CBSE Class 10 Mathematics Quadratic Equations Assignment Set 15

Question 1. Find the value of k so that sum of the roots of the equation kx^2+2x+3k=0 is equal to their product of roots.
Answer: Let \( \alpha \) and \( \beta \) be the roots of the quadratic equation \( kx^2 + 2x + 3k = 0 \). Comparing this with the general quadratic equation \( ax^2 + bx + c = 0 \), we get the coefficients as \( a = k \), \( b = 2 \), and \( c = 3k \). The sum of the roots is calculated as \( \alpha + \beta = -\frac{b}{a} = -\frac{2}{k} \), and the product of the roots is \( \alpha\beta = \frac{c}{a} = \frac{3k}{k} = 3 \) (for \( k \neq 0 \)). According to the given condition, the sum of these roots equals their product:
\( \implies -\frac{2}{k} = 3 \)
\( \implies 3k = -2 \)
\( \implies k = -\frac{2}{3} \)
In simple words: We find the sum and product of the roots using the coefficient formulas, set them equal to each other, and then solve for \( k \).
Exam Tip: Always state the non-zero condition \( k \neq 0 \). In any quadratic equation, the coefficient of \( x^2 \) can never be zero, and this step is crucial when simplifying fractions.

Question 2. Find k so that the equation x^2- (k+6)x + 2(2k-1) = 0 has sum of roots is half the product of roots.
Answer: Let the roots of the equation \( x^2 - (k+6)x + 2(2k-1) = 0 \) be \( \alpha \) and \( \beta \). The coefficients of this equation are \( a = 1 \), \( b = -(k+6) \), and \( c = 2(2k-1) \). The sum of the roots is \( \alpha + \beta = -\frac{b}{a} = k+6 \), and the product of the roots is \( \alpha\beta = \frac{c}{a} = 2(2k-1) \). We are given that the sum of the roots is half of the product:
\( \alpha + \beta = \frac{1}{2} (\alpha\beta) \) Substituting the expressions:
\( k+6 = \frac{1}{2} [2(2k-1)] \)
\( \implies k+6 = 2k-1 \) Rearranging the terms to solve for \( k \):
\( \implies 2k - k = 6 + 1 \)
\( \implies k = 7 \)
In simple words: Write out the expressions for the sum and product of the roots, plug them into the given relationship, and solve the equation to find \( k \).

Exam Tip: Be careful with the negative sign when calculating the sum of the roots. Since the coefficient \( b \) is already negative, \( -b \) becomes positive.

Question 3. If the equation ax^2-7x+c = 0 has 14 as the sum of the roots and also the product of roots. Find a & c
Answer: Let \( \alpha \) and \( \beta \) be the roots of the equation \( ax^2 - 7x + c = 0 \). The coefficients of this quadratic equation are \( a = a \), \( b = -7 \), and \( c = c \). We determine the sum of the roots as \( \alpha + \beta = -\frac{b}{a} = \frac{7}{a} \), and the product of the roots as \( \alpha\beta = \frac{c}{a} \). According to the problem, both the sum and the product of the roots are equal to 14. First, we use the sum of the roots to find \( a \):
\( \frac{7}{a} = 14 \)
\( \implies a = \frac{7}{14} = \frac{1}{2} \) Next, we substitute this value of \( a \) into the product equation to solve for \( c \):
\( \frac{c}{a} = 14 \)
\( \implies \frac{c}{1/2} = 14 \)
\( \implies 2c = 14 \)
\( \implies c = 7 \) Therefore, the required values are \( a = \frac{1}{2} \) and \( c = 7 \).
In simple words: Use the given sum of roots to calculate \( a \) first. Then, use that value of \( a \) in the product formula to find \( c \).

Exam Tip: Solving for \( a \) first makes the entire calculation much simpler. Always look for the variable that can be determined independently to prevent algebraic errors.

Question 4. Find the quadratic equation whose roots are the reciprocal of the roots of the equation 3x^2-20x+17=0
Answer: Let \( \alpha \) and \( \beta \) be the roots of the quadratic equation \( 3x^2 - 20x + 17 = 0 \). The sum of the roots is \( \alpha + \beta = \frac{20}{3} \), and the product of the roots is \( \alpha\beta = \frac{17}{3} \). We need to find a new quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \). The sum of these new reciprocal roots is: \( \text{Sum of new roots} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \) Substituting the values we found: \( \text{Sum of new roots} = \frac{20/3}{17/3} = \frac{20}{17} \) The product of the new roots is: \( \text{Product of new roots} = \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} \) Substituting the product value: \( \text{Product of new roots} = \frac{1}{17/3} = \frac{3}{17} \) The formula for a quadratic equation is \( x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0 \). Substituting our values: \( x^2 - \frac{20}{17}x + \frac{3}{17} = 0 \) Multiplying the entire equation by 17 to clear the denominators:
\( \implies 17x^2 - 20x + 3 = 0 \)
In simple words: Find the sum and product of the new reciprocal roots by rewriting them using the old roots. Then, substitute these into the standard quadratic equation formula.

