Solve the following quadratic equation by factorization:

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

Given:

Given quadratic equation is $\frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2}, x ≠ 1, -2, 2$.

To do:

We have to solve the given quadratic equation.

Solution:

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

$\frac{(x+1)(x+2)+(x-1)(x-2)}{(x-1)(x+2)}=\frac{4(x-2)-(2x+3)}{x-2}$

$\frac{x^2+x+2x+2+x^2-x-2x+2}{x^2+2x-x-2}=\frac{4x-8-2x-3}{x-2}$

$\frac{2x^2+4}{x^2+x-2}=\frac{2x-11}{x-2}$

$(x-2)(2x^2+4)=(2x-11)(x^2+x-2)$

$2x^3+4x-4x^2-8=2x^3+2x^2-4x-11x^2-11x+22$

$4x-4x^2-8=-9x^2-15x+22$

$(9-4)x^2+(15+4)x-8-22=0$

$5x^2+19x-30=0$

$5x^2+25x-6x-30=0$

$5x(x+5)-6(x+5)=0$

$(5x-6)(x+5)=0$

$5x-6=0$ or $x+5=0$

$5x=6$ or $x=-5$

$x=\frac{6}{5}$ or $x=-5$

The values of $x$ are $\frac{6}{5}$ and $-5$.

Tutorialspoint

e

Updated on: 10-Oct-2022

51 Views

Kickstart Your Career

Get certified by completing the course

Get Started