# Solve the Quadratic Equation by Substitution

Solve the following quadratic equation by substitution

\begin{aligned} 4\sqrt{\frac{3-x}{3+x}}-&\sqrt{\frac{3+x}{3-x}}= \sqrt{2}\\ \end{aligned}

x=??

## Solving the quadratic equation by substitution

let

u=\sqrt{\frac{3-x}{3+x}}

then

\frac{1}{u}=\sqrt{\frac{3+x}{3-x}}

Now we can right

\begin{aligned} 4\sqrt{\frac{3-x}{3+x}}-&\sqrt{\frac{3+x}{3-x}}=4u-\frac{1}{u}=\sqrt{2}\\ \end{aligned}

So

\begin{aligned} &4u^2-1=\sqrt{2}u \\ \\ \Rightarrow&4u^2-\sqrt{2}u-1=0 \\ \end{aligned}

Which is a quadratic equation in u, the solutions of this equation are found from the quadratic formula

\begin{aligned} &u=\frac{\sqrt2 \pm \sqrt{2-4 \times 4\times (-1)}}{2\times4} \\ \\ \Rightarrow \ &u=\frac{\sqrt2 \pm \sqrt{2+16}}{8} \\ \\ \Rightarrow \ &u=\frac{\sqrt2 \pm \sqrt{18}}{8} \\ \\ \Rightarrow \ &u=\frac{\sqrt2 \pm 3\sqrt{2}}{8} \\ \\ \Rightarrow \ &u=\frac{1}{\sqrt2} \ \ \ or \ \ u=-\frac{1}{2\sqrt2} \\ \\ \end{aligned}

We know that

u=\sqrt{\frac{3-x}{3+x}}

from the substitution of u, we can clearly see that u can be imaginary of a positive real number but not a negative real number so,

u=\frac{1}{\sqrt2}

Then

\sqrt{\frac{3-x}{3+x}} = \frac{1}{\sqrt2}

Square both sides

\begin{aligned} &{\frac{3-x}{3+x}} = \frac{1}{2}\\ \\ \Rightarrow \ &3+x =6-2x\\ \\ \Rightarrow \ &3x =3\\ \\ \Rightarrow \ &x =1\\ \\ \end{aligned}

That is the solution to the math equation is x = 1