# Question #e1353

Nov 19, 2016

$x = \setminus \pm \setminus \sqrt{14}$

## Equation given

${x}^{2} - 2 = 12$

## Method 1

Isolate the ${x}^{2}$ and solve by square root.
${x}^{2} \setminus \cancel{- 2} \setminus \cancel{\setminus \textcolor{b l u e}{+ 2}} = 12 \setminus \textcolor{b l u e}{+ 2}$
${x}^{2} = 14$
$x = \setminus \pm \setminus \sqrt{14}$

## Method 2

Form a quadratic equation by making one side equal to 0, then solve by factoring said quadratic.
${x}^{2} - 2 \setminus \textcolor{red}{- 12} = \setminus \cancel{12} \setminus \cancel{\setminus \textcolor{red}{- 12}}$
${x}^{2} - 14 = 0$ $\Leftarrow$ the factor of variable $x$ is 0x, so...
$\setminus \Rightarrow {x}^{2} + 0 x - 14 = 0$ apply quadratic formula

$x = \frac{- 0 \setminus \pm \setminus \sqrt{{0}^{2} - 4 \left(1\right) \left(- 14\right)}}{2 \left(1\right)} = \frac{\setminus \pm \setminus \sqrt{56}}{2} = \frac{\setminus \pm 2 \setminus \sqrt{14}}{2}$
$\setminus \Rightarrow \frac{\setminus \pm \setminus \cancel{2} \setminus \sqrt{14}}{\setminus} \cancel{2} = \setminus \pm \setminus \sqrt{14}$