# How do you solve 4 /( x^2 + 3x - 10) - 1 / (x^2 - x -6) = 3 / (x^2 - x -12) ?

Sep 25, 2016

Solutions are:

${x}_{n} = \frac{1}{24} \left(7 + 2 \sqrt{1513} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{10531 \sqrt{1513}}{2289169}\right) + \frac{2 n \pi}{3}\right)\right)$

for $n = 0 , 1 , 2$

#### Explanation:

Let us first look at the factors of the denominators:

${x}^{2} + 3 x - 10 = \left(x + 5\right) \left(x - 2\right)$

${x}^{2} - x - 6 = \left(x - 3\right) \left(x + 2\right)$

${x}^{2} - x - 12 = \left(x - 4\right) \left(x + 3\right)$

Note that there are no linear factors in common, so we might as well leave the quadratics unfactorised for now:

Subtract the right hand side of the given equation from the left to get:

$0 = \frac{4}{{x}^{2} + 3 x - 10} - \frac{1}{{x}^{2} - x - 6} - \frac{3}{{x}^{2} - x - 12}$

$\textcolor{w h i t e}{0} = \frac{4 \left({x}^{2} - x - 6\right) \left({x}^{2} - x - 12\right) - \left({x}^{2} + 3 x - 10\right) \left({x}^{2} - x - 12\right) - 3 \left({x}^{2} + 3 x - 10\right) \left({x}^{2} - x - 6\right)}{\left({x}^{2} + 3 x - 10\right) \left({x}^{2} - x - 6\right) \left({x}^{2} - x - 12\right)}$

$\textcolor{w h i t e}{0} = \frac{4 \left({x}^{4} - 2 {x}^{3} - 17 {x}^{2} + 18 x + 72\right) - \left({x}^{4} + 2 {x}^{3} - 25 {x}^{2} - 26 x + 120\right) - 3 \left({x}^{4} + 2 {x}^{3} - 19 {x}^{2} - 8 x + 60\right)}{\left({x}^{2} + 3 x - 10\right) \left({x}^{2} - x - 6\right) \left({x}^{2} - x - 12\right)}$

$\textcolor{w h i t e}{0} = \frac{\left(4 {x}^{4} - 8 {x}^{3} - 68 {x}^{2} + 72 x + 288\right) - \left({x}^{4} + 2 {x}^{3} - 25 {x}^{2} - 26 x + 120\right) - \left(3 {x}^{4} + 6 {x}^{3} - 57 {x}^{2} - 24 x + 180\right)}{\left({x}^{2} + 3 x - 10\right) \left({x}^{2} - x - 6\right) \left({x}^{2} - x - 12\right)}$

$\textcolor{w h i t e}{0} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4 {x}^{4}}}} - 8 {x}^{3} - 68 {x}^{2} + 72 x + 288 - \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{4}}}} - 2 {x}^{3} + 25 {x}^{2} + 26 x - 120 - \textcolor{red}{\cancel{\textcolor{b l a c k}{3 {x}^{4}}}} - 6 {x}^{3} + 57 {x}^{2} + 24 x - 180}{\left({x}^{2} + 3 x - 10\right) \left({x}^{2} - x - 6\right) \left({x}^{2} - x - 12\right)}$

$\textcolor{w h i t e}{0} = \frac{- 16 {x}^{3} + 14 {x}^{2} + 122 x - 12}{\left({x}^{2} + 3 x - 10\right) \left({x}^{2} - x - 6\right) \left({x}^{2} - x - 12\right)}$

$\textcolor{w h i t e}{0} = \frac{- 2 \left(8 {x}^{3} - 7 {x}^{2} - 61 x + 6\right)}{\left({x}^{2} + 3 x - 10\right) \left({x}^{2} - x - 6\right) \left({x}^{2} - x - 12\right)}$

So the solutions of the original equation are the roots of:

$8 {x}^{3} - 7 {x}^{2} - 61 x + 6 = 0$

To find how to solve this, see: https://socratic.org/s/aydGmACm

The roots are:

${x}_{n} = \frac{1}{24} \left(7 + 2 \sqrt{1513} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{10531 \sqrt{1513}}{2289169}\right) + \frac{2 n \pi}{3}\right)\right)$

for $n = 0 , 1 , 2$