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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]


Sagot :

To produce an equivalent fraction with a rational denominator, we need to rationalize the denominator of the given fraction [tex]\(\frac{3}{\sqrt{17} - \sqrt{2}}\)[/tex].

Rationalizing the denominator involves multiplying both 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].

So, we multiply [tex]\(\frac{3}{\sqrt{17} - \sqrt{2}}\)[/tex] by [tex]\(\frac{\sqrt{17} + \sqrt{2}}{\sqrt{17} + \sqrt{2}}\)[/tex]:

[tex]\[ \frac{3}{\sqrt{17} - \sqrt{2}} \times \frac{\sqrt{17} + \sqrt{2}}{\sqrt{17} + \sqrt{2}} = \frac{3(\sqrt{17} + \sqrt{2})}{(\sqrt{17} - \sqrt{2})(\sqrt{17} + \sqrt{2})} \][/tex]

When we multiply the denominators, we use the difference of squares formula:

[tex]\[ (\sqrt{17} - \sqrt{2})(\sqrt{17} + \sqrt{2}) = 17 - 2 = 15 \][/tex]

Thus, the expression becomes:

[tex]\[ \frac{3(\sqrt{17} + \sqrt{2})}{15} \][/tex]

Which simplifies to:

[tex]\[ \frac{3\sqrt{17} + 3\sqrt{2}}{15} \][/tex]

This shows that the original fraction's denominator has been rationalized. Hence, the fraction we need to multiply by to rationalize the denominator is:

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