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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 $7$.