# How do you solve 4n ^ { 2} + 4n + 1= 0 using the quadratic formula?

Apr 11, 2017

$n = - \frac{1}{2}$

#### Explanation:

The standard form of a $\textcolor{b l u e}{\text{quadratic equation}}$ is.

• ax^2+bx+c=0;a!=0

To find the roots using the $\textcolor{b l u e}{\text{quadratic formula}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{here " a=4,b=4" and } c = 1$

$\Rightarrow n = \frac{- 4 \pm \sqrt{{4}^{2} - \left(4 \times 4 \times 1\right)}}{2 \times 4}$

$\textcolor{w h i t e}{\Rightarrow n} = \frac{- 4 \pm \sqrt{0}}{8} = \frac{- 4}{8} = - \frac{1}{2} \leftarrow \textcolor{red}{\text{ repeated root}}$

Normally a quadratic equation has 2 solutions ( roots) but in this case has only 1. This informs us that the quadratic touches the x-axis at only 1 point
graph{4x^2+4x+1 [-10, 10, -5, 5]}