# Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$

I am asked to simplify $$\frac{\sqrt{8}}{1-\sqrt{3x}}$$. The solution is provided as $$\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$$ and I am unable to arrive at this. I was able to arrive at $$\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$$

Here is my working: $$\frac{\sqrt{8}}{1-\sqrt{3x}}$$ = $$\frac{\sqrt{8}}{1-\sqrt{3x}}$$ * $$\frac{1+\sqrt{3x}}{1+\sqrt{3x}}$$ = $$\frac{1+\sqrt{8}\sqrt{3x}}{1-3x}$$ = $$\frac{1+\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{3x}}{1-3x}$$ = $$\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$$

Is $$\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$$ correct and part of the way? How can I arrive at the provided solution $$\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$$?

• Please review your computations. There is a least 2 errors. – mathcounterexamples.net Jan 5 '19 at 18:10

$$\frac{\sqrt{8}}{1-\sqrt{3x}} = \frac{\sqrt{8}}{1-\sqrt{3x}} \cdot \frac{1+\sqrt{3x}}{1+\sqrt{3x}} = \color{blue}{\frac{\sqrt{8}\cdot\left({1+\sqrt{3x}}\right)}{1-3x}} = \frac{\sqrt{8}+\sqrt{24x}}{1-3x}$$

$$= \frac{\sqrt{2^3}+\sqrt{2^3\cdot3x}}{1-3x} = \frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$$

Notice the step I highlighted in blue, which is where you made an error. You have to multiply $$\sqrt{8}$$ to $$\left(1+\sqrt{3x}\right)$$ completely. You multiplied it by only $$\sqrt{3x}$$ and added $$1$$, which wasn’t correct.

Observe that $$\sqrt{8}\times (1+\sqrt{3x})=\sqrt{8}+\sqrt{8}\times \sqrt{3x}$$ and $$\sqrt{8}\times \sqrt{3}=\sqrt{24}=\sqrt{4\times 6}=2\sqrt{6}.$$

Well, in rationalizing the denominator, we arrive at the intermediate step $$\frac{\sqrt{8}+\sqrt{24x}}{1-3x}$$ which simplifies to the provided solution.

Your error comes in multiplying by the conjugate of the denominator. $$\sqrt{8}*(1+\sqrt{3x})$$ becomes $$\sqrt{8}+\sqrt{24x}$$.

Write $$\frac{\sqrt{8}(1+\sqrt{3x})}{(1-\sqrt{3x})(1+\sqrt{3x})}$$ and this is $$\frac{2\sqrt{2}(1+\sqrt{3x})}{1-3x}$$

Hints:$$\sqrt8=\sqrt{4×2}=2\sqrt{2}.$$

$$\sqrt24=2\sqrt6$$. (Why?)

$$(1+\sqrt{3x})×(1-\sqrt{3x})=1-3x$$.

The second equality is where you mess up - note that $$\sqrt{8}\cdot(1+\sqrt{3x})=\sqrt{8}+\sqrt{8}\cdot\sqrt{3x}$$ by the distributive property of real numbers