Question

If -5 is a root of the quadratic equation $2x^2+px-15=0$ and the equation $4x^{2} -2px+k = 0$ has equal roots. Find the value of x

Answer 1

CORRECT QUESTION :–

▪︎If -5 is a root of the quadratic equation $2x^2+px-15=0$ and the equation $4x^{2} -2px+k = 0$ has equal roots. Find the value of 'k' –

ANSWER :–

▪︎$\\{ \boxed { \bold{ \: \: k = { \dfrac{49}{4} } } }}$

EXPLANATION :–

GIVEN :–

• -5 is a root of the quadratic equation $2x^2+px-15=0$

• Equation $4x^{2} -2px+k = 0$ has equal roots.

TO FIND :–

Value of 'k' .

SOLUTION :–

☞ -5 is a root of the quadratic equation $2x^2+px-15=0$

• So that , x = -5 will satisfy the equation –

=> 2(-5)² + p(-5) -15 = 0

=> 2(25) - 5p - 15 = 0

=> 50 - 15 = 5p

=> 5p = 35

=> p = 7

☞ Equation $4x^{2} -2px+k = 0$ has equal roots.

$\\ \implies 4x^{2} -2(7)x+k = 0$

$\\ \implies 4x^{2} -14x+k = 0$

• Equation have equal roots .

• Let the roots are a and a

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

$\\{ \bold{ \implies 2 a = - \dfrac{( - 14)}{4} }}$

$\\{ \bold{ \implies a = \dfrac{7}{4} }}$

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

$\\{ \bold{ \implies \: \: a \times a = \frac{k}{4} }}$

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

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

$\\{ \bold{ \implies \: \: k = 4{( \dfrac{7}{4} )}^{2} }}$

$\\{ \bold{ \implies \: \: k = { \dfrac{49}{4} } }}$