# How to use the discriminant to find out what type of solutions the equation has for 4x^2 + 1 = 0?

May 25, 2015

Written in the standard form for a quadratic $a {x}^{2} + b x + c = 0$

$4 {x}^{2} + 1 = 0$
becomes
$4 {x}^{2} + \left(0\right) x + 1 = 0$

The discriminant $\Delta = {b}^{2} - 4 a c$
becomes
$\Delta = - 16$

$\Delta \left\{\begin{matrix}< 0 \rightarrow \text{no Real roots (onlt Complex roots)" \\ =0 rarr "1 Real root" \\ >0 rarr "2 Real roots}\end{matrix}\right.$

Therefore
$4 {x}^{2} + 1 = 0$ has only Complex roots