Solve the following quadratic equation by factorization:

$\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x}, x ≠0, -1, 2$

Given:

Given quadratic equation is $\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x}, x ≠0, -1, 2$.

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

$\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x}$

$\frac{2\times2(x-2)+3(x+1)}{(x+1)(x-2)}=\frac{23}{5x}$

$\frac{4x-8}{x^2-2x+x-2}=\frac{23}{5x}$

$\frac{4x-8}{x^2-x-2}=\frac{23}{5x}$

$(5x)(4x-8)=23(x^2-x-2)$   (on cross multiplication)

$20x^2-40x=23x^2-23x-46$

$(23-20)x^2-23x+40x-46=0$

$3x^2+17x-46=0$

$3x^2+23x-6x-46=0$

$3x(x-2)+23(x-2)=0$

$(3x+23)(x-2)=0$

$3x+23=0$ or $x-2=0$

$3x+23=0$ or $x-2=0$

$3x=-23$ or $x=2$

$x=\frac{-23}{3}$ or $x=2$

The values of $x$ are $-\frac{23}{3}$ and $2$. 

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Updated on: 10-Oct-2022

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