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

Aug 19, 2015

$x = 3$

#### Explanation:

Since you're dealing with a radical term, recognize that you need to have

$x + 1 \ge 0 \text{ }$ and $\text{ } x - 1 \ge 0$

This is the case because, for real numbers, you can only take the square root of a positive number; moreover, the result of this operation will always be a positive number.

If $x - 1 \ge 0$, then you automatically have $x + 1 \ge 0$, so you can combine these two conditions

$x - 1 \ge 0 \implies x \ge 1$

Square both sides of the equation to get rid of the radical term

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

$x + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} = {x}^{2} - 2 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}}$

This is equivalent to

${x}^{2} - 3 x = 0$

$x \left(x - 3\right) = 0$

This equation can be euqal to zero if either $x = 0$ or $x - 3 = 0$, which would imply

$x - 3 = 0 \implies x = 3$

Since $x = 0$ does not satisfy the initial contidion $x \ge 1$, it will be an extraneous solution. The only valid solution to your original equation will thus be $x = \textcolor{g r e e n}{3}$.

Do a quick check to make sure that you got the calculations right

$\sqrt{3 + 1} = 3 - 1$

$\sqrt{4} = 2$

$2 = 2 \text{ } \textcolor{g r e e n}{\sqrt{}}$