# How do you solve x^2 + 2 = x + 4 and find any extraneous solutions?

May 20, 2017

$x = 2 , - 1$

#### Explanation:

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

${x}^{2} + 2 - x - 4 = 0$
Rearrange Equation

${x}^{2} + 2 - x - 4 = 0$
Collect like terms so $2 - 4 = - 2$

${x}^{2} - 2 - x = 0$
Plug values into the Quadratic Formula

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \times 1 \times \left(- 2\right)}}{2 \times 1}$
Then solve
$x = \frac{1 \pm \sqrt{1 + 8}}{2}$

$x = \frac{1 \pm \sqrt{9}}{2}$

$x = \frac{1 \pm 3}{2}$

$x = \frac{1 + 3}{2}$

$x = \frac{4}{2}$

$x = \frac{1 - 3}{2}$

$x = \frac{- 2}{2}$

$x = 2 , - 1$