Question

The positive value of k for which the equation x² + kx + 64 = 0 and x² – 8x + k = 0 will both have real roots, is ______.

विकल्प

• 4

• 8

• 12

• 16

Answer 1

The positive value of k for which the equation x² + kx + 64 = 0 and x² – 8x + k = 0 will both have real roots, is 16.

Explanation:

The given quadric equation are x² + kx + 64 = 0 and x² – 8x + k = 0 roots are real.

Then find the value of a.

Here, x² + kx + 64 = 0   ...(1)

x² – 8x + k = 0   ...(2)

a₁ = 1, b₁ = k and c₁ = 64

a₂ = 1, b₂ = –8 and c₂ =  k

As we know that D₁ = b² – 4ac

Putting the value of a₁ = 1, b₁ = k and c₁ = 64

= (k)² – 4 × 1 × 64

= k² – 256

The given equation will have real and distinct roots, if D > 0

k² – 256 = 0

k² = 256

`k = sqrt256`

k = ±16

Therefore, putting the value of k = 16 in equation (2), we get

x² – 8x + 16 = 0

(x – 4)² = 0

x – 4 = 0

x = 4

The value of k = 16 satisfying to both equations.