Question

Q5) Find the value of k such that sum of zeroes is equal to the product of the zeroes of the
following quadratic polynomial (k+1) x² + (2k+1)x - 9.​

Answer 1

GIVEN :–

• Sum of zeroes is equal to the product of the zeroes of the quadratic polynomial (k+1) x² + (2k+1)x - 9.

TO FIND :–

• Value of 'k' = ?

SOLUTION :–

$\\ \implies { \bold{(k + 1) {x}^{2} + (2k + 1)x - 9 = 0 }}$

• We know that –

$\\ \longrightarrow { \bold{sum \: \: of \: \: roots = - \dfrac{coffieciant \: \: of \: \: x}{coffieciant \: \: of \: \: {x}^{2} } }}$

$\\ \implies{ \bold{sum \: \: of \: \: roots = - \dfrac{(2k + 1)}{(k + 1)} }}$

$\\ \implies{ \bold{sum \: \: of \: \: roots = - \left[ \: \dfrac{2k + 1}{k + 1}\right] }} \:$

$\\ \longrightarrow { \bold{ \: product \: \: of \: \: roots = \dfrac{constant \: \: term}{coffieciant \: \: of \: \: {x}^{2} } }}$

$\\ \implies{ \bold{product \: \: of \: \: roots = \dfrac{( - 9)}{(k + 1)} }}$

$\\ \implies{ \bold{product \: \: of \: \: roots = - \left[ \: \dfrac{9}{k + 1}\right] }} \:$

• Now According to the question –

$\\ \implies{ \bold{sum \: \: of \: \: roots = product \: \: of \: \: roots}} \:$

$\\ \implies{ \bold{- \left[ \: \dfrac{2k + 1}{k + 1}\right] = - \left[ \: \dfrac{9}{k + 1}\right]}} \:$

$\\ \implies{ \bold{(2k + 1)(k + 1) = (k + 1)9}} \:$

$\\ \implies{ \bold{(2k + 1)(k + 1) - (k + 1)9 = 0}} \:$

$\\ \implies{ \bold{(2k + 1 - 9)(k + 1) = 0}} \:$

$\\ \implies{ \bold{(2k - 8)(k + 1) = 0}} \:$

$\\ \implies{ \bold{(k - 4)(k + 1) = 0}} \:$

$\\ \implies \large{ \boxed{ \bold{k = 4 \: , \: - 1}}} \:$

∵ At k = -1 Leading coffieciant of quadratic equation will be zero , but we know that coffieciant of x² can't be zero.

• Hence , The value of k is 4.

Answer 2

Sum of roots = product of roots

So, -b/a = c/a

Also, a cannot be equal to zero as this is a quadratic equation.

So, From Question, we have

k cannot be equal to -1.

So, -b = c

Therefore, from Question, we have

-(2k+1) = -9

2k = 8

k = 4

So, k will be equal to 4.

Thanks you!