Question

If 2 is a root of the quadratic equation $3x^{2} +px-8=0$ and the quadratic equation $4x^{2} -2px+k=0$ has equal roots. Find the value of k

Answer 1

ANSWER :–

$\\ { \boxed{\bold{ { k = 1 }}}}$

EXPLANATION :–

GIVEN :–

▪︎ 2 is a root of the quadratic equation $3x^{2} + px - 8 = 0$

▪︎ And the quadratic equation $4x^{2} -2px+k=0$ has equal roots.

TO FIND :–

▪︎ Value of 'x'

SOLUTION :–

☞ 2 is a root of the quadratic equation ${ \bold { 3x^{2} + px - 8 = 0}}$

• Now x = 2 will satisfy the equation –

=> 3(2)² + p(2) - 8 = 0

=> 3(4) + 2p - 8 = 0

=> 12 - 8 + 2p = 0

=> 2p + 4 = 0

=> p = -2

☞ Quadratic equation ${ \bold{4x^{2} -2px+k=0}}$ has equal roots .

• Now put p = -2 in given equation –

$\\ \implies{ \bold{4x^{2} + 4x+k=0}}$

• Let the roots are a and a.

$\\ \implies{ \bold{sum \: \: of \: \: roots = - \frac{b}{a} }}$

$\\ \implies{ \bold{a + a = - \frac{4}{4} }}$

$\\ \implies{ \bold{2a = - 1 }}$

$\\ \implies{ \bold{a = - \frac{1}{2} }}$

$\\ { \bold{ \: \: \: \: . \: \: product \: \: of \: \: roots \: = \frac{c}{a} }}$

$\\ \implies { \bold{ {a}^{2} = \frac{k}{4} }}$

• Now put the value of 'a' –

$\\ \implies { \bold{ { (- \frac{1}{2}) }^{2} = \frac{k}{4} }}$

$\\ \implies { \bold{ { \frac{1}{4}}= \frac{k}{4} }}$

$\\ \implies { \boxed{\bold{ { k = 1 }}}}$