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
