Solve the following quadratic equation by factorization:

$\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1}, x ≠1, -1, \frac{1}{4}$

Given:

Given quadratic equation is $\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1}, x ≠1, -1, \frac{1}{4}$

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

$\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1}$

$\frac{3(x-1)+4(x+1)}{(x+1)(x-1)}=\frac{29}{4x-1}$

$\frac{3x-3+4x+4}{x^2-1^2}=\frac{29}{4x-1}$

$\frac{7x+1}{x^2-1}=\frac{29}{4x-1}$

$(7x+1)(4x-1)=29(x^2-1)$   (on cross multiplication)

$28x^2-7x+4x-1=29x^2-29$

$(29-28)x^2+7x-4x-29+1=0$

$x^2+3x-28=0$

$x^2+7x-4x-28=0$

$x(x+7)-4(x+7)=0$

$(x+7)(x-4)=0$

$x+7=0$ or $x-4=0$

$x+7=0$ or $x-4=0$

$x=-7$ or $x=4$

The values of $x$ are $-7$ and $4$.  

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Updated on: 10-Oct-2022

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