CBSE Class 10 Mathematics Quadratic Equations Assignment Set 16

Question. If one root of the equation \( 2x^2+x+k=0 \) is \(-2\) , Find the value of \( k \) and other root.
Answer: Since \(-2\) is a root of the equation, substituting \( x = -2 \) gives: \[ 2(-2)^2 + (-2) + k = 0 \] \[ 2(4) - 2 + k = 0 \] \[ 8 - 2 + k = 0 \implies k = -6 \] To find the other root (let it be \( \beta \)), we can use the sum of roots formula: \[ -2 + \beta = -\frac{1}{2} \] \[ \beta = 2 - \frac{1}{2} = \frac{3}{2} \] Thus, \( k = -6 \) and the other root is \( \frac{3}{2} \).

Question. Find the Condition that one root of the equation \( ax^2+bx+c=0 \) , \( a \neq 0 \) is the reciprocal of other.
Answer: Let the roots of the quadratic equation be \( \alpha \) and \( \frac{1}{\alpha} \). The product of the roots is: \[ \alpha \cdot \frac{1}{\alpha} = \frac{c}{a} \] \[ 1 = \frac{c}{a} \implies c = a \] Thus, the required condition is \( c = a \).

Question. Find the Condition that the two roots of the equation \( px^2+qx+r=0 \) , \( p \neq 0 \) are equal in magnitude but opposite in sign.
Answer: Let the roots be \( \alpha \) and \( -\alpha \). The sum of the roots is: \[ \alpha + (-\alpha) = -\frac{q}{p} \] \[ 0 = -\frac{q}{p} \implies q = 0 \] Thus, the required condition is \( q = 0 \).

Question. Find \( p \) and \( q \) So that the sum and product of roots of equation \( px^2-2x+q=0 \) are \( \frac{2}{3} \) and \( -\frac{5}{3} \) respectively.
Answer: For the equation \( px^2-2x+q=0 \): The sum of the roots is given by \( -\frac{-2}{p} = \frac{2}{p} \). We are given that the sum is \( \frac{2}{3} \): \[ \frac{2}{p} = \frac{2}{3} \implies p = 3 \] The product of the roots is given by \( \frac{q}{p} \). We are given that the product is \( -\frac{5}{3} \): \[ \frac{q}{p} = -\frac{5}{3} \implies \frac{q}{3} = -\frac{5}{3} \implies q = -5 \] Thus, \( p = 3 \) and \( q = -5 \).

Question. If \( \alpha \) and \( \beta \) are the roots of the equation \( x^2-8x+p=0 \) , Find \( p \) if \( \alpha^2+\beta^2=40 \).
Answer: For the equation \( x^2-8x+p=0 \), we have: \[ \alpha + \beta = 8 \] \[ \alpha\beta = p \] Using the given condition: \[ \alpha^2 + \beta^2 = 40 \] \[ (\alpha + \beta)^2 - 2\alpha\beta = 40 \] \[ (8)^2 - 2p = 40 \] \[ 64 - 2p = 40 \implies 2p = 24 \implies p = 12 \] Thus, \( p = 12 \).

Question. If \( \alpha \) and \( \beta \) are the roots of the equation \( x^2-5x+k=0 \) . Find the value of \( k \) such that \( 2\alpha+3\beta=9 \).
Answer: For the equation \( x^2-5x+k=0 \), we have: \[ \alpha + \beta = 5 \implies \alpha = 5 - \beta \] \[ \alpha\beta = k \] Given the relation: \[ 2\alpha + 3\beta = 9 \] Substitute \( \alpha = 5 - \beta \): \[ 2(5 - \beta) + 3\beta = 9 \] \[ 10 - 2\beta + 3\beta = 9 \implies \beta = -1 \] Substitute \( \beta = -1 \) back to find \( \alpha \): \[ \alpha = 5 - (-1) = 6 \] Now, calculate \( k \): \[ k = \alpha\beta = 6 \times (-1) = -6 \] (Note: The printed answer key in the worksheet lists \( K = 6 \)).

Question. If \( \alpha,\beta \) are the roots of a quadratic equation such that \( \alpha+\beta=24 \wedge \alpha-\beta=8 \) Form the equation.
Answer: Solve the system of equations for \( \alpha \) and \( \beta \): \[ \alpha + \beta = 24 \] \[ \alpha - \beta = 8 \] Adding both equations: \[ 2\alpha = 32 \implies \alpha = 16 \] Subtracting the second equation from the first: \[ 2\beta = 16 \implies \beta = 8 \] The product of the roots is: \[ \alpha\beta = 16 \times 8 = 128 \] The quadratic equation is given by: \[ x^2 - (\alpha + \beta)x + \alpha\beta = 0 \] \[ x^2 - 24x + 128 = 0 \]

Question. Solve for \( x \) by the method of completing the square \( 6x^2-5x+1=0 \)
Answer: Divide the entire equation by 6: \[ x^2 - \frac{5}{6}x + \frac{1}{6} = 0 \] \[ x^2 - \frac{5}{6}x = -\frac{1}{6} \] Add the square of half the coefficient of \( x \), which is \( \left(\frac{5}{12}\right)^2 \), to both sides: \[ x^2 - \frac{5}{6}x + \left(\frac{5}{12}\right)^2 = -\frac{1}{6} + \frac{25}{144} \] \[ \left(x - \frac{5}{12}\right)^2 = \frac{-24 + 25}{144} \] \[ \left(x - \frac{5}{12}\right)^2 = \frac{1}{144} \] Taking the square root on both sides: \[ x - \frac{5}{12} = \pm \frac{1}{12} \] Case 1: \[ x = \frac{5}{12} + \frac{1}{12} = \frac{6}{12} = \frac{1}{2} \] Case 2: \[ x = \frac{5}{12} - \frac{1}{12} = \frac{4}{12} = \frac{1}{3} \] Thus, \( x = \frac{1}{2}, \frac{1}{3} \).

