2 - x + 7 = 0 \):
Sum of roots: \( \alpha + \beta = -(-1)/1 = 1 \).
Product of roots: \( \alpha\beta = 7/1 = 7 \).

(i) To find the numerical value of \( \frac{\alpha}{\beta+3}+\frac{\beta}{\alpha+3} \):
Let \( S' = \frac{\alpha}{\beta+3}+\frac{\beta}{\alpha+3} \).
Combine the fractions by finding a common denominator:
\( S' = \frac{\alpha(\alpha+3)+\beta(\beta+3)}{(\beta+3)(\alpha+3)} \)
Expand the numerator:
\( S' = \frac{\alpha^2+3\alpha+\beta^2+3\beta}{(\beta+3)(\alpha+3)} = \frac{(\alpha^2+\beta^2)+3(\alpha+\beta)}{(\beta+3)(\alpha+3)} \)
Expand the denominator:
\( (\beta+3)(\alpha+3) = \alpha\beta+3\alpha+3\beta+9 = \alpha\beta+3(\alpha+\beta)+9 \)
We know \( \alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \).
Substitute the values of \( \alpha+\beta \) and \( \alpha\beta \):
Numerator: \( (\alpha+\beta)^2 - 2\alpha\beta + 3(\alpha+\beta) = (1)^2 - 2(7) + 3(1) = 1 - 14 + 3 = -10 \).
Denominator: \( \alpha\beta + 3(\alpha+\beta) + 9 = 7 + 3(1) + 9 = 7 + 3 + 9 = 19 \).
So, the numerical value of \( S' = \frac{-10}{19} \).

(ii) To find an equation whose roots are \( \frac { \alpha }{ \beta+3 } \) and \( \frac { \beta }{ \alpha +3 } \):
We already found the sum of these roots, \( S' = -\frac{10}{19} \).
Now, let's find the product of these new roots, \( P' \):
\( P' = \left(\frac{\alpha}{\beta+3}\right)\left(\frac{\beta}{\alpha+3}\right) = \frac{\alpha\beta}{(\beta+3)(\alpha+3)} \)
The numerator is \( \alpha\beta = 7 \).
The denominator is the same as calculated for \( S' \), which is 19.
So, \( P' = \frac{7}{19} \).
The required quadratic equation is \( x^2 - S'x + P' = 0 \).
\( x^2 - \left(-\frac{10}{19}\right)x + \frac{7}{19} = 0 \)
\( x^2 + \frac{10}{19}x + \frac{7}{19} = 0 \)
To remove the fractions, multiply the entire equation by 19:
\( 19x^2 + 10x + 7 = 0 \).
In simple words: First, we find the sum and product of the original roots from the given equation. Then, for part (i), we add the two new roots together, using a common bottom number, and replace the sum and product of original roots to get a single number. For part (ii), we use this sum and also find the product of the new roots. Finally, we put these into the standard form of a quadratic equation.

🎯 Exam Tip: When forming a new quadratic equation from given roots, always express the sum and product of the new roots in terms of the sum and product of the original roots. This avoids needing to solve for individual roots which can be complex.

Question 22. Given that \( \alpha \) and \( \beta \) are the roots of the equation \( 2x^2 - 3x + 4 = 0 \), find an equation whose roots are \( \alpha + \frac { 1 }{ \alpha } \) and \( \beta + \frac { 1 }{ \beta } \).
Answer:
From the given equation \( 2x^2 - 3x + 4 = 0 \), first divide by 2 to get it in the standard form \( x^2 - \frac{3}{2}x + 2 = 0 \).
Sum of original roots: \( \alpha + \beta = -(-\frac{3}{2})/1 = \frac{3}{2} \).
Product of original roots: \( \alpha\beta = 2/1 = 2 \).

Let the new roots be \( r_1 = \alpha + \frac{1}{\alpha} \) and \( r_2 = \beta + \frac{1}{\beta} \).
Sum of new roots (\( S' \)):
\( S' = r_1 + r_2 = \left(\alpha + \frac{1}{\alpha}\right) + \left(\beta + \frac{1}{\beta}\right) \)
Group the terms:
\( S' = (\alpha + \beta) + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) \)
Combine the fractions in the second part:
\( S' = (\alpha + \beta) + \left(\frac{\beta + \alpha}{\alpha\beta}\right) \)
Substitute the values of \( (\alpha+\beta) \) and \( \alpha\beta \):
\( S' = \frac{3}{2} + \frac{\frac{3}{2}}{2} \)
\( S' = \frac{3}{2} + \frac{3}{4} \)
Find a common denominator:
\( S' = \frac{6}{4} + \frac{3}{4} = \frac{9}{4} \).

