1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Solve the following quadratic equations by factorization: $\frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2};x\ne 1,-2,2\phantom{\rule{0ex}{0ex}}$

Open in App
Solution

## $\frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2}\phantom{\rule{0ex}{0ex}}⇒\frac{\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x+2\right)}=\frac{4\left(x-2\right)-\left(2x+3\right)}{x-2}\phantom{\rule{0ex}{0ex}}⇒\frac{\left({x}^{2}+2x+x+2\right)+\left({x}^{2}-2x-x+2\right)}{{x}^{2}+2x-x-2}=\frac{4x-8-2x-3}{x-2}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}+3x+2+{x}^{2}-3x+2}{{x}^{2}+x-2}=\frac{2x-11}{x-2}$ $⇒\frac{2{x}^{2}+4}{{x}^{2}+x-2}=\frac{2x-11}{x-2}\phantom{\rule{0ex}{0ex}}⇒\left(2{x}^{2}+4\right)\left(x-2\right)=\left(2x-11\right)\left({x}^{2}+x-2\right)\phantom{\rule{0ex}{0ex}}⇒2{x}^{3}-4{x}^{2}+4x-8=2{x}^{3}+2{x}^{2}-4x-11{x}^{2}-11x+22\phantom{\rule{0ex}{0ex}}⇒2{x}^{3}-4{x}^{2}+4x-8=2{x}^{3}-9{x}^{2}-15x+22\phantom{\rule{0ex}{0ex}}⇒2{x}^{3}-2{x}^{3}-4{x}^{2}+9{x}^{2}+4x+15x-8-22=0$ $⇒5{x}^{2}+19x-30=0\phantom{\rule{0ex}{0ex}}⇒5{x}^{2}+25x-6x-30=0\phantom{\rule{0ex}{0ex}}⇒5x\left(x+5\right)-6\left(x+5\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(5x-6\right)\left(x+5\right)=0\phantom{\rule{0ex}{0ex}}⇒5x-6=0,x+5=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{6}{5},x=-5$

Suggest Corrections
59
Join BYJU'S Learning Program
Related Videos
Quadratic Equations
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program