# How do you solve and check for extraneous solutions in sqrt(x+1) + 5 = x?

Aug 1, 2015

The only valid solution is $x = 8$
A second candidate solution $x = 3$ can be eliminated by checking for the validity of the given equation with $x = 3$ (and noting that it is not valid).

#### Explanation:

Given $\sqrt{x + 1} + 5 = x$

Subtract $5$ from both sides
$\textcolor{w h i t e}{\text{XXXX}}$$\sqrt{x + 1} = x - 5$

Square both sides (possibly generating an extraneous root at this point)
$\textcolor{w h i t e}{\text{XXXX}}$$x + 1 = {x}^{2} - 10 x + 25$

Subtract $\left(x + 1\right)$ from both sides (and flip sides)
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 11 x + 24 = 0$

Factor
$\textcolor{w h i t e}{\text{XXXX}}$$\left(x - 3\right) \left(x - 8\right) = 0$

$\Rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$x = 3$$\textcolor{w h i t e}{\text{XXXX}}$or$\textcolor{w h i t e}{\text{XXXX}}$$x = 8$

Substituting $3$ for $x$ in the Left Side of original equation
$\textcolor{w h i t e}{\text{XXXX}}$$\sqrt{3 + 1} + 5 = 2 + 3$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\ne 3$
The solution $x = 3$ is extraneous.

Substituting $8$ for $x$ in the Left Side of original equation
$\textcolor{w h i t e}{\text{XXXX}}$$\sqrt{8 + 1} + 5 = 3 + 5$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= 8$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= x$
The solution $x = 8$ is valid.