2014/076) Solve for largest x

 $\dfrac{6}{x-6}+ \dfrac{8}{x-8}+\dfrac{20}{x-20}+\dfrac{22}{x-22}=x^2-14x-4$


add 1 to each term onLHS and so 4 to RHS to get

$\dfrac{x}{x-6} + \dfrac{x}{x-8} + \dfrac{x}{x-20} + \dfrac{x}{x-22} = x^2-14x$

so one solution is x = 0 and further we deviding by x we get
$\dfrac{1}{x-6} + \dfrac{1}{x-8} + \dfrac{1}{x-20} + \dfrac{1}{x-22} = x-14$

put y = x - 14 to get


$\dfrac{1}{y+8} + \dfrac{1}{y+6} + \dfrac{1}{y-6} + \dfrac{1}{y-8} = y$

or 

$\dfrac{1}{y+8} + \dfrac{1}{y-8} + \dfrac{1}{y+6} + \dfrac{1}{y-6} = y$

or

$\dfrac{2y}{y^2-64} + \dfrac{2y}{y^2-36} = y$

so y = 0

or 
$\dfrac{2}{y^2-64} + \dfrac{2}{y^2-36} = 1$

or
$2((y^2-36) + y^2-64))= (y^2-36)(y^2-64)$
or $2((2y^2-100))= (y^2-36)(y^2-64)= y^4-100y^2+ 36 *64$
or $y^4- 104y^2+48^2+200=0$
or $(y^2-52)^2 = 200$
$y^2 = 52 \pm 10\sqrt{2}$
we should take the higher of the 2 and add 14 to get the largest x or x = 14 + $\sqrt{52+10\sqrt{2}}$