The positive value of k for which the equation x2 + kx + 64 = 0 and x2 − 8x + k = 0

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

Then find the value of a.

Here, $x^{2}+k x+64=0$.......(1)

$x^{2}-8 x+k=0 \cdots(2)$

$a_{1}=1, b_{1}=k$ and, $c_{1}=64$

$a_{2}=1, b_{2}=-8$ and,$c_{2}=k$

As we know that $D_{1}=b^{2}-4 a c$

Putting the value of $a_{1}=1, b_{1}=k$ and, $c_{1}=64$

$=(k)^{2}-4 \times 1 \times 64$

$=k^{2}-256$

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

$k^{2}-256=0$

$k^{2}=256$

$k=\sqrt{256}$

$k=\pm 16$

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

$x^{2}-8 x+16=0$

$(x-4)^{2}=0$

$x-4=0$

$x=4$

The value of $k=16$ satisfying to both equations

Thus, the correct answer is $(d)$