Question

Find the roots of the equation $\begin{array}{c}\frac{1}{x+1}+\frac{2}{x+2}\\ =\frac{4}{x+4}\end{array}$

Answer 1

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

$(x+2)(x+4)+2(x+1)(x+4)=4(x+1)(x+2)$

$x^2+6x+8+2(x^2+5x+4)=4(x^2+3x+2)$

$x^2+6x+8+2x^2+10x+8=4x^2+12x+8$

$3x^2+16x+16=4x^2+12x+8$

$4x+8=x^2$

$-x^2+4x+8=0$

$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-4+\sqrt{4^2-4(-1)8}}{2(-1)}:-2(\sqrt{3}-1)$

$x=\frac{-4-\sqrt{4^2-4(-1)8}}{2(-1)}:2(1+\sqrt{3})$

$x=-2(\sqrt{3}-1),x=2(1+\sqrt{3})$