# Completing the square but in different situation [duplicate]

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.

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• Start by multiplying the equation with $(x-2)^2$. – md2perpe Jun 12 '17 at 7:26
• 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. – CY Aries Jun 12 '17 at 7:28
• 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 – 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}

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

Well, we have:

$$x^2+4\cdot\left(\frac{x}{x-2}\right)^2=45\tag1$$

Bring together using a common denominator:

$$\frac{x^2\cdot\left(x^2-4x+8\right)}{\left(x-2\right)^2}=45\tag2$$

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

$$x^2\cdot\left(x^2-4x+8\right)=45\cdot\left(x-2\right)^2\tag3$$

Expand out terms of the right hand side:

$$x^2\cdot\left(x^2-4x+8\right)=45x^2-180x+180\tag4$$

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

$$x^2\cdot\left(x^2-4x+8\right)-45x^2+180x-180=0\tag5$$

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

$$\left(x-6\right)\cdot\left(x-3\right)\cdot\left(x^2+5x-10\right)=0\tag6$$