# How do you solve 4x^2 + 8x + 4 = 0?

Mar 14, 2018

The solution is $x = - 1$.

#### Explanation:

First, divide both sides by $4$. Then, factor. Lastly, square root both sides and solve for $x$. Here's what that looks like:

$4 {x}^{2} + 8 x + 4 = 0$

$\frac{4 {x}^{2} + 8 x + 4}{4} = \frac{0}{4}$

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

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

$\textcolor{red}{x} \cdot x + \textcolor{red}{x} \cdot 1 + \textcolor{b l u e}{1} \cdot x + \textcolor{b l u e}{1} \cdot 1 = 0$

$\textcolor{red}{x} \left(x + 1\right) + \textcolor{b l u e}{1} \left(x + 1\right) = 0$

$\left(\textcolor{red}{x} + \textcolor{b l u e}{1}\right) \left(x + 1\right) = 0$

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

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

$x + 1 = 0$

$x = - 1$

This is the answer. Hope this helped!