# How do you solve the equation and identify any extraneous solutions for sqrt(x+7) = x + 1?

Jun 11, 2015

The term under the root must be $\ge 0$ so $x \ge - 7$
The outcome must also be $> 0$ so $x \ge - 1$

#### Explanation:

Now we can square both sides:
$x + 7 = {\left(x + 1\right)}^{2} = {x}^{2} + 2 x + 1 \to$
${x}^{2} + 2 x - x + 1 - 7 = {x}^{2} + x - 6 = 0$
$\left(x + 3\right) \left(x - 2\right) = 0 \to x = - 3 \mathmr{and} x = 2$

The only allowed solution is $x = 2$ as the other one does not meet the conditions at the start.