# How do you solve sqrt(4 - x) + sqrt(6 + x) = sqrt[6(x+1)?

Aug 10, 2015

$x = 2$

#### Explanation:

Since you're dealing with raedical terms, start by writing the conditions that a possible solution must satisfy.

• $4 - x \ge 0 \implies x \le 4$
• $6 + x \ge 0 \implies x \ge - 6$
• $x + 1 \ge 0 \implies x \ge - 1$

Since you have two positive terms on the left side of the equation, the right side cannot be equal to zero, since that would imply that one of the two square roots on the left side would produce a negative number.

The last condition will thus be $x + 1 > 0 \implies x > - 1$. If you combine these conditions, you'll get that $x \in \left(- 1 , 4\right]$.

Next, square both sides of the equation. This will reduce the number of radical terms from three to one.

${\left(\sqrt{4 - x} + \sqrt{6 + x}\right)}^{2} = {\left(\sqrt{6 \left(x + 1\right)}\right)}^{2}$

${\left(\sqrt{4 - x}\right)}^{2} + 2 \sqrt{\left(4 - x\right) \left(6 + x\right)} + {\left(\sqrt{6 + x}\right)}^{2} = 6 \left(x + 1\right)$

$4 - \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 2 \sqrt{\left(4 - x\right) \left(6 + x\right)} + \cancel{6} + \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} = 6 x + \cancel{6}$

$4 + 2 \sqrt{\left(4 - x\right) \left(6 + x\right)} = 6 x$

Next, isolate the remaining radical on one side of the equation and square both sides again

$\cancel{4} - \cancel{4} + 2 \sqrt{\left(4 - x\right) \left(6 + x\right)} = 6 x - 4$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \sqrt{\left(4 - x\right) \left(6 + x\right)}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} = \frac{6 x - 4}{2}$

$\sqrt{\left(4 - x\right) \left(6 + x\right)} = 3 x - 2$

${\left(\sqrt{\left(4 - x\right) \left(6 + x\right)}\right)}^{2} = {\left(3 x - 2\right)}^{2}$

$\left(4 - x\right) \left(6 + x\right) = 9 {x}^{2} - 12 x + 4$

$24 + 4 x - 6 x - {x}^{2} = 9 {x}^{2} - 12 x + 4$

Rearrange to get

$10 {x}^{2} - 10 x - 20 = 0$

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

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

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

${x}_{1 , 2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2} = \left\{\begin{matrix}{x}_{1} = \frac{1 + 3}{2} = 2 \\ {x}_{2} = \frac{1 - 3}{2} = - 1\end{matrix}\right.$

Notice that ${x}_{2} = - 1$ does not satisfy the condition $x \in \left(- 1 , 4\right]$, which means that ${x}_{2}$ is an extraneous solution.

The original equation will thus have one valid solution, $x = \textcolor{g r e e n}{2}$.

Do a quick check to make sure that the calculations are correct

$\sqrt{4 - 2} + \sqrt{6 + 2} = \sqrt{6 \cdot 3}$

$\sqrt{2} + 2 \sqrt{2} = \sqrt{18}$

$3 \sqrt{2} = 3 \sqrt{2} \textcolor{g r e e n}{\sqrt{}}$