# How do you find the discriminant and how many solutions does x^2 + 8x + 16 = 0 have?

Apr 29, 2015

Equations of the form $A {x}^{2} + B x + C = 0$

Discriminant=$D = \sqrt{{B}^{2} - 4 A C} = {8}^{2} - 4 \cdot 1 \cdot 16 = 0$
which means there is one solution
For $D > 0$ there would be two solutions,
for $D < 0$ there would be no (real) solution.

Without working out the discriminant, it's fairly easy to see that:
$\to {\left(x + 4\right)}^{2} = 0 \to x = - 4$
In the graph you see the standard ${x}^{2}$-graph moved by $- 4$
graph{x^2+8x+16 [-15.5, 4.5, -2.4, 7.6]}