Product of new roots (\( P' \)):
\( P' = r_1 \cdot r_2 = \left(\alpha + \frac{1}{\alpha}\right)\left(\beta + \frac{1}{\beta}\right) \)
Expand the product:
\( P' = \alpha\beta + \alpha\left(\frac{1}{\beta}\right) + \left(\frac{1}{\alpha}\right)\beta + \left(\frac{1}{\alpha}\right)\left(\frac{1}{\beta}\right) \)
\( P' = \alpha\beta + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + \frac{1}{\alpha\beta} \)
To simplify \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \), we use \( \frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} \).
Substitute the values into this part:
\( \frac{(\frac{3}{2})^2 - 2(2)}{2} = \frac{\frac{9}{4} - 4}{2} = \frac{\frac{9-16}{4}}{2} = \frac{-\frac{7}{4}}{2} = -\frac{7}{8} \).
Now substitute all parts back into the expression for \( P' \):
\( P' = 2 + \left(-\frac{7}{8}\right) + \frac{1}{2} \)
Find a common denominator of 8:
\( P' = \frac{16}{8} - \frac{7}{8} + \frac{4}{8} = \frac{16 - 7 + 4}{8} = \frac{13}{8} \).

The required quadratic equation is \( x^2 - S'x + P' = 0 \).
\( x^2 - \frac{9}{4}x + \frac{13}{8} = 0 \)
To eliminate fractions, multiply the entire equation by 8:
\( 8x^2 - 18x + 13 = 0 \).
In simple words: First, we find the sum and product of the given roots. Then, we find the sum of the new roots by adding them and combining terms using the original sum and product. Next, we find the product of the new roots by multiplying them out and simplifying, again using the original sum and product. Finally, we use these new sum and product values to write down the new quadratic equation.

🎯 Exam Tip: When new roots are in a complex form, remember to expand them fully and rearrange terms so that you can substitute \( (\alpha+\beta) \) and \( \alpha\beta \) which are usually easier to find.

Question 23. The roots of the quadratic equation \( x^2 + px + 8 = 0 \) are \( \alpha \) and \( \beta \). Obtain the values of p, if
(i) \( \alpha = \beta^2 \)
(ii) \( \alpha - \beta = 2 \)
Answer:
From the given equation \( x^2 + px + 8 = 0 \):
Sum of roots: \( \alpha + \beta = -p \).
Product of roots: \( \alpha\beta = 8 \).

(i) If \( \alpha = \beta^2 \):
Substitute \( \alpha = \beta^2 \) into the product of roots equation:
\( (\beta^2)\beta = 8 \)
\( \beta^3 = 8 \)
\( \implies \beta = 2 \)
Now, find \( \alpha \) using \( \alpha = \beta^2 \):
\( \alpha = (2)^2 = 4 \)
Substitute \( \alpha = 4 \) and \( \beta = 2 \) into the sum of roots equation:
\( \alpha + \beta = -p \)
\( 4 + 2 = -p \)
\( 6 = -p \)
\( \implies p = -6 \).

(ii) If \( \alpha - \beta = 2 \):
We know the identity \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \). This formula helps connect the difference of roots to their sum and product.
Substitute the given values into this identity:
\( (2)^2 = (-p)^2 - 4(8) \)
\( 4 = p^2 - 32 \)
Add 32 to both sides:
\( p^2 = 36 \)
Take the square root of both sides:
\( \implies p = \pm 6 \).
In simple words: For both parts, we first write down how the sum and product of roots relate to 'p' from the given equation. For part (i), we use the condition that one root is the square of the other, which helps us find the exact values of the roots first, and then 'p'. For part (ii), we use a special formula that links the difference of roots to their sum and product, which lets us find the possible values for 'p' directly.

