Let $m$ be the largest real root of the equation $\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4$ find $m$ 
Let $m$ be the largest real root of the equation $$\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4.$$ Find $m$.

do we literally add all the fractions or do we do something else 
I have no clue how to solve this 
 A: HINT:
Write the LHS as $$\left(\frac x{x-3}-1\right)+\left(\frac x{x-5}-1\right)+\left(\frac x{x-17}-1\right)+\left(\frac x{x-19}-1\right)$$ and this gives a constant term of $-4$. Equating this with the RHS, we have$$x\left(\frac1{x-3}+\frac1{x-5}+\frac1{x-17}+\frac1{x-19}\right)=x(x-11)$$ and note that $3,5$ are 'symmetrical' around $11$; that is, $17-11=11-5$ and $19-11=11-3$.
A: We need to solve
$$\frac3{x-3}+1+ \frac5{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1 =x^2 - 11x $$ or
$$x\left(\frac{1}{x-3} + \frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19}\right) =x^2 - 11x$$ or
$$2x(x-11)\left(\frac{1}{(x-3)(x-19)}+\frac{1}{(x-5)(x-17)}\right)=x(x-11),$$ which gives $x_1=0$, $x_2=11$ or
$$\frac{1}{x^2-22x+57}+\frac{1}{x^2-22x+85}=\frac{1}{2}.$$
Let $x^2-22x+57=a$.
Thus, $$\frac{1}{a}+\frac{1}{a+28}=\frac{1}{2}$$ or
$$a^2+24a-56=0$$ or
$$a^2+24a+144=200,$$ which gives
$$a=-12+10\sqrt2$$ or
$$a=-12-10\sqrt2,$$
which gives 
$$x^2-22x+69\pm10\sqrt2=0$$ or
$$x^2-22x+121=52\pm10\sqrt2$$ or 
$$x_{3,4,5,6}=11\pm\sqrt{52\pm10\sqrt2}$$ and we got a maximal root:
$$11+\sqrt{52+10\sqrt2}.$$