Exam Tip: A quick way to verify your answer is to replace \( x \) with \( \frac{1}{x} \) in the original equation and simplify. This method directly yields the reciprocal equation with minimal steps.

Question 5. Find K for real linear factors
(i) \( kx^2-2x+2 \)
(ii) \( x^2-kx+4 \)
Answer: For any quadratic expression \( Ax^2 + Bx + C \) to have real linear factors, its corresponding equation must have real roots. This requires the discriminant \( D = B^2 - 4AC \) to be greater than or equal to zero (\( D \geq 0 \)). (i) For \( kx^2 - 2x + 2 \): Here, \( A = k \), \( B = -2 \), and \( C = 2 \). \( D = (-2)^2 - 4(k)(2) = 4 - 8k \) For real linear factors, set \( D \geq 0 \): \( 4 - 8k \geq 0 \)
\( \implies 4 \geq 8k \)
\( \implies 8k \leq 4 \)
\( \implies k \leq \frac{4}{8} \)
\( \implies k \leq \frac{1}{2} \) (ii) For \( x^2 - kx + 4 \): Here, \( A = 1 \), \( B = -k \), and \( C = 4 \). \( D = (-k)^2 - 4(1)(4) = k^2 - 16 \) For real linear factors, set \( D \geq 0 \): \( k^2 - 16 \geq 0 \)
\( \implies (k - 4)(k + 4) \geq 0 \) This inequality holds true when: \( k \leq -4 \) or \( k \geq 4 \) (which can also be represented as \( |k| \geq 4 \)).
In simple words: A quadratic expression has real factors when its discriminant is not negative. We calculate the discriminant for each expression and solve the inequality.

Exam Tip: Remember to include the equal-to sign in your inequality (\( D \geq 0 \)) because equal roots (\( D = 0 \)) also yield valid real linear factors.

Question 6. If (-4) is a root of the quadratic equation x^2+px-4=0 and the equation x^2+px+k=0 has equal roots, find K.
Answer: Since \( -4 \) is a root of the equation \( x^2 + px - 4 = 0 \), it must satisfy it. Substituting \( x = -4 \) into the equation: \( (-4)^2 + p(-4) - 4 = 0 \)
\( \implies 16 - 4p - 4 = 0 \)
\( \implies 12 - 4p = 0 \)
\( \implies 4p = 12 \)
\( \implies p = 3 \) Now, the second quadratic equation is \( x^2 + px + k = 0 \). Substituting our calculated value \( p = 3 \): \( x^2 + 3x + k = 0 \) Since this equation is given to have equal roots, its discriminant \( D \) must be zero. Here, the coefficients are \( a = 1 \), \( b = 3 \), and \( c = k \). \( D = b^2 - 4ac = 0 \) \( 3^2 - 4(1)(k) = 0 \)
\( \implies 9 - 4k = 0 \)
\( \implies 4k = 9 \)
\( \implies k = \frac{9}{4} \)
In simple words: Substitute the root \( -4 \) into the first equation to find the value of \( p \). Then, use this value of \( p \) in the second equation and set its discriminant to zero to find \( k \).

Exam Tip: Be structured when dealing with multi-equation problems. First find the value of the shared coefficient, and then proceed to the equal-roots condition.

Question 7. If the roots of the equation px^2+qx+r=0 are equal then show that q^2=4pr.
Answer: For any standard quadratic equation \( ax^2 + bx + c = 0 \), the roots are equal when the discriminant is zero (\( D = 0 \)). The formula for the discriminant is: \( D = b^2 - 4ac \) For the given quadratic equation \( px^2 + qx + r = 0 \), the coefficients are \( a = p \), \( b = q \), and \( c = r \). Substituting these coefficients into the discriminant formula: \( D = q^2 - 4pr \) Since the roots of the equation are equal, we set \( D = 0 \): \( q^2 - 4pr = 0 \)
\( \implies q^2 = 4pr \) Hence, it is shown that \( q^2 = 4pr \).
In simple words: When roots are identical, the discriminant must equal zero. Writing the formula for the discriminant with our coefficients gives the proof directly.

Exam Tip: Clearly write down the condition "For equal roots, \( D = 0 \)" before showing the step-by-step algebraic manipulation to score full presentation marks.

