# How do you solve the equation and identify any extraneous solutions for  sqrt(x^2+5)=3-x?

Jun 11, 2015

We can square both sides of the equation. But we have to remember that the outcome of a radical is allways $\ge 0$

#### Explanation:

This means that $3 - x \ge 0 \to x \le 3$

Now for the squaring (which annulls the root at the left):
${x}^{2} + 5 = {\left(3 - x\right)}^{2} \to$
${x}^{2} + 5 = {x}^{2} - 6 x + 9 \to$
$5 = - 6 x + 9 \to 6 x = 4 \to x = \frac{2}{3}$

And this fits with the condition that $x \le 3$
(it also checks if you put the number back into the equation)