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 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)

 `a_1 = 1,b_1 = k and ,c_1 = 64`

 `a_2 = 1,b_2 = -8 and ,c_2 =  k`

As we know that `D_1 = b^2 - 4ac`

Putting the value of `a_1 = 1,b_1 = k and ,c_1 = 64`

`=(k)^2 - 4 xx 1 xx 64`

`= k^2 - 256`

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

`k^2 - 256 = 0`

            `k^2 = 256`

              `k = sqrt256`

             ` k  = ± 16`

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

` x^2 - 8x + 16 = 0`

         `(x - 4)^2 = 0`

                  x - 4 = 0

                       x = 4

The value of  k = 16 satisfying to both equations