Extraneous Solutions

Example 1:

Solve for $x$ , $\frac{1}{x-2}+\frac{1}{x+2}=\frac{4}{\left(x-2\right)\left(x+2\right)}$ .

$\frac{1}{x-2}+\frac{1}{x+2}=\frac{4}{\left(x-2\right)\left(x+2\right)}$

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

$\left(x-2\right)+\left(x+2\right)=4$

$2x=4$

$x=2$

But $2$ is excluded from the domain of the original equation because it would make the denominator of one of the fractions zero--and division by zero is not allowed!  .  Therefore, it cannot be a root of the original equation.  So, $2$ is an extraneous solution. So, the equation has no solutions.

Example 2:

Solve for $x$ , $\sqrt{x+4}=x-2$

$\sqrt{x+4}=x-2$

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

$x+4=x^2-4x+4$

$0=x^2-5x$

$0=x\left(x-5\right)$

$x=0\text{ or }x=5$

Check your solutions in the original equation.

Let $x=5$ .

$\sqrt{5+4}\stackrel{?}{=}5-2$

$3=3$

So, $5$ is a solution.

Let $x=0$ .

$\sqrt{0+4}\stackrel{?}{=}0-2$

$2\ne -2$

So, $0$ is an extraneous solution.