# Solve for ‘x’ From an Algebra Math Equation

Solve for ‘*x*‘ from the algebra equation 4 times square root of (3 *−* *x*) by (3 + *x*) minus square root of (3 + *x*) by (3 *−* *x*) equal to root 2

4 \sqrt{\dfrac{3-x}{3+x}} - \sqrt{\dfrac{3+x}{3-x}} = \sqrt2

x = ??

## Solution

\begin{aligned} 4 \sqrt{\dfrac{3-x}{3+x}} - \sqrt{\dfrac{3+x}{3-x}} &= \sqrt2\\ \\ 4 \sqrt{\dfrac{3-x}{3+x}} - \sqrt{\dfrac{3+x}{3-x}} &={\dfrac{4\sqrt{3-x}}{\sqrt{3+x}}} - {\dfrac{\sqrt{3+x}}{\sqrt{3-x}}} \\ \\ &={\dfrac{4({3-x})-({3+x})}{\sqrt{9-x^2}}} \\ \\ &={\dfrac{12-4x-3-x}{\sqrt{9-x^2}}} \\ \\ &={\dfrac{9-5x}{\sqrt{9-x^2}}} \\ \\ \Rightarrow {\dfrac{9-5x}{\sqrt{9-x^2}}} &= \sqrt{2}\\ \\ \Rightarrow {9-5x}&= \sqrt{18-2x^2}\\ \\ \end{aligned}

Square both sides, then

\begin{aligned} ({9-5x})^2&={18-2x^2}\\ \\ \Rightarrow 81-90x+25x^2&={18-2x^2}\\ \\ \Rightarrow 27x^2+63-90x &=0\\ \\ \Rightarrow 27x^2-90x+63 &=0\\ \\ \Rightarrow 3x^2-10x+7 &=0\\ \\ \end{aligned}

### This is a quadratic math equation so

\begin{aligned} x &=\frac{10 \pm \sqrt{10^2-4 \times 3 \times 7}}{2 \times 3}\\ \\ &=\frac{10 \pm \sqrt{100-84}}{6}\\ \\ &=\frac{10 \pm 4}{6}\\ \\ \Rightarrow x&=\frac{7}{3}, 1\\ \\ \end{aligned}

So the solution to this math problem is *x* = 7/3 & 1