Question

multiplying \\(\frac{3}{\sqrt{17}-\sqrt{2}}\\) by which fraction will produce an equivalent fraction with a rational denominator?
options:
\\(\frac{\sqrt{2}-\sqrt{17}}{\sqrt{2}-\sqrt{17}}\\)
\\(\frac{\sqrt{17}-\sqrt{2}}{\sqrt{17}-\sqrt{2}}\\)
\\(\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}\\)
\\(\frac{\sqrt{15}}{\sqrt{15}}\\)

Explanation:

Step1: Recall Rationalizing Denominators

To rationalize a denominator of the form \(\sqrt{a} - \sqrt{b}\), we multiply by the conjugate \(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}}\) (since \((x - y)(x + y)=x^{2}-y^{2}\), which eliminates square roots in the denominator). Here, the denominator is \(\sqrt{17}-\sqrt{2}\), so its conjugate is \(\sqrt{17}+\sqrt{2}\).

Step2: Identify the Correct Fraction

We need to multiply \(\frac{3}{\sqrt{17}-\sqrt{2}}\) by a fraction equal to 1 (so numerator and denominator are equal) to rationalize the denominator. The conjugate of \(\sqrt{17}-\sqrt{2}\) is \(\sqrt{17}+\sqrt{2}\), so the fraction should be \(\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}\) (looking at the options, this is the top - left option: \(\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}\)).

Answer:

The fraction \(\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}\) (the top - left option among the given choices)