# How do you solve x = 3 +sqrt(x^2 - 9)?

Aug 5, 2015

Isolate the radical; square; use normal solution techniques; then check for extraneous results.
Giving: $x = 3$

#### Explanation:

Given: $x = 3 + \sqrt{{x}^{2} - 9}$

Isolate radical by subtracting 3 from both sides:
$\textcolor{w h i t e}{\text{XXXX}}$$x - 3 = \sqrt{{x}^{2} - 9}$
Square both sides
$\textcolor{w h i t e}{\text{XXXX}}$${\left(x - 3\right)}^{2} = x - 9$
Factor (both sides)
$\textcolor{w h i t e}{\text{XXXX}}$$\left(x - 3\right) \left(x - 3\right) = \left(x + 3\right) \left(x - 3\right)$
Note potential solution $\left(x - 3 = 0 \rightarrow x = 3\right)$
Divide both sides by $\left(x - 3\right)$ assuming $x \ne 3$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} - 3 = \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 3$

$\textcolor{w h i t e}{\text{XXXX}}$$- 3 = + 3$

So the only possible solution we have found is $x = 3$

Checking to ensure that this is not an extraneous root introduced by squaring; replace $x$ with $3$ in the original equation:
$\textcolor{w h i t e}{\text{XXXX}}$(3) ?=? 3 +sqrt((3)^2-9)?
$\textcolor{w h i t e}{\text{XXXX}}$Yes; equality holds
$\textcolor{w h i t e}{\text{XXXX}}$$x = 3$ is a valid solution