Question 8. Factorise
(i) \( 2\sqrt{2}x^2+4x+\sqrt{2} \)
(ii) \( x^2+4\sqrt{2}x+6 \)
Answer: (i) To factorize \( 2\sqrt{2}x^2 + 4x + \sqrt{2} \), we split the middle term. We need two numbers that add up to 4 and have a product of: \( (2\sqrt{2}) \times (\sqrt{2}) = 4 \) These numbers are 2 and 2 because \( 2 + 2 = 4 \) and \( 2 \times 2 = 4 \). Splitting the middle term: \( 2\sqrt{2}x^2 + 2x + 2x + \sqrt{2} \) Grouping the terms: \( = 2x(\sqrt{2}x + 1) + \sqrt{2}(\sqrt{2}x + 1) \) Factoring out the common binomial term: \( = (2x + \sqrt{2})(\sqrt{2}x + 1) \) We can pull out a factor of \( \sqrt{2} \) from \( (2x + \sqrt{2}) \): \( = \sqrt{2}(\sqrt{2}x + 1)(\sqrt{2}x + 1) \) \( = \sqrt{2}(\sqrt{2}x + 1)^2 \) This can also be written as: \( = 2\sqrt{2}\left(x + \frac{1}{\sqrt{2}}\right)^2 \) (ii) To factorize \( x^2 + 4\sqrt{2}x + 6 \), we split the middle term. We need two numbers that add up to \( 4\sqrt{2} \) and multiply to 6. Let the two numbers be of the form \( a\sqrt{2} \) and \( b\sqrt{2} \). Their sum is \( (a + b)\sqrt{2} = 4\sqrt{2} \implies a + b = 4 \). Their product is \( (a\sqrt{2}) \times (b\sqrt{2}) = 2ab = 6 \implies ab = 3 \). The values of \( a \) and \( b \) that satisfy this are 3 and 1. So, the two parts are \( 3\sqrt{2}x \) and \( \sqrt{2}x \). Splitting the middle term: \( x^2 + 3\sqrt{2}x + \sqrt{2}x + 6 \) Grouping the terms: \( = x(x + 3\sqrt{2}) + \sqrt{2}(x + 3\sqrt{2}) \) Factoring out the common binomial term: \( = (x + \sqrt{2})(x + 3\sqrt{2}) \)
In simple words: Break down the middle coefficient into two parts that sum up to the middle term and multiply to the product of the outer coefficients, then group and factor.

Exam Tip: For expressions with square roots, expressing the middle split terms using the radical sign (like \( 3\sqrt{2} \) and \( \sqrt{2} \)) makes grouping and factoring much simpler.

Question 9. If \alpha, \beta are the roots of the equation px^2+qx+r=0, p\neq0, Find the value of \alpha^2+\beta^2
Answer: Let \( \alpha \) and \( \beta \) be the roots of the quadratic equation \( px^2 + qx + r = 0 \). The relations between the roots and coefficients are: Sum of roots: \( \alpha + \beta = -\frac{q}{p} \) Product of roots: \( \alpha\beta = \frac{r}{p} \) We want to evaluate \( \alpha^2 + \beta^2 \). Using the algebraic identity: \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \) Substituting our root-coefficient formulas into the identity: \( \alpha^2 + \beta^2 = \left(-\frac{q}{p}\right)^2 - 2\left(\frac{r}{p}\right) \) \( = \frac{q^2}{p^2} - \frac{2r}{p} \) Taking the common denominator \( p^2 \): \( = \frac{q^2 - 2pr}{p^2} \)
In simple words: Rewrite the term \( \alpha^2 + \beta^2 \) using the sum and the product of the roots. Then substitute the coefficient formulas to get the final answer.

Exam Tip: Master standard symmetric expressions like \( \alpha^2 + \beta^2 \) and \( \alpha^3 + \beta^3 \), as they are very common in exams and always rely on these exact algebraic expansions.

