# How do you solve 3 - x =sqrt(x^2 + 15)?

Aug 11, 2015

$x = - 1$

#### Explanation:

Start by taking a look at the expression that's under the square root.

${x}^{2} + 15 > 0 , \left(\forall\right) x \in \mathbb{R}$

In other words, as far as the radical term is concerned, $x$ can take any value in $\mathbb{R}$. On the other hand, the possible values of $x$ are restricted by the expression on the left side of the equation.

Since the square root of a positive number is always a positive number, you have

• $3 - x \ge 0 \implies x \le 3$

Now, square both sides of the equation to get rid of the radical term

${\left(3 - x\right)}^{2} = {\left(\sqrt{{x}^{2} + 15}\right)}^{2}$

$9 - 6 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} + 15$

This is equivalent to

$- 6 x = 6 \implies x = \frac{6}{- 6} = \textcolor{g r e e n}{- 1}$

Since $x = - 1$ satisfies the condition $x \le 3$, this will be the solution to the original equation.

Do a quick check to make sure that everything came out right

$3 - \left(- 1\right) = \sqrt{{\left(- 1\right)}^{2} + 15}$

$3 + 1 = \sqrt{16}$

$4 = 4 \textcolor{g r e e n}{\sqrt{}}$