# How do you solve 4(x+1)^2=100?

Aug 1, 2016

$x = 4 \text{ }$ or $\text{ } x = - 6$

#### Explanation:

The key to this problem is the fact that for a given number $a \in \mathbb{R}$, you have

${a}^{2} = a \cdot a \text{ }$ and $\text{ } {a}^{2} = \left(- a\right) \cdot \left(- a\right)$

so right from the start you know that you're looking for two values of $x$, since for $a \in \mathbb{R}$, you have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sqrt{{a}^{2}} = | a | = \left\{\begin{matrix}+ a & \text{ for "a >=0 \\ -a & " for } a < 0\end{matrix}\right.} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The first thing to do here is divide both sides of the equation by $4$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \cdot {\left(x + 1\right)}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} = \frac{100}{4}$

${\left(x + 1\right)}^{2} = 25$

If you take the square root of both sides of the equation

$\sqrt{{\left(x + 1\right)}^{2}} = \sqrt{25}$

you will end up with

$\left(x + 1\right) = + 5 \text{ }$ or $\text{ } \left(x + 1\right) = - 5$

This will get you

$x + 1 = 5 \implies x = 4 \text{ }$ or $\text{ } x + 1 = - 5 \implies x = - 6$

Therefore, the original equation has two possible solutions

$x = 4 \text{ }$ or $\text{ } x = - 6$

Do a quick check to make sure that the calculations are correct

$x = 4 : \text{ } 4 \cdot {\left(4 + 1\right)}^{2} = 100$

$4 \cdot {5}^{2} = 100 \text{ } \textcolor{g r e e n}{\sqrt{}}$

$x = - 6 : \text{ } 4 \cdot {\left(- 6 + 1\right)}^{2} = 100$

$4 \cdot {\left(- 5\right)}^{2} = 100 \text{ } \textcolor{g r e e n}{\sqrt{}}$