# For which value of k will the roots of 2x^2+kx+1=0 be real?

Dec 2, 2016

The answer is k in ] -oo,-sqrt8 ] uu [sqrt8, oo[

#### Explanation:

In order for the roots of a quadratic equation, to de real, the dircriminant $\Delta \ge 0$

If the equation is $a {x}^{2} + b x + c = 0$

the discriminant is $\Delta = {b}^{2} - 4 a c$

$2 {x}^{2} + k x + 1 = 0$

$\Delta = {k}^{2} - 4 \cdot 2 \cdot 1 \ge 0$

${k}^{2} - 8 \ge 0$

$\left(k - \sqrt{8}\right) \left(k + \sqrt{8}\right) \ge 0$

We do a sign chart

$\textcolor{w h i t e}{a a a a}$$k$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- \sqrt{8}$$\textcolor{w h i t e}{a a a a}$$\sqrt{8}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$k + \sqrt{8}$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$k - \sqrt{8}$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$\Delta$$\textcolor{w h i t e}{a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

So, for $\Delta \ge 0$

k in ] -oo,-sqrt8 ] uu [sqrt8, oo[