How do you solve ｜x-2｜ =｜x^2-4｜?

Jun 10, 2016

$x = - 1$, $x = - 3$ or $x = 2$

Explanation:

Given:

$\left\mid x - 2 \right\mid = \left\mid {x}^{2} - 4 \right\mid$

We must have one of the following:

a) $\textcolor{w h i t e}{0} \left(x - 2\right) = \left({x}^{2} - 4\right)$

b) $\textcolor{w h i t e}{0} \left(x - 2\right) = - \left({x}^{2} - 4\right)$

$\textcolor{w h i t e}{}$
Case a)

$x - 2 = {x}^{2} - 4$

Subtract $\left(x - 2\right)$ from both sides to get:

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

So $x = - 1$ or $x = 2$

$\textcolor{w h i t e}{}$
Case b)

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

Add $\left({x}^{2} - 4\right)$ to both sides and transpose to get:

$0 = {x}^{2} + x - 6 = \left(x + 3\right) \left(x - 2\right)$

So $x = - 3$ or $x = 2$

$\textcolor{w h i t e}{}$
Check potential solutions

Trying each of these values of $x$ as possible solutions of the original equation:

$\left\mid \left(- 1\right) - 2 \right\mid = \left\mid - 3 \right\mid = \left\mid {\left(- 1\right)}^{2} - 4 \right\mid$

$\left\mid \left(2\right) - 2 \right\mid = 0 = \left\mid {\left(2\right)}^{2} - 4 \right\mid$

$\left\mid \left(- 3\right) - 2 \right\mid = 5 = \left\mid {\left(- 3\right)}^{2} - 4 \right\mid$

So all the possible solutions are solutions of the original equation.