# How do you solve and check for extraneous solutions in sqrt(x+7) = x + 1?

Aug 10, 2015

$x = 2$

#### Explanation:

Any solution that you will find must satisfy two conditions

• $x + 7 \ge 0 \implies x \ge - 7$
• $x + 1 \ge 0 \implies x \ge - 1$

Overall, the solution(s) to this equation must satisfy the condition $x \ge - 1$.

Start by squaring both sides of the equation to get rid of the radical term

${\left(\sqrt{x + 7}\right)}^{2} = {\left(x + 1\right)}^{2}$

$x + 7 = {x}^{2} + 2 x + 1$

Rearrange this equation into classic quadratic form

${x}^{2} + x - 6 = 0$

Use the quadratic formula to find the two solutions to this equation

${x}_{1 , 2} = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 1 \cdot \left(- 6\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{- 1 \pm \sqrt{25}}{2}$

${x}_{1 , 2} = \frac{- 1 \pm 5}{2} = \left\{\begin{matrix}{x}_{1} = \frac{- 1 - 5}{2} = - 3 \\ {x}_{2} = \frac{- 1 + 5}{2} = 2\end{matrix}\right.$

Only one of these two solutions, $x = 2$, satisfies the condition $x \ge - 1$, which means that $x = - 3$ will be an extraneous solution.

Therefore, the solution to this equation is $x = \textcolor{g r e e n}{2}$.

Check to see if this is the case

$\sqrt{2 + 7} = 2 + 1$

$\sqrt{9} = 3$

$3 = 3$ $\textcolor{g r e e n}{\sqrt{}}$