question_answer

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

Answer:

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

\[(x+2)(x+4)+2(x+1)(x+4)=4(x+1)(x+2)\]

\[{{x}^{2}}+2x+4x+8+2({{x}^{2}}+x+4x+4)=4({{x}^{2}}+x+2x+2)\]

\[{{x}^{2}}+6x+8+2{{x}^{2}}+10x+8=4{{x}^{2}}+12x+8\]

\[3{{x}^{2}}+16x+16=4{{x}^{2}}+12x+8\]

\[{{x}^{2}}-4x-8=0\]

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

\[x=\frac{4\pm \sqrt{48}}{2}=2\pm 4\sqrt{2}\]

\[\therefore x=2\pm 2\sqrt{3}\]