🎯 Exam Tip: Remember the fundamental relationships between roots and coefficients: for \( ax^2+bx+c=0 \), sum of roots is \( -b/a \) and product is \( c/a \). Also, the identity \( (\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta \) is very useful for problems involving the difference of roots.

Question 24. If the roots of \( x^2 - bx + c = 0 \) be two consecutive integers, then find the value of \( b^2 - 4c \).
Answer:
Let the quadratic equation be \( x^2 - bx + c = 0 \).
Let the two consecutive integer roots be \( \alpha \) and \( \alpha + 1 \). Consecutive integers always differ by 1.
From the relationships between roots and coefficients:
Sum of roots: \( \alpha + (\alpha + 1) = -(-b)/1 \)
\( 2\alpha + 1 = b \) ---(Equation 1)
Product of roots: \( \alpha(\alpha + 1) = c/1 \)
\( \alpha^2 + \alpha = c \) ---(Equation 2)
We need to find the value of \( b^2 - 4c \). This expression is actually the discriminant (D) of the quadratic equation \( x^2 - bx + c = 0 \).
Substitute the expressions for \( b \) and \( c \) from Equation 1 and Equation 2 into \( b^2 - 4c \):
\( b^2 - 4c = (2\alpha + 1)^2 - 4(\alpha^2 + \alpha) \)
Expand the term \( (2\alpha + 1)^2 \):
\( b^2 - 4c = (4\alpha^2 + 4\alpha + 1) - (4\alpha^2 + 4\alpha) \)
Now, simplify by canceling out common terms:
\( b^2 - 4c = 4\alpha^2 + 4\alpha + 1 - 4\alpha^2 - 4\alpha \)
\( b^2 - 4c = 1 \).
In simple words: We let the roots be any two numbers that are one after the other. Then, we use how the sum and product of roots are connected to the 'b' and 'c' in the equation. When we put these connections into the expression \( b^2 - 4c \), all the tricky parts cancel out, and we are left with a simple number, which is 1.

🎯 Exam Tip: For any quadratic equation \( ax^2+bx+c=0 \), the discriminant \( b^2-4ac \) tells you about the nature of the roots. If the roots are consecutive integers, the discriminant will always be 1.

Question 25. The roots of the equation \( px^2 - 2(p + 2)x + 3p = 0 \) are \( \alpha \) and \( \beta \). If \( \alpha - \beta = 2 \), calculate the value of \( \alpha, \beta \) and \( p \).
Answer:
Given the quadratic equation \( px^2 - 2(p + 2)x + 3p = 0 \).
From the relationships between roots and coefficients:
Sum of roots: \( \alpha + \beta = - \frac{-2(p+2)}{p} = \frac{2(p+2)}{p} \) ---(Equation A)
Product of roots: \( \alpha\beta = \frac{3p}{p} = 3 \) ---(Equation B)
We are also given that the difference of the roots is:
\( \alpha - \beta = 2 \) ---(Equation C)

We use the identity that links the sum, product, and difference of roots: \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \).
Substitute the values from Equations A, B, and C into this identity:
\( (2)^2 = \left(\frac{2(p+2)}{p}\right)^2 - 4(3) \)
\( 4 = \frac{4(p+2)^2}{p^2} - 12 \)
Add 12 to both sides of the equation:
\( 16 = \frac{4(p+2)^2}{p^2} \)
Divide both sides by 4:
\( 4 = \frac{(p+2)^2}{p^2} \)
Take the square root of both sides. Remember to consider both positive and negative roots:
\( \pm \sqrt{4} = \sqrt{\frac{(p+2)^2}{p^2}} \)
\( \pm 2 = \frac{p+2}{p} \)
We now have two possible cases for \( p \):

**Case 1: \( 2 = \frac{p+2}{p} \)**
Multiply both sides by \( p \):
\( 2p = p + 2 \)
Subtract \( p \) from both sides:
\( p = 2 \)
Now that we have \( p = 2 \), we can find \( \alpha \) and \( \beta \).
Substitute \( p = 2 \) into Equation A:
\( \alpha + \beta = \frac{2(2+2)}{2} = \frac{2(4)}{2} = 4 \) ---(Equation D)
We also have \( \alpha - \beta = 2 \) ---(Equation E, same as C)
Add Equation D and Equation E:
\( (\alpha + \beta) + (\alpha - \beta) = 4 + 2 \)
\( 2\alpha = 6 \)
\( \implies \alpha = 3 \)
Substitute \( \alpha = 3 \) into Equation D:
\( 3 + \beta = 4 \)
\( \implies \beta = 1 \)
Check with Product of roots (Equation B): \( \alpha\beta = 3 \times 1 = 3 \). This is consistent.
So, for this case, \( p = 2, \alpha = 3, \beta = 1 \).

