#### Question

The positive value of k for which the equation x^{2}^{ }+ kx + 64 = 0 and x^{2} − 8x + k = 0 will both have real roots, is

4

8

12

16

#### Solution

The given quadric equation are x^{2}^{ }+ kx + 64 = 0 and x^{2} − 8x + k = 0 roots are real.

Then find the value of *a.*

Here, x^{2}^{ }+ kx + 64 = 0 ….. (1)

x^{2} − 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