How do you solve $1/(x^2 -3) = 8/x$ ?

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Answer 1 · 2016-04-21T20:41:36

$x=1/16+-sqrt(769)/16$

Multiply both sides by $x(x^2-3)$ to get:

$x = 8(x^2-3) = 8x^2-24$

Subtract $x$ from both sides and transpose to get:

$8x^2-x-24 = 0$

This is in the form $ax^2+bx+c = 0$ with $a=8$ , $b=-1$ and $c=-24$ , which has roots given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$=(1+-sqrt((-1)^2-4(8)(-24)))/(2*8)$

$=1/16+-sqrt(769)/16$

How do you solve 1/(x^2 -3) = 8/x 1/(x^2 -3) = 8/x?