# How do you solve and check for extraneous solutions in x^2 + 2 = x + 4?

Aug 7, 2015

Express the given equation in standard form, factor for solution values, and check each resulting value to ensure that it is not extraneous.
$x = 2$ or $x = - 1$

#### Explanation:

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

Subtract $\left(x + 4\right)$ from both sides to get into standard form
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - x - 2 = 0$

Factor
$\textcolor{w h i t e}{\text{XXXX}}$$\left(x - 2\right) \left(x + 1\right) = 0$

Either $\left(x - 2\right) = 0$ or $\left(x + 1\right) = 0$
$\Rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$x = 2$ or $x = - 1$

Checking $x = 2$ by substituting into ${x}^{2} + 2$ and $x + 4$
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + 2 = {\left(2\right)}^{2} + 2 = 6$
$\textcolor{w h i t e}{\text{XXXX}}$$x + 4 = \left(2\right) + 4 = 6$
${x}^{2} + 2 = x + 4$ so $x = 2$ is not extraneous.

Similarly, checking $x = - 1$ by substituting into ${x}^{2} + 2$ and $x + 4$
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + 2 = {\left(- 1\right)}^{2} + 2 = 3$
$\textcolor{w h i t e}{\text{XXXX}}$$x + 4 = \left(- 1\right) + 4 = 3$
${x}^{2} + 2 = x + 4$ so $x = - 1$ is not extraneous.