Question

Solve this equation $\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4}$, where $x+1\ne ,x+2\ne =0$ and $x+4\ne 0$ using quadratic formula.

Answer 1

Note that the given equation is not in the standard form of a quadratic equation.

Consider $\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4}$

That is, $\frac{1}{x+1}=2[\frac{2}{x+4}-\frac{1}{x+2}]=2\left[\frac{2x+4-x-4}{(x+4)(x+2)}\right]$

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

$x^2+6x+8=2x^2+2x$

Thus, we have $x^2-4x-8=0$, which is a quadratic equation.

(The above equation can also be obtained by taking LCM )

Using the quadratic formula we get,

$x=\frac{4\pm \sqrt{16-4\left(1\right)(-8)}}{2\left(1\right)}=\frac{4\pm \sqrt{48}}{2}$

Thus, $x=2+2\sqrt{3}$ or $2-2\sqrt{3}$

Hence, the solution set is $\{2-2\sqrt{3},2+2\sqrt{3}\}$