Question #14172

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Question #14172

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Answer 1 · 2018-01-23T08:11:55

$\frac{2 \sqrt{2} \cdot (1 + \sqrt{3 x})}{1 - 3 x}$ $(2sqrt2 *(1+sqrt(3x)))/(1-3x)$

Explanation:

This is very confusing first of, the way you asked the question. So here is what I interpreted:

Simplify, $\frac{\sqrt{8}}{1 - \sqrt{3 x}}$ $sqrt8 / (1-sqrt(3x))$ .

When simplifying we want to remove roots from the denominator. We can do this by multiplying the fraction as a whole by the denominator's conjugate: $1 + \sqrt{3 x}$ $1+sqrt(3x)$ .

So, $\frac{\sqrt{8}}{1 - \sqrt{3 x}} \cdot \frac{1 + \sqrt{3 x}}{1 + \sqrt{3 x}} = \frac{\sqrt{8} \cdot (1 + \sqrt{3 x})}{1 - 3 x}$ $sqrt8 / (1-sqrt(3x)) * (1+sqrt(3x))/(1+sqrt(3x)) = (sqrt8 *(1+sqrt(3x)))/(1-3x)$

Further simplification, $\frac{2 \sqrt{2} \cdot (1 + \sqrt{3 x})}{1 - 3 x}$ $(2sqrt2 *(1+sqrt(3x)))/(1-3x)$

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Question #14172

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