# How do you solve sqrt (x+3)= x-3 and find any extraneous solutions?

Jul 22, 2016

$x = 6 \to$ valid solution
$x = 1 \to$ extraneous solution

#### Explanation:

You're dealing with a square root here, so right from the start you know two things

• $x + 3 \ge 0 \implies x \ge - 3$

When working with real numbers, i.e. $x \in \mathbb{R}$, the square root can only be taken from positive numbers. This of course implies that $x + 3$ must be positive or at best equal to zero.

• $x - 3 \ge 0 \implies x \ge 3$

Similarly, when working in $\mathbb{R}$, the square root of a number is always positive. This implies that $x - 3$ must be positive or at best equal to zero.

Combine these two restrictions to get

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x \ge 3} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ any value of $x$ that acts as solution to the original equation must satisfy this condition

Now, your goal here is to get rid of the square root. You can do that by squaring both sides of the equation

${\left(\sqrt{x + 3}\right)}^{2} = {\left(x - 3\right)}^{2}$

This will get you

$x + 3 = {x}^{2} - 6 x + 9$

${x}^{2} - 7 x + 6 = 0$

Now, two numbers that add up to give $- 7$ and multiply to give $6$ are $- 1$ and $- 6$. This means that you can factor the quadratic as

$\left(x - 1\right) \left(x - 6\right) = 0 \implies \left\{\begin{matrix}{x}_{1} = 1 \\ {x}_{2} = 6\end{matrix}\right.$

Now, notice that the first possible solution does not satisfy the required condition

$x = 1 \textcolor{red}{\cancel{\textcolor{b l a c k}{>}}} = 3 \to$ this means that $x = 1$ is an extraneous solution

As a result, the only valid solution to the original equation is $x = 6$. Do a quick check to make sure that you got the calculations right

$x = 6 : \text{ "sqrt(6+3) = 6 -3 <=> 3 = 3 " } \textcolor{g r e e n}{\sqrt{}}$

As you can see, $x = 1$ is not a solution to the original equation

$x = 1 : \text{ " sqrt(1 + 3) = 1 - 3 <=> 2 != -2" } \textcolor{red}{\times}$