Question

Solve for x: $\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4},x\ne -1,-2,-4$

Answer 1

We have $\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4},x\ne -1,-2,-4$

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

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

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

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

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

$\therefore x=\frac{4\pm \sqrt{16+32}}{2}$

$x=\frac{4\pm \sqrt{48}}{2}=2\pm 4\sqrt{2}$

$\therefore x=2\pm 2\sqrt{3}$