**Case 2: \( -2 = \frac{p+2}{p} \)**
Multiply both sides by \( p \):
\( -2p = p + 2 \)
Add \( 2p \) to both sides:
\( 0 = 3p + 2 \)
\( -2 = 3p \)
\( \implies p = -\frac{2}{3} \)
Now that we have \( p = -\frac{2}{3} \), we can find \( \alpha \) and \( \beta \).
Substitute \( p = -\frac{2}{3} \) into Equation A:
\( \alpha + \beta = \frac{2(-\frac{2}{3}+2)}{-\frac{2}{3}} = \frac{2(\frac{-2+6}{3})}{-\frac{2}{3}} = \frac{2(\frac{4}{3})}{-\frac{2}{3}} = \frac{\frac{8}{3}}{-\frac{2}{3}} = -4 \) ---(Equation F)
We also have \( \alpha - \beta = 2 \) ---(Equation G, same as C)
Add Equation F and Equation G:
\( (\alpha + \beta) + (\alpha - \beta) = -4 + 2 \)
\( 2\alpha = -2 \)
\( \implies \alpha = -1 \)
Substitute \( \alpha = -1 \) into Equation F:
\( -1 + \beta = -4 \)
\( \implies \beta = -3 \)
Check with Product of roots (Equation B): \( \alpha\beta = (-1) \times (-3) = 3 \). This is consistent.
So, for this case, \( p = -\frac{2}{3}, \alpha = -1, \beta = -3 \).
In simple words: We use the basic formulas that link the sum and product of roots to the 'p' value in the equation. We are also given how the roots differ. We use a special rule that connects the sum, product, and difference of the roots to set up an equation that helps us find 'p'. Since it involves squaring, we get two possible values for 'p'. For each 'p' value, we then find the specific values for \( \alpha \) and \( \beta \).

🎯 Exam Tip: When solving for parameters and roots, always check both positive and negative square roots to ensure all possible solutions are found. Consistency checks (like using the product of roots at the end) are also crucial for verifying your results.

Question 26. The roots of the equation \( ax^2 + bx + c = 0 \) are \( \alpha \) and \( \beta \). Form the quadratic equation whose roots are \( \alpha + \frac { 1 }{ \beta } \) and \( \beta + \frac { 1 }{ \alpha } \).
Answer:
From the given equation \( ax^2 + bx + c = 0 \):
Sum of original roots: \( \alpha + \beta = -\frac{b}{a} \).
Product of original roots: \( \alpha\beta = \frac{c}{a} \).

Let the new roots be \( r_1 = \alpha + \frac{1}{\beta} \) and \( r_2 = \beta + \frac{1}{\alpha} \).

**Step 1: Find the sum of the new roots (\( S' \)).**
\( S' = r_1 + r_2 = \left(\alpha + \frac{1}{\beta}\right) + \left(\beta + \frac{1}{\alpha}\right) \)
Group the terms together:
\( S' = (\alpha + \beta) + \left(\frac{1}{\beta} + \frac{1}{\alpha}\right) \)
To combine the fractions \( \frac{1}{\beta} + \frac{1}{\alpha} \), find a common denominator:
\( S' = (\alpha + \beta) + \left(\frac{\alpha + \beta}{\alpha\beta}\right) \)
Factor out \( (\alpha + \beta) \):
\( S' = (\alpha + \beta)\left(1 + \frac{1}{\alpha\beta}\right) \)
Now, substitute the values of \( (\alpha + \beta) \) and \( \alpha\beta \):
\( S' = \left(-\frac{b}{a}\right)\left(1 + \frac{1}{\frac{c}{a}}\right) \)
\( S' = \left(-\frac{b}{a}\right)\left(1 + \frac{a}{c}\right) \)
Combine the terms inside the second bracket:
\( S' = \left(-\frac{b}{a}\right)\left(\frac{c+a}{c}\right) \)
\( S' = -\frac{b(c+a)}{ac} \).

