# What is x if -4(x+2)^2=-20?

Jul 21, 2018

$\text{ }$
Approximate values of $\textcolor{red}{x}$ are:

color(blue)(x~~0.236

color(blue)(x~~-4.236

#### Explanation:

$\text{ }$
How do we find the value of color(red)(x using

color(red)([-4(x+2)^2]=(-20)

$\Rightarrow \left(- 4\right) \left({x}^{2} + 4 x + 4\right) = \left(- 20\right)$

Divide both sides of the equation by color(red)((-4)

rArr [(-4)(x^2+4x+4)]/((-4))=((-20))/((-4)

rArr [cancel (-4)(x^2+4x+4)]/(cancel ((-4)))=((-20))/((-4)

$\Rightarrow {x}^{2} + 4 x + 4 = 5$

Subtract $\textcolor{red}{5}$ from both sides of the equation.

$\Rightarrow {x}^{2} + 4 x + 4 - 5 = 5 - 5$

$\Rightarrow {x}^{2} + 4 x + 4 - 5 = \cancel{5} - \cancel{5}$

$\Rightarrow {x}^{2} + 4 x - 1 = 0$ ...Eqn.1

We now have a quadratic equation.

Use the quadratic formula to find the values of color(red)(x

color(blue)(x=[-b+-sqrt(b^2-4ac)]/(2a)

Using Eqn.1, we get

a=1; b=4 and c= (-1)

Substitute the values in the quadratic formula above:

color(blue)(x=[-4+-sqrt(4^2-4*1*(-1))]/(2*1)

$x = \frac{- 4 \pm \sqrt{16 + 4}}{2}$

$x = \frac{- 4 \pm \sqrt{20}}{2}$

$x = \frac{- 4 \pm \sqrt{4 \cdot 5}}{2}$

$x = \frac{- 4 \pm \sqrt{4} \cdot \sqrt{5}}{2}$

$x = \frac{- 4 \pm 2 \cdot \sqrt{5}}{2}$

$x = \left(- \frac{4}{2}\right) \pm \frac{2 \cdot \sqrt{5}}{2}$

$x = \left(- 2\right) \pm \frac{\cancel{2} \cdot \sqrt{5}}{\cancel{2}}$

$x = - 2 \pm \sqrt{5}$

$x = - 2 - \sqrt{5}$, $x = - 2 + \sqrt{5}$

Using a spreadsheet software or a calculator, we get

Hence, approximate values of $\textcolor{red}{x}$ are:

color(blue)(x~~0.236

color(blue)(x~~-4.236

Hope it helps.

Jul 21, 2018

$x = - 2 + \sqrt{5}$ and $- 2 - \sqrt{5}$

#### Explanation:

Let's start by dividing both sides by $- 4$. We now have

${\left(x + 2\right)}^{2} = 5$

Let's take the square root of both sides to get

$x + 2 = \sqrt{5}$ and $x + 2 = - \sqrt{5}$

Subtracting $2$ from both sides of both equations, we get

$x = - 2 + \sqrt{5}$ and $- 2 - \sqrt{5}$

Hope this helps!

Jul 21, 2018

$x = - 2 \pm \sqrt{5}$

#### Explanation:

$- 4 {\left(x + 2\right)}^{2} = - 20$

First, divide both sides by $\textcolor{b l u e}{- 4}$:
$\frac{- 4 {\left(x + 2\right)}^{2}}{\textcolor{b l u e}{- 4}} = \frac{- 20}{\textcolor{b l u e}{- 4}}$

${\left(x + 2\right)}^{2} = 5$

Expand/simplify the left hand side:
${x}^{2} + 4 x + 4 = 5$

Subtract $\textcolor{b l u e}{5}$ from both sides:
${x}^{2} + 4 x + 4 \quad \textcolor{b l u e}{- \quad 5} = 5 \quad \textcolor{b l u e}{- \quad 5}$

${x}^{2} + 4 x - 1 = 0$

This is now in standard form, $a {x}^{2} + b x + c$.

Use the quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ to find the value of $x$:
$x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(1\right) \left(- 1\right)}}{2 \left(1\right)}$

$x = \frac{- 4 \pm \sqrt{16 + 4}}{2}$

$x = \frac{- 4 \pm \sqrt{20}}{2}$

$x = \frac{- 4 \pm 2 \sqrt{5}}{2}$

$x = - 2 \pm \sqrt{5}$

Hope this helps!