Determine the positive values of $k$ for which the equation $x^{2} + k x + 64 = 0$ and $x^{2} - 8 x + k = 0$ will both have real roots.

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Solution

Given equations are $x^{2} + k x + 64 = 0..... (i)$ and
$x^{2} - 8 x + k = 0...... (i i)$

Let $D_{1}$ and $D_{2}$ be the discriminants of equations (i) and (ii). Then

$D_{1} = k^{2} - 4 \times 64 = k^{2} - 256$
$D_{2} = {(- 8)}^{2} - 4 k = 64 - 4 k$

Both the equations will have real roots, if
$D_{1} \geq 0; D_{2} \geq 0$

$\Rightarrow k^{2} - 256 \geq 0; 64 - 4 k \geq 0$

$\Rightarrow k \geq 16 (∵ k > 0); k \leq 16 \Rightarrow k = 16$

Hence, both equations will have real roots. when $k = 16$

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