# Is there another way to solve this quadratic equation?

$$\frac { 4 }{ x^{ 2 }-2x+1 } +\frac { 7 }{ x^{ 2 }-2x+4 } =2$$

Steps I took:

$$\frac { 4(x^{ 2 }-2x+4) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } +\frac { 7(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } =\frac { 2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) }$$

$$4(x^{ 2 }-2x+4)+7(x^{ 2 }-2x+1)=2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1)$$

$$11x^{ 2 }-22x+23=2x^{ 4 }-8x^{ 3 }+18x^{ 2 }-20x+8$$

I can keep going with all the steps I took, but is there a more elegant way to arrive at the solution for this equation? It seems as if I keep going the way I am, I will hit a dead end. No actual solution, please. Hints are much better appreciated.

To simplify computations a little bit, I would put $y=(x-1)^2$, so that the equation becomes $$\frac{4}{y}+\frac{7}{y+3}=2$$ which you can easily solve to find the values of $y$ and finally the corresponding values of $x$.

Setting $t=x^2-2x+1$ gives $$\frac{4}{t}+\frac{7}{t+3}=2$$ $$4(t+3)+7t=2t(t+3)$$ $$2t^2+6t-4t-7t-12=0$$ $$2t^2-5t-12=0$$ $$(2t+3)(t-4)=0$$

By inspection, the two denominators differ by $3$, just as the numerators do, hinting that to establish $$\frac44+\frac77=2,$$ we may set

$$x^2-2x+1=(x-1)^2=4,$$ i.e. $$\color{green}{x=-1\lor x=3}.$$

The other two roots are a little more elusive. But we can observe that when expanding

$$\frac4z+\frac7{z+3}-2=0,$$ we will get terms $-2z^2$ and $12$, so that the product of the $z$ roots is $-6$, and

$$(x-1)^2=-\frac32,$$ i.e. $$\color{green}{x=1-i\sqrt{\frac32}\lor x=1+i\sqrt{\frac32}}.$$