What is the value of x in this equation?

[Albert1]

$(\dfrac {13x-x^2}{x+1})(x+\dfrac{13-x}{x+1})=42$

$find \,\, real \,\, x$

 

[lfdahl]

My solution:

Spoiler

Using polynomial division, we get:
\[\left ( \frac{13x-x^2}{x+1} \right )\left ( x+\frac{13-x}{x+1} \right )=\left ( \frac{13x-x^2}{x+1} \right ) \left ( \frac{13+x^2}{x+1} \right )=42 \\\\ \Rightarrow \left (14-\left (x+\frac{14}{x+1} \right ) \right )\left ( -1+\left ( x+\frac{14}{x+1} \right ) \right )=42\]Let $x+\frac{14}{x+1} = \alpha$ and solve the quadratic equation:\[(14-\alpha )(-1+\alpha )=42\Rightarrow -\alpha^2+15\alpha -56 = 0\Rightarrow \alpha \in \left \{ \frac{15\pm 1}{2} \right \}=\left \{ 7,8 \right \}\]Using the $\alpha$-expression, we get:


\[x+\frac{14}{x+1}=7\Rightarrow x^2-6x+7 = 0\Rightarrow x\in \left \{ 3\pm \sqrt{2} \right \} \\\\ x+\frac{14}{x+1}=8 \Rightarrow x^2-7x+6 = 0 \Rightarrow x\in \left \{ 1,6 \right \}\]

- thus the set of solutions is: \[S = \left \{ 1,3-\sqrt{2},3+\sqrt{2},6 \right \}\]