Question

Solve the following quadratic equations by factorization:

$\frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2};x\ne 1,-2,2$

Answer 1

$$\begin{gathered} \frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2} \\ \Rightarrow \frac{(x+1)(x+2)+(x-1)(x-2)}{(x-1)(x+2)}=\frac{4(x-2)-(2x+3)}{x-2} \\ \Rightarrow \frac{(x^2+2x+x+2)+(x^2-2x-x+2)}{x^2+2x-x-2}=\frac{4x-8-2x-3}{x-2} \\ \Rightarrow \frac{x^2+3x+2+x^2-3x+2}{x^2+x-2}=\frac{2x-11}{x-2} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{2x^2+4}{x^2+x-2}=\frac{2x-11}{x-2} \\ \Rightarrow (2x^2+4)(x-2)=(2x-11)(x^2+x-2) \\ \Rightarrow 2x^3-4x^2+4x-8=2x^3+2x^2-4x-11x^2-11x+22 \\ \Rightarrow 2x^3-4x^2+4x-8=2x^3-9x^2-15x+22 \\ \Rightarrow 2x^3-2x^3-4x^2+9x^2+4x+15x-8-22=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 5x^2+19x-30=0 \\ \Rightarrow 5x^2+25x-6x-30=0 \\ \Rightarrow 5x(x+5)-6(x+5)=0 \\ \Rightarrow (5x-6)(x+5)=0 \\ \Rightarrow 5x-6=0,x+5=0 \\ \Rightarrow x=\frac{6}{5},x=-5 \end{gathered}$$