Multiplying [tex]\frac{3}{\sqrt{17}-\sqrt{2}}[/tex] by which fraction will produce an equivalent fraction with a rational denominator?

A. [tex]\frac{\sqrt{17}-\sqrt{2}}{\sqrt{17}-\sqrt{2}}[/tex]

B. [tex]\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}[/tex]

C. [tex]\frac{\sqrt{2}-\sqrt{17}}{\sqrt{2}-\sqrt{17}}[/tex]

D. [tex]\frac{\sqrt{15}}{\sqrt{15}}[/tex]



Answer :

To solve this problem, we need to rationalize the denominator of the given fraction [tex]\(\frac{3}{\sqrt{17} - \sqrt{2}}\)[/tex]. Rationalizing the denominator means getting rid of the square roots in the denominator.

To achieve this, we can multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\(\sqrt{17} - \sqrt{2}\)[/tex] is [tex]\(\sqrt{17} + \sqrt{2}\)[/tex].

Let's multiply the fraction by [tex]\(\frac{\sqrt{17} + \sqrt{2}}{\sqrt{17} + \sqrt{2}}\)[/tex]:

[tex]\[ \frac{3}{\sqrt{17} - \sqrt{2}} \cdot \frac{\sqrt{17} + \sqrt{2}}{\sqrt{17} + \sqrt{2}} \][/tex]

Now, we calculate the numerator and the denominator separately.

1. Numerator:
[tex]\[ 3 \cdot (\sqrt{17} + \sqrt{2}) = 3\sqrt{17} + 3\sqrt{2} \][/tex]

2. Denominator:
[tex]\[ (\sqrt{17} - \sqrt{2}) \cdot (\sqrt{17} + \sqrt{2}) \][/tex]
Using the difference of squares formula, [tex]\((a - b)(a + b) = a^2 - b^2\)[/tex], we get:
[tex]\[ (\sqrt{17})^2 - (\sqrt{2})^2 = 17 - 2 = 15 \][/tex]

Thus, the numerator is:
[tex]\[ 3(\sqrt{17} + \sqrt{2}) \approx 3 \cdot 4.355 \approx 16.61195756397227 \][/tex]

and the denominator is:
[tex]\[ 15 \][/tex]

So, after rationalizing the denominator, the fraction becomes:
[tex]\[ \frac{16.61195756397227}{15} \][/tex]

Therefore, the fraction we choose to multiply by is:
[tex]\[ \frac{\sqrt{17} + \sqrt{2}}{\sqrt{17} + \sqrt{2}} \][/tex]

This corresponds to the second option.

Thus, the answer is [tex]\(\boxed{\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}}\)[/tex].