**Step 2: Find the product of the new roots (\( P' \)).**
\( P' = r_1 \cdot r_2 = \left(\alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right) \)
Expand this product:
\( P' = \alpha\beta + \alpha\left(\frac{1}{\alpha}\right) + \left(\frac{1}{\beta}\right)\beta + \left(\frac{1}{\beta}\right)\left(\frac{1}{\alpha}\right) \)
\( P' = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} \)
\( P' = \alpha\beta + 2 + \frac{1}{\alpha\beta} \)
Substitute the values of \( \alpha\beta \):
\( P' = \frac{c}{a} + 2 + \frac{1}{\frac{c}{a}} \)
\( P' = \frac{c}{a} + 2 + \frac{a}{c} \)
To combine these terms, find a common denominator, which is \( ac \):
\( P' = \frac{c^2}{ac} + \frac{2ac}{ac} + \frac{a^2}{ac} \)
\( P' = \frac{c^2 + 2ac + a^2}{ac} \)
Recognize the numerator as a perfect square: \( (a+c)^2 \).
\( P' = \frac{(a+c)^2}{ac} \).

**Step 3: Form the quadratic equation.**
The required quadratic equation is \( x^2 - S'x + P' = 0 \).
Substitute the expressions for \( S' \) and \( P' \):
\( x^2 - \left(-\frac{b(c+a)}{ac}\right)x + \frac{(a+c)^2}{ac} = 0 \)
\( x^2 + \frac{b(c+a)}{ac}x + \frac{(a+c)^2}{ac} = 0 \)
To clear the denominators, multiply the entire equation by \( ac \):
\( acx^2 + b(c+a)x + (a+c)^2 = 0 \).
In simple words: First, we find the sum and product of the given roots using the 'a', 'b', and 'c' from the original equation. Then, we find the sum of the new roots by adding them and simplifying the expression so that it only contains the original sum and product of roots. We do the same for the product of the new roots. Finally, we use these new sum and product expressions to write the new quadratic equation in its standard form.

🎯 Exam Tip: When dealing with general coefficients like 'a', 'b', 'c', be careful with algebraic manipulation and common denominators. Always simplify expressions for the new sum and product as much as possible before forming the final equation.

Question 27. Two candidates attempt to solve a quadratic equation of the form \( x^2 + px + q = 0 \). One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2 and – 9. Find the correct roots and the equation.
Answer:
The standard form of the quadratic equation is \( x^2 + px + q = 0 \).
For this equation:
Sum of roots = \( -p \)
Product of roots = \( q \)

**From the first candidate's information:**
This candidate used a wrong value of \( p \) but a correct value of \( q \).
The roots found were 2 and 6.
The product of these roots is \( 2 \times 6 = 12 \).
Since \( q \) was correct for this candidate, the correct value of \( q = 12 \).

**From the second candidate's information:**
This candidate used a wrong value of \( q \) but a correct value of \( p \).
The roots found were 2 and -9.
The sum of these roots is \( 2 + (-9) = -7 \).
Since \( p \) was correct for this candidate, the correct value of \( -p = -7 \), which means \( p = 7 \).

**Now we have the correct values for \( p \) and \( q \):**
Correct \( p = 7 \).
Correct \( q = 12 \).

**Form the correct quadratic equation:**
Using the correct \( p \) and \( q \), the equation is \( x^2 + 7x + 12 = 0 \).

**Find the correct roots of this equation:**
We can factor the quadratic equation \( x^2 + 7x + 12 = 0 \). We need two numbers that multiply to 12 and add up to 7. These numbers are 3 and 4.
So, \( (x + 3)(x + 4) = 0 \).
This gives us the roots \( x = -3 \) or \( x = -4 \).

Therefore, the correct quadratic equation is \( x^2 + 7x + 12 = 0 \), and its correct roots are -3 and -4.
In simple words: The first person got the "q" part of the equation right, so we find "q" from their roots. The second person got the "p" part right, so we find "p" from their roots. Once we have the correct "p" and "q", we can write down the correct equation. Then, we solve this correct equation to find the true roots.

