Solve this equation.

Question

$$ \dfrac{x^2-1}{x^2+9-6x-1}=\dfrac{x+2}{x-4}-\dfrac5{(x-2)^2} $$

Can you tell me what should I factorize the denominator? I thought to put $$x^2+9-6x-1=x^2-6x+8$$But I suppose they gave it in in form for a purpose. Am I right or should I put as denominator $$(x^2-6x+8)(x-4)(x-2)^2$$

Answer 1

$$x^2+9-6x-1=(x-3)^2-1^2=(x-3-1)(x-3+1)=(x-4)(x-2).$$

This exercise clearly aims at doing smart calculations. The equation can be rewritteb=n as: $$\frac{x^2-1}{(x-2)(x-4)}=\dfrac{x+2}{x-4}-\dfrac5{(x-2)^2}.$$ Multiply both sides with $(x-2)^2(x-4)$ and get: \begin{alignat*}{2} &&&(x^2-1)(x-2)=(x+2)(x-2)^2 -5(x-4)=(x^2-4)(x-2)-5(x-4)\\ &\iff\enspace&&\\ &&&\Bigl((x^2-1)-(x^2-4)\Bigr)(x-2)=-5(x-4)\iff 3(x-2)+5(x-4)=0 \end{alignat*} Thus $\,x=\dfrac{13}4$.

Answer 2

$$x^2 \color{red}{- 6}x + \color{blue}{ 8} = (x - 2)(x - 4)$$

Note that we have $\color{blue}{ 8} = (-2)(-4)$ and $\color{red}{-6} = -2+-4$.

Answer 3

$$x^2+9-6x-1=0 \Rightarrow x^2-6x+8=0$$

$$\Delta=36-32=4$$

$$x_{1,2}=\frac{6 \pm \sqrt{4}}{2}=3 \pm 1=4,2$$

So, $x^2+9-6x-1=(x-2)(x-4)$