Question 10. If \alpha, \beta are the roots of the equation 3x^2-6x-5=0. Find the value of the following:
(i) \( \alpha - \beta \)
(ii) \( \alpha^2 + \beta^2 \)
(iii) \( \alpha^3 + \beta^3 \)
(iv) \( \alpha^3 - \beta^3 \)
(v) \( \alpha^4 + \beta^4 \)
(vi) \( \alpha^4 - \beta^4 \)
(vii) \( \alpha^2\beta + \beta^2\alpha \)
(viii) \( \left(\alpha + \frac{1}{\beta}\right)\left(\frac{1}{\alpha} + \beta\right) \)
(ix) \( \alpha^2 - \beta^2 \)
(x) \( \frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta \)
Answer: First, we find the sum and the product of the roots for the quadratic equation \( 3x^2 - 6x - 5 = 0 \). Here, the coefficients are \( a = 3 \), \( b = -6 \), and \( c = -5 \). Sum of roots: \( \alpha + \beta = -\frac{b}{a} = -\frac{-6}{3} = 2 \) Product of roots: \( \alpha\beta = \frac{c}{a} = -\frac{5}{3} \) We now solve each of the sub-parts:
(i) \( \alpha - \beta \) Using the identity \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \): \( (\alpha - \beta)^2 = (2)^2 - 4\left(-\frac{5}{3}\right) = 4 + \frac{20}{3} = \frac{12 + 20}{3} = \frac{32}{3} \) Taking the square root (considering the positive value): \( \alpha - \beta = \sqrt{\frac{32}{3}} = \frac{4\sqrt{2}}{\sqrt{3}} = \frac{4\sqrt{6}}{3} \)
(ii) \( \alpha^2 + \beta^2 \) Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \): \( \alpha^2 + \beta^2 = (2)^2 - 2\left(-\frac{5}{3}\right) = 4 + \frac{10}{3} = \frac{22}{3} \)
(iii) \( \alpha^3 + \beta^3 \) Using the factorization formula \( \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = (\alpha + \beta)[(\alpha^2 + \beta^2) - \alpha\beta] \): \( \alpha^3 + \beta^3 = (2)\left[\frac{22}{3} - \left(-\frac{5}{3}\right)\right] = 2\left[\frac{22 + 5}{3}\right] = 2\left[\frac{27}{3}\right] = 2(9) = 18 \)
(iv) \( \alpha^3 - \beta^3 \) Using the factorization formula \( \alpha^3 - \beta^3 = (\alpha - \beta)(\alpha^2 + \alpha\beta + \beta^2) = (\alpha - \beta)[(\alpha^2 + \beta^2) + \alpha\beta] \): \( \alpha^3 - \beta^3 = \left(\frac{4\sqrt{6}}{3}\right)\left[\frac{22}{3} + \left(-\frac{5}{3}\right)\right] = \left(\frac{4\sqrt{6}}{3}\right)\left[\frac{17}{3}\right] = \frac{68\sqrt{6}}{9} \)
(v) \( \alpha^4 + \beta^4 \) Using the identity \( \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 \): \( \alpha^4 + \beta^4 = \left(\frac{22}{3}\right)^2 - 2\left(-\frac{5}{3}\right)^2 = \frac{484}{9} - 2\left(\frac{25}{9}\right) = \frac{484 - 50}{9} = \frac{434}{9} \)
(vi) \( \alpha^4 - \beta^4 \) Using the difference of squares \( \alpha^4 - \beta^4 = (\alpha^2 - \beta^2)(\alpha^2 + \beta^2) = (\alpha - \beta)(\alpha + \beta)(\alpha^2 + \beta^2) \): \( \alpha^4 - \beta^4 = \left(\frac{4\sqrt{6}}{3}\right)(2)\left(\frac{22}{3}\right) = \frac{176\sqrt{6}}{9} \)
(vii) \( \alpha^2\beta + \beta^2\alpha \) Factoring out \( \alpha\beta \): \( \alpha^2\beta + \beta^2\alpha = \alpha\beta(\alpha + \beta) \) Substituting the values: \( = \left(-\frac{5}{3}\right)(2) = -\frac{10}{3} \)
(viii) \( \left(\alpha + \frac{1}{\beta}\right)\left(\frac{1}{\alpha} + \beta\right) \) Expanding the brackets: \( = 1 + \alpha\beta + \frac{1}{\alpha\beta} + 1 = 2 + \alpha\beta + \frac{1}{\alpha\beta} \) Substituting the product of roots: \( = 2 + \left(-\frac{5}{3}\right) + \left(-\frac{3}{5}\right) = 2 - \frac{5}{3} - \frac{3}{5} = \frac{30 - 25 - 9}{15} = -\frac{4}{15} \)
(ix) \( \alpha^2 - \beta^2 \) Using the identity \( \alpha^2 - \beta^2 = (\alpha - \beta)(\alpha + \beta) \): \( \alpha^2 - \beta^2 = \left(\frac{4\sqrt{6}}{3}\right)(2) = \frac{8\sqrt{6}}{3} \)
(x) \( \frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta \) Combining the first two terms: \( = \frac{\alpha + \beta}{\alpha\beta} - 2\alpha\beta \) Substituting our values: \( = \frac{2}{-5/3} - 2\left(-\frac{5}{3}\right) = -\frac{6}{5} + \frac{10}{3} = \frac{-18 + 50}{15} = \frac{32}{15} \)
In simple words: We first find the numerical values of the sum and product of the roots. Then, we write each algebraic expression in terms of sum and product, and substitute these values to compute the results.

Exam Tip: Keep your calculations neat and structured. Calculating basic components like \( \alpha - \beta \) and \( \alpha^2 + \beta^2 \) correctly in the initial steps will prevent errors in more complex parts like \( \alpha^4 - \beta^4 \).