Solve for $x$:
$\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}, x≠3, -5$

Given:

Given quadratic equation is $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}, x≠3, -5$.

To do:

We have to solve the given quadratic equation.

Solution:

$\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}, x≠3, -5$

$\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}$

$\frac{1(x+5)-1(x-3)}{(x-3)(x+5)}=\frac{1}{6}$

$\frac{x+5-x+3}{x^2-3x+5x-15}=\frac{1}{6}$

$\frac{8}{x^2+2x-15}=\frac{1}{6}$

$(8)(6)=1(x^2+2x-15)$   (On cross multiplication)

$48=x^2+2x-15$

$x^2+2x-15-48=0$

$x^2+2x-63=0$

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=1, b=2$ and $c=-63$.

Therefore, the roots of the given equation are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(2)\pm \sqrt{(2)^2-4(1)(-63)}}{2(1)}$ 

$x=\frac{-2\pm \sqrt{4+252}}{2}$ 

$x=\frac{-2\pm \sqrt{256}}{2}$ 

$x=\frac{-2\pm 16)}{2}$ 

$x=\frac{-2+16}{2}$ or $x=\frac{-2-16}{2}$

$x=\frac{14}{2}$ or $x=\frac{-18}{2}$

$x=7$ or $x=-9$

The values of $x$ are $-9$ and $7$.

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Updated on: 10-Oct-2022

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