🎯 Exam Tip: Remember that for an equation \( x^2+px+q=0 \), the coefficient 'p' is related to the sum of roots (sum = -p), and the constant 'q' is related to the product of roots (product = q). This distinction is key for solving such problems.

Question 28. Given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 = 7x + 4 \). (i) Show that (a) \( \alpha^3 = 53\alpha + 28 \) and (b) Find the numerical value of \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \).
Answer:
The given quadratic equation is \( x^2 = 7x + 4 \). This can be rewritten as \( x^2 - 7x - 4 = 0 \).
From this equation:
Sum of roots: \( \alpha + \beta = -(-7)/1 = 7 \).
Product of roots: \( \alpha\beta = -4/1 = -4 \).

(i) (a) To show \( \alpha^3 = 53\alpha + 28 \):
Since \( \alpha \) is a root of the equation, it satisfies \( \alpha^2 = 7\alpha + 4 \).
To find \( \alpha^3 \), multiply both sides of this equation by \( \alpha \):
\( \alpha \cdot (\alpha^2) = \alpha \cdot (7\alpha + 4) \)
\( \alpha^3 = 7\alpha^2 + 4\alpha \)
Now, substitute \( \alpha^2 = 7\alpha + 4 \) back into the equation for \( \alpha^3 \):
\( \alpha^3 = 7(7\alpha + 4) + 4\alpha \)
Distribute the 7:
\( \alpha^3 = 49\alpha + 28 + 4\alpha \)
Combine the terms with \( \alpha \):
\( \alpha^3 = 53\alpha + 28 \). This proves the statement.

(b) To find the numerical value of \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \):
First, find a common denominator for the fractions:
\( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} \)
We know that \( \alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \).
Substitute this identity into the expression:
\( \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} \)
Now, substitute the values of \( (\alpha+\beta) = 7 \) and \( \alpha\beta = -4 \):
\( \frac{(7)^2 - 2(-4)}{-4} \)
\( = \frac{49 + 8}{-4} \)
\( = \frac{57}{-4} = -\frac{57}{4} \).
In simple words: For part (a), we use the fact that \( \alpha \) is a root, so it fits the given equation. We then multiply by \( \alpha \) to get \( \alpha^3 \) and replace \( \alpha^2 \) again to show the required result. For part (b), we use the sum and product of the roots from the equation. Then we combine the fractions and use a known formula to simplify, putting in the numbers for the sum and product to get the final answer.

🎯 Exam Tip: To prove relationships involving powers of roots, repeatedly substitute the basic quadratic equation to reduce higher powers of \( \alpha \) or \( \beta \) into linear terms. This method is often simpler than finding the actual roots first.

Question 29. The ratio of the roots of the equation \( x^2 + ax + x + 2 = 0 \) is 2. Find the values of the parameter a.
Answer:
First, rewrite the given quadratic equation \( x^2 + ax + x + 2 = 0 \) in the standard form \( Ax^2 + Bx + C = 0 \):
\( x^2 + (a+1)x + 2 = 0 \)
Here, \( A=1, B=(a+1), C=2 \).
Let the roots of this equation be \( \alpha \) and \( \beta \).
We are given that the ratio of the roots is 2. So, we can write \( \alpha = 2\beta \).

From the relationships between roots and coefficients:
1. Sum of roots: \( \alpha + \beta = -B/A \)
\( \alpha + \beta = -(a+1) \)
Substitute \( \alpha = 2\beta \):
\( 2\beta + \beta = -(a+1) \)
\( 3\beta = -(a+1) \) ---(Equation 1)

2. Product of roots: \( \alpha\beta = C/A \)
\( \alpha\beta = 2 \)
Substitute \( \alpha = 2\beta \):
\( (2\beta)\beta = 2 \)
\( 2\beta^2 = 2 \)
Divide by 2:
\( \beta^2 = 1 \)
Taking the square root of both sides gives two possible values for \( \beta \):
\( \beta = \pm 1 \).

