# How do you find the discriminant and how many and what type of solutions does x^2 +4x +4=0 have?

May 10, 2015

discriminant $D = {b}^{2} - 4 a c$

we have equation :
${x}^{2} + 4 x + 4 = 0$

here:
$a = 1$
$b = 4$
$c = 4$
(the coefficients of ${x}^{2}$ , $x$ and the constant term respectively)

finding $D$:

$D = {b}^{2} - 4 a c = \left({4}^{2}\right) - \left(4 \times 1 \times 4\right)$
$D = 16 - 16 = 0$

formula for roots :
$x = \frac{- b \pm \sqrt{D}}{2 a}$
$x = \frac{- 4 \pm \sqrt{0}}{2 \times 1}$
$x = \frac{- 4 + 0}{2} \mathmr{and} \frac{- 4 - 0}{2}$
$x$ has two equal solutions:
$x = - 2$

the solutions are real and equal as $D = 0$