Question

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

Answer 1

Given: expression $\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4}$

To solve the given expression

Sol: $\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4}⟹\frac{(x+2)+2(x+1)}{(x+1)(x+2)}=\frac{4}{x+4}⟹\frac{3x+4}{x^2+2x+x+2}=\frac{4}{x+4}⟹(3x+4)(x+4)=4(x^2+3x+2)⟹3x^2+12x+4x+16=4x^2+12x+8⟹4x^2-3x^2+12x-16x+8-16=0⟹x^2-4x-8=0$

This is form $a{x}^2+bx+c=0$, hence $a=1,b=-4,c=-8$

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

i.e., $x=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(1\left)\right(-8)}}{2\left(1\right)}⟹x=\frac{4\pm \sqrt{16+32}}{2}⟹x=\frac{4\pm 6.93}{2}⟹x_1=\frac{4+6.93}{2},x_2=\frac{4-6.93}{2}⟹x_1=5.5,x_2=-1.5$