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


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

(a) 4

(b) 8

(c) 12

(d) 16


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, $c_{1}=64$

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


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




$k=\pm 16$

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

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




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

Thus, the correct answer is $(d)$

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