Solve the following quadratic equation by factorization:

$\frac{4}{x}-3=\frac{5}{2x+3}, x ≠0,\frac{-3}{2}$

Given:

Given quadratic equation is $\frac{4}{x}-3=\frac{5}{2x+3}, x ≠0,\frac{-3}{2}$.

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

$\frac{4}{x}-3=\frac{5}{2x+3}$

$\frac{4-3(x)}{x}=\frac{5}{2x+3}$

$\frac{4-3x}{x}=\frac{5}{2x+3}$

$(2x+3)(4-3x)=5(x)$   (on cross multiplication)

$8x-6x^2+12-9x=5x$

$-6x^2-x+12=5x$

$6x^2+5x+x-12=0$

$6x^2+6x-12=0$

$6(x^2+x-2=0)$

$x^2+x-2=0$

$x^2-2x+x-2=0$

$x(x-2)+1(x-2)=0$

$(x-2)(x+1)=0$

$x-2=0$ or $x+1=0$

$x-2=0$ or $x+1=0$

$x=2$ or $x=-1$

The values of $x$ are $-1$ and $2$. 

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Updated on: 10-Oct-2022

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