Solve the equation $$x^2+4\left(\frac{x}{x-2}\right)^2=45$$

My attempt,

I decided to use completing the square method, so I change it to $$x^2+\left(\frac{2x}{x-2}\right)^2=45$$

But I never encounter this before. Normally, for example $x^2+4x=5$, we can change it to $x^2+4x+(\frac{4}{2})^2=45+(\frac{4}{2})^2$. But in this question is different. Could someone give me some hints for it? Thanks in advance.

  • $\begingroup$ Start by multiplying the equation with $(x-2)^2$. $\endgroup$ – md2perpe Jun 12 '17 at 7:26
  • $\begingroup$ The equation is not quadratic. It can be rewritten as $x^4-4x^3-37x^2+180x-180=0$. You can solve it with Factor Theorem. $\endgroup$ – CY Aries Jun 12 '17 at 7:28
  • $\begingroup$ But the suggested solution wrote $x^2+(\frac{2x}{x-2})^2+\frac{4x^2}{x-2}=45+\frac{4x^2}{x-2}$. How? @CYAries $\endgroup$ – Mathxx Jun 12 '17 at 7:38

The suggested solution posted by OP provide a clever method.

\begin{align} x^2+4\left(\frac{x}{x-2}\right)^2&=45\\ x^2+2(x)\left(\frac{2x}{x-2}\right)+\left(\frac{2x}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\ \left(x+\frac{2x}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\ \left(\frac{x^2}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\ \left(\frac{x^2}{x-2}\right)^2-4\left(\frac{x^2}{x-2}\right)+4&=49\\ \left(\frac{x^2}{x-2}-2\right)^2&=49\\ \end{align}

  • $\begingroup$ Yes. This I what I mean. Thanks a lot $\endgroup$ – Mathxx Jun 12 '17 at 7:47

Well, we have:


Bring together using a common denominator:


Multiply both sides by $\left(x-2\right)^2$:


Expand out terms of the right hand side:


Subtract $45x^2-180x+180$ from both sides:


The left hand side factors into a product with three terms:



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