PEEWSRIGWETRYqx

Answered

2021-12-18

Solve the following quadratic equation by factorization.

$\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0$

Answer & Explanation

Elaine Verrett

Expert

2021-12-19Added 41 answers

Given

$\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0$

Step by step solution

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

$\Rightarrow 4{x}^{2}+6x+x-3+3x+9=0$

$\Rightarrow 4{x}^{2}+10x+6=0$

$\Rightarrow 2(2{x}^{2}+5x+3)=0$

$\Rightarrow 2{x}^{2}+5x+3=0$

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

$\Rightarrow x=-1$

or

2x+3=0

$x=\frac{-3}{2}$

Step by step solution

Complete Solution

2x(x+1)+3(x+1)=0

(x+1)(2x+3)=0

or

2x+3=0

Paul Mitchell

Expert

2021-12-20Added 40 answers

Step 1

Given equation is$\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0$

Upon simplification this can be written as:

$\frac{2x(2x+3)+1(x-3)+3x+9}{(x-3)(2x+3)}=0$

Taking denominator to the right hand side

2x(2x+3)+1(x-3)+3x+9=0

Step 2

$\Rightarrow 4{x}^{2}+6x+x-3+3x+9=0$

$\Rightarrow 4{x}^{2}+10x+6=0$

$\Rightarrow 4{x}^{2}+4x+6x+6=0$

$\Rightarrow 4x(x+1)+6(x+1)=0$

$\Rightarrow (4x+6)(x+1)=0$

$\Rightarrow 4x+6=0$ or x+1=0

$\Rightarrow 4x=-6$ or x=-1

$\Rightarrow x=\frac{-6}{4}$ or x=-1

$\Rightarrow x=\frac{-3}{2}$ or x=-1

Therefore x=-1.5 or -1

Given equation is

Upon simplification this can be written as:

Taking denominator to the right hand side

2x(2x+3)+1(x-3)+3x+9=0

Step 2

Therefore x=-1.5 or -1

nick1337

Expert

2021-12-28Added 573 answers

Step 1

Factorization method:-

But

Step 2

Thus, x=-1 is the solution of the given equation

Most Popular Questions