Question

If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.

Answer 1

Given: -5 is a root of the quadratic equation 2x² + px – 15 = 0

Substitute the value of x = -5

2(-5)² + p(-5) – 15 = 0

50 – 5p – 15 = 0

35 – 5p = 0

p = 7

Again,

In quadratic equation p(x² + x) + k = 0

7 (x² + x) + k = 0 (put value of p = 7)

7x² + 7x + k = 0

Compare given equation with the general form of quadratic equation, which is ax² + bx + c = 0

a = 7, b = 7, c = k

Find Discriminant:

D = b² – 4ac

= (7)² – 4 x 7 x k

= 49 – 28k

Since roots are real and equal, put D = 0

49 – 28k = 0

28k = 49

k = 7 / 4

The value of k is 7/4.

Answer 2

-5 is a root of quadratic equation 2x² + px - 15 = 0

So, 2(-5)² + p(-5) - 15 = 0

⇒ 50 - 5p - 15 = 0

⇒ 35 - 5p = 0

p = 7

Now, put p = 7 in second quadratic equation,

p(x² + x ) + k = 0

⇒ 7(x² + x ) + k = 0

⇒ 7x² + 7x + k = 0

Above equation has equal roots

So, D = b² - 4ac = 0

⇒ 7² - 4 × 7 × k = 0

⇒ 7 - 4k = 0

⇒ k = 7/4 = 1.75

Hence, the value of k = 1.75.