# How do you solve 10 - x =sqrt(3x + 24)?

Aug 10, 2015

$x = 4$

#### Explanation:

Right from the start, you know that any solution you find for this equation must satisfy two conditions

• $3 x + 24 \ge 0 \implies x \ge - 8$
• $10 - x \ge 0 \implies x \le 10$

This means that you must for an $x$ that satisfies the overall condition $x \in \left[- 8 , 10\right]$.

Since the radical is already isolated on one side of the equation, square both sides to get rid of the square root

${\left(\sqrt{3 x + 24}\right)}^{2} = {\left(10 - x\right)}^{2}$

$3 x + 24 = 100 - 20 x + {x}^{2}$

Rearrange this equation into classic quadratic form

${x}^{2} - 23 x + 76 = 0$

Use the quadratic formula to find the two solutions to this equation

${x}_{1 , 2} = \frac{- \left(- 23\right) \pm \sqrt{{\left(- 23\right)}^{2} - 4 \cdot 1 \cdot 76}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{23 \pm \sqrt{225}}{2}$

${x}_{1 , 2} = \frac{23 \pm 15}{2} = \left\{\begin{matrix}{x}_{1} = \frac{23 + 15}{2} = 19 \\ {x}_{2} = \frac{23 - 15}{2} = 4\end{matrix}\right.$

Now, since $x = 19$ $\cancel{\in} \left[- 8 , 10\right]$, this is not a valid solution solution.

As a result, the only solution to this equation is $x = \textcolor{g r e e n}{4}$.

Check to see if this is the case

$10 - 4 = \sqrt{3 \cdot \left(4\right) + 24}$

$6 = \sqrt{36}$

$6 = 6$ $\textcolor{g r e e n}{\sqrt{}}$