# Common denominator with radicals

Can you make these 2 fractions into 1?

$$2\sqrt{9-2x} - \dfrac{2x}{\sqrt{9-2x}}$$

I thought you could make them into $\dfrac{-2x+18}{\sqrt{9-2x}}$

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rather $18-4x-2x$ in the nominator –  Berci Oct 13 '12 at 11:35
You're right, thank you. –  Tittyboy Oct 13 '12 at 11:38
Else it's maximally ok –  Berci Oct 13 '12 at 11:39

## 1 Answer

You're almost correct.

\begin{align} 2\sqrt{9-2x} - \frac{2x}{\sqrt{9-2x}} &= 2\sqrt{9-2x} \times \frac{\sqrt{9-2x}}{\sqrt{9-2x}} - \frac{2x}{\sqrt{9-2x}}\\ &= \frac{2(\sqrt{9-2x})^2}{\sqrt{9-2x}} - \frac{2x}{\sqrt{9-2x}}\\ &= \frac{2(9-2x)}{\sqrt{9-2x}} - \frac{2x}{\sqrt{9-2x}}\\ &= \frac{2(9-2x) - 2x}{\sqrt{9-2x}}\\ &= \frac{18 - 4x - 2x}{\sqrt{9-2x}}\\ &= \frac{18-6x}{\sqrt{9-2x}} \end{align}

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I assume that we should make the statement that $x$ is not equal 9/2. –  Emmad Kareem Oct 13 '12 at 12:38
That is implicit as soon as the initial expression is given. In fact, you need $x < \frac{9}{2}$. –  Michael Albanese Oct 13 '12 at 12:44
OK, thanks, may be the OP should take notice. –  Emmad Kareem Oct 13 '12 at 12:45