Question

Find the value of $K$, for which the eq $x^2+kx+64=0$ and $x^2-8x+k=0$ with both have real root.

Answer 1

(1) $x^2+kx+64=0$ has real roots; (2)$x^2-8x+k=0$ has real roots.

$D\ge 0⟹b^2-4ac\ge 0$

After comparing the above equations with the standard quadratic equation $a{x}^2+bx+c=0$

$\Rightarrow k^2-4\left(64\right)\ge 0$; $(8^2-4k)\ge 0$

$\Rightarrow k^2\ge 4\left(64\right)$; $k\le \frac{64}{4}$

$\Rightarrow k\in (-\infty ,-16]\cup [16,\infty )$ ……….$\left(1\right)$

$k\le 16$ ……$\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$

$k\in (-\infty ,-16]\cup \left\{16\right\}$.