Fractional Equations #1

$\frac{1}{x}(\frac{1}{ x - \frac{8}{x} + 11}) + \frac{1}{x}(\frac{1}{ x - \frac{8}{x} + 2}) + \frac{1}{x}(\frac{1}{ x - \frac{8}{x} - 13}) = 0$

now $\frac{1}{x}$ can't be zero

therefore solving the equation we get $x = ( 8 , -1) or ( -8 , 1)$

$\dfrac{1}{x^2+11x-8} + \dfrac{1}{x^2+2x-8} + \dfrac{1}{x^2-13x-8} \\ = \dfrac{1}{ (x^2+2x-8)+9x} + \dfrac{1}{x^2+2x-8} + \dfrac{1}{(x^2-2x-8)-15x} \\ = \dfrac{1}{(x+4)(x-2)+9x} + \dfrac{1}{(x+4)(x-2)} + \dfrac{1}{(x+4)(x-2)-15x}$

Let $t= \boxed{(x+4)(x-2)}$ , the equation can be written as

$\dfrac{1}{t-9x} + \dfrac{1}{t} + \dfrac{1}{t-15x} = 0$

Multilply the equation by $t(t-9x)(t-15x)$ , we get

$t(t-15x) + (t+9x)(t-15x) + t(t+9x) = 0 \\ \Rightarrow 3t^2-12xt-135x^2 = 0 \\ t^2 - 4xt -45x^2 = 0 \\ (t-9x)(t+5x) =0 \\ t = 9x \quad or \quad t=-5x$

When $t=9x$ ,

$(x+4)(x-2)=9x \Rightarrow (x-8)(x+1)=0 \Rightarrow x= \boxed {8, -1}$

When $t=-5x$ ,

$(x+4)(x-2) = -5x \Rightarrow (x+8)(x-1)=0 \Rightarrow x= \boxed {-8, 1}$

$\therefore$ , Sum $= \boxed{8+(-1)+(-8)+1=0}$

nice solution

using the coefficients to find the sum of zeroes of polynomial, we will get a four degree equation and we need to find out the coefficient of x^4 and x^3 to get the sum of roots, which are 3( 1+1+1) and 0 ( -13 2 + 2 2 + 11*2) respectively .. ! we have our answer i.e. 0 .