Solve the following quadratic equation by factorization:

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

Given:

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

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

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

$\frac{(5+x)(5+x)-(5-x)(5-x)}{(5-x)(5+x)}=\frac{4\times3+3}{4}$

$\frac{x^2+5x+5x+25-(x^2-5x-5x+25)}{-x^2-5x+5x+25}=\frac{15}{4}$

$\frac{x^2-x^2+10x+10x+25-25}{-x^2+25}=\frac{15}{4}$

$\frac{20x}{-x^2+25}=\frac{15}{4}$

$4(20x)=15(-x^2+25)$   (on cross multiplication)

$4\times5(4x)=15(-x^2+25)$

$16x=-3x^2+75$

$3x^2+16x-75=0$

$3x^2+25x-9x-75=0$

$3x(x-3)+25(x-3)=0$

$(3x+25)(x-3)=0$

$3x+25=0$ or $x-3=0$

$3x+25=0$ or $x-3=0$

$3x=-25$ or $x=3$

$x=-\frac{25}{3}$ or $x=3$

The values of $x$ are $-\frac{25}{3}$ and $3$.  

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Updated on: 10-Oct-2022

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