PEEWSRIGWETRYqx

2021-12-18

Solve the following quadratic equation by factorization.
$\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{\left(x-3\right)\left(2x+3\right)}=0$

Elaine Verrett

Expert

Given
$\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{\left(x-3\right)\left(2x+3\right)}=0$
Step by step solution
$⇒\frac{2x\left(2x+3\right)+1\left(x-3\right)+3x+4}{\left(x-3\right)\left(2x+3\right)}$
$⇒4{x}^{2}+6x+x-3+3x+9=0$
$⇒4{x}^{2}+10x+6=0$
$⇒2\left(2{x}^{2}+5x+3\right)=0$
$⇒2{x}^{2}+5x+3=0$
Complete Solution
$2{x}^{2}+5x+3=0$
$2{x}^{2}+2x+3x+3=0$
2x(x+1)+3(x+1)=0
(x+1)(2x+3)=0
$⇒x=-1$
or
2x+3=0
$x=\frac{-3}{2}$

Paul Mitchell

Expert

Step 1
Given equation is $\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{\left(x-3\right)\left(2x+3\right)}=0$
Upon simplification this can be written as:
$\frac{2x\left(2x+3\right)+1\left(x-3\right)+3x+9}{\left(x-3\right)\left(2x+3\right)}=0$
Taking denominator to the right hand side
2x(2x+3)+1(x-3)+3x+9=0
Step 2
$⇒4{x}^{2}+6x+x-3+3x+9=0$
$⇒4{x}^{2}+10x+6=0$
$⇒4{x}^{2}+4x+6x+6=0$
$⇒4x\left(x+1\right)+6\left(x+1\right)=0$
$⇒\left(4x+6\right)\left(x+1\right)=0$
$⇒4x+6=0$ or x+1=0
$⇒4x=-6$ or x=-1
$⇒x=\frac{-6}{4}$ or x=-1
$⇒x=\frac{-3}{2}$ or x=-1
Therefore x=-1.5 or -1

nick1337

Expert

Step 1
Factorization method:-

But $x\ne -\frac{3}{2}$
Step 2
Thus, x=-1 is the solution of the given equation

Do you have a similar question?