Multiplying $\frac{3}{\sqrt{17} - \sqrt{2}}$ by which fraction will produce an equivalent fraction with a rational denominator.

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Solution

Finding the fraction:
The given expression is $\frac{3}{\sqrt{17} - \sqrt{2}}$
Rationalizing the denominator by multiplying and dividing $\sqrt{17} + \sqrt{2}$
$= \frac{3}{\sqrt{17} - \sqrt{2}} \times (\frac{\sqrt{17} + \sqrt{2}}{\sqrt{17} + \sqrt{2}}) = \frac{3 (\sqrt{17} + \sqrt{2})}{{(\sqrt{17})}^{2} - {(\sqrt{2})}^{2}} [∵ (a - b) (a + b) = a^{2} - b^{2}] = \frac{3 (\sqrt{17} + \sqrt{2})}{17 - 2} = \frac{3 (\sqrt{17} + \sqrt{2})}{15} = \frac{(\sqrt{17} + \sqrt{2})}{5}$
So, we have to multiply $\frac{\sqrt{17} + \sqrt{2}}{\sqrt{17} + \sqrt{2}}$ to the given expression to get a rational denominator.

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