Now, we use Equation 1 for each possible value of \( \beta \):

**Case 1: If \( \beta = 1 \).**
Substitute \( \beta = 1 \) into Equation 1:
\( 3(1) = -(a+1) \)
\( 3 = -a - 1 \)
Add \( a \) and subtract 3 from both sides:
\( a = -1 - 3 \)
\( a = -4 \).
If \( \beta = 1 \), then \( \alpha = 2\beta = 2(1) = 2 \). The roots are 1 and 2.
Check the equation with \( a=-4 \): \( x^2 + (-4+1)x + 2 = 0 \implies x^2 - 3x + 2 = 0 \). The roots are \( (x-1)(x-2)=0 \), which gives \( x=1, x=2 \). This is correct.

**Case 2: If \( \beta = -1 \).**
Substitute \( \beta = -1 \) into Equation 1:
\( 3(-1) = -(a+1) \)
\( -3 = -a - 1 \)
Add \( a \) to both sides and add 3 to the right side:
\( a = -1 + 3 \)
\( a = 2 \).
If \( \beta = -1 \), then \( \alpha = 2\beta = 2(-1) = -2 \). The roots are -1 and -2.
Check the equation with \( a=2 \): \( x^2 + (2+1)x + 2 = 0 \implies x^2 + 3x + 2 = 0 \). The roots are \( (x+1)(x+2)=0 \), which gives \( x=-1, x=-2 \). This is also correct.

Therefore, the possible values of the parameter \( a \) are -4 and 2.
In simple words: First, we write the equation clearly. We are told one root is twice the other, so we set them up as \( \alpha \) and \( 2\alpha \). Then, we use the sum and product formulas for roots to create two equations with \( \alpha \) and 'a'. We solve these to find the possible values for \( \alpha \), which then helps us find the possible values for 'a'.

🎯 Exam Tip: When the ratio of roots is given, always represent the roots as \( k\alpha \) and \( \alpha \) (or \( \alpha \) and \( \beta \) with \( \alpha = k\beta \)) to simplify the expressions for sum and product of roots. This often leads to direct values for the roots or 'a'.

Question 30. If \( (1 - p) \) is a root of the quadratic equation \( x^2 + px + (1 - p) = 0 \), then its roots are
(a) 0, – 1
(b) – 1, 1
(c) 0, 1
(d) – 1, 2
Answer: (a) 0, – 1
Given the quadratic equation: \( x^2 + px + (1 - p) = 0 \).
We are told that \( (1 - p) \) is one of its roots.
If a value is a root of an equation, it means that when you substitute that value for \( x \), the equation becomes true. So, substitute \( x = (1 - p) \) into the given equation:
\( (1 - p)^2 + p(1 - p) + (1 - p) = 0 \)
Notice that \( (1 - p) \) is a common factor in all three terms. We can factor it out:
\( (1 - p) [ (1 - p) + p + 1 ] = 0 \)
Now, simplify the expression inside the square brackets:
\( (1 - p) [ 1 - p + p + 1 ] = 0 \)
\( (1 - p) [ 2 ] = 0 \)
This equation simplifies to \( 2(1 - p) = 0 \).
For this product to be zero, one of the factors must be zero. Since 2 is not zero, \( (1 - p) \) must be zero:
\( 1 - p = 0 \)
\( \implies p = 1 \)
Now that we have the value of \( p \), substitute \( p = 1 \) back into the original quadratic equation:
\( x^2 + (1)x + (1 - 1) = 0 \)
\( x^2 + x + 0 = 0 \)
\( x^2 + x = 0 \)
To find the roots, factor out \( x \):
\( x(x + 1) = 0 \)
This gives two possible solutions for \( x \):
\( x = 0 \) or \( x + 1 = 0 \implies x = -1 \).
So, the roots of the equation are 0 and -1.
Comparing this with the given options, option (a) is 0, -1.
In simple words: If a number is a root of an equation, it means the equation works when you put that number in place of 'x'. So, we put \( (1 - p) \) into the equation for 'x'. This helps us find that 'p' must be 1. Once we know 'p', we put it back into the main equation to get a simple equation. Then we solve that simple equation to find its roots, which are 0 and -1.

🎯 Exam Tip: When a root of a polynomial is given in terms of a variable (like \( 1-p \)), substitute it into the equation to solve for that variable first. This often simplifies the problem significantly before you need to find all the roots.