Question

The value of k, so that the quadratic equation x^2 + kx + 1 = 0 has real roots, is​

Answer 1

Answer:

$solve\:for\:x,\:x^2+kx+1=0\quad :\quad x=\frac{-k+\sqrt{k^2-4}}{2},\:x=\frac{-k-\sqrt{k^2-4}}{2}$

Step-by-step explanation:

$x^2+kx+1=0$

$x_{1,\:2}=\frac{-k\pm \sqrt{k^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}$

$x_{1,\:2}=\frac{-k\pm \sqrt{k^2-4}}{2\cdot \:1}$

$\mathrm{Separate\:the\:solutions}$

$x_1=\frac{-k+\sqrt{k^2-4}}{2\cdot \:1},\:x_2=\frac{-k-\sqrt{k^2-4}}{2\cdot \:1}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{-k+\sqrt{k^2-4}}{2},\:x=\frac{-k-\sqrt{k^2-4}}{2}$