Question. Solve for \( x \) :- \( \frac{x-1}{x-2} - \frac{x-2}{x-3} = \frac{x-5}{x-6} - \frac{x-6}{x-7} \)
Answer: Rewrite each fraction: \[ \left(1 + \frac{1}{x-2}\right) - \left(1 + \frac{1}{x-3}\right) = \left(1 + \frac{1}{x-6}\right) - \left(1 + \frac{1}{x-7}\right) \] \[ \frac{1}{x-2} - \frac{1}{x-3} = \frac{1}{x-6} - \frac{1}{x-7} \] Combine fractions on both sides: \[ \frac{(x-3) - (x-2)}{(x-2)(x-3)} = \frac{(x-7) - (x-6)}{(x-6)(x-7)} \] \[ \frac{-1}{(x-2)(x-3)} = \frac{-1}{(x-6)(x-7)} \] This implies: \[ (x-2)(x-3) = (x-6)(x-7) \] \[ x^2 - 5x + 6 = x^2 - 13x + 42 \] \[ -5x + 13x = 42 - 6 \] \[ 8x = 36 \implies x = \frac{36}{8} = \frac{9}{2} \]

Question. Find the Condition that Sum of roots, of \( mx^2+nx+p=0 \) , \( m \neq 0 \) , \( n \neq 0 \) , is equal to the product of roots
Answer: For the equation \( mx^2+nx+p=0 \): Sum of roots \( = -\frac{n}{m} \) Product of roots \( = \frac{p}{m} \) Equating the sum and product of the roots: \[ -\frac{n}{m} = \frac{p}{m} \] Since \( m \neq 0 \), we can multiply both sides by \( m \): \[ -n = p \implies p + n = 0 \] Thus, the required condition is \( p + n = 0 \).

Question. Check whether the polynomial \( 2x^2-4x+7=0 \) can be factorized into two real linear factors
Answer: A quadratic expression can be factorized into two real linear factors if its discriminant \( D \ge 0 \). Here, \( a = 2 \), \( b = -4 \), \( c = 7 \). \[ D = b^2 - 4ac = (-4)^2 - 4(2)(7) = 16 - 56 = -40 \] Since \( D < 0 \), the polynomial cannot be factorized into two real linear factors.

Question. Find \( p \) so that the quadratic polynomial \( px^2+4x+1 \) can be factorised into two real linear factors
Answer: For the quadratic expression to have real linear factors, its discriminant \( D \) must satisfy \( D \ge 0 \). Here, \( a = p \), \( b = 4 \), \( c = 1 \). \[ D = b^2 - 4ac = 4^2 - 4(p)(1) = 16 - 4p \] Setting the discriminant \( D \ge 0 \): \[ 16 - 4p \ge 0 \implies 4p \le 16 \implies p \le 4 \] Thus, \( p \le 4 \) (with \( p \neq 0 \)).

Question. If \( x=2 \) and \( x=3 \) are roots of the equation \( 3x^2-2mx+2n=0 \) , Find the values of \( m \) and \( n \)
Answer: Using the relationship between roots and coefficients for \( 3x^2-2mx+2n=0 \): Sum of roots: \[ 2 + 3 = -\frac{-2m}{3} \] \[ 5 = \frac{2m}{3} \implies 2m = 15 \implies m = \frac{15}{2} \] Product of roots: \[ 2 \times 3 = \frac{2n}{3} \] \[ 6 = \frac{2n}{3} \implies 2n = 18 \implies n = 9 \] Thus, the values are \( m = \frac{15}{2} \) and \( n = 9 \).

Question. Find the roots of \( 2x^2-9x+10=0 \) if
(i) \( x \in N \)
(ii) \( x \in Q \)
Answer: Solve the quadratic equation \( 2x^2-9x+10=0 \) by splitting the middle term: \[ 2x^2 - 5x - 4x + 10 = 0 \] \[ x(2x - 5) - 2(2x - 5) = 0 \] \[ (2x - 5)(x - 2) = 0 \] The roots of the equation are \( x = 2 \) and \( x = \frac{5}{2} \). (i) For \( x \in N \) (Natural numbers): The only natural root is \( x = 2 \). (ii) For \( x \in Q \) (Rational numbers): Both roots are rational, so \( x = 2, \frac{5}{2} \).

Question. Solve the equation \( 10ax^2-6x+15ax-9=0 \)
Answer: Grouping terms to factorize: \[ 10ax^2 + 15ax - 6x - 9 = 0 \] \[ 5ax(2x + 3) - 3(2x + 3) = 0 \] \[ (5ax - 3)(2x + 3) = 0 \] This yields: \[ 5ax - 3 = 0 \implies x = \frac{3}{5a} \] or \[ 2x + 3 = 0 \implies x = -\frac{3}{2} \] Thus, the solutions are \( \left(\frac{3}{5a}, -\frac{3}{2}\right) \).