Answer: The correct option is (B) [tex]\dfrac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}.[/tex]
Step-by-step explanation: We are given to select the correct fraction by which the following be multiplied to produce an equivalent fraction with a rational denominator.
[tex]F=\dfrac{3}{\sqrt{17}-\sqrt2}~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We know that
to rationalize the denominator having radical terms, we multiply both numerator and denominator by a common term which contains the opposite sign before the radical term as in the original denominator.
In the given denominator, both terms are radical ones, so we can change the sign of any one of them and then multiply to the denominator and numerator simultaneously.
So, we will multiply both numerator and denominator by the term [tex]\sqrt{17}+\sqrt2.[/tex]
From (i), we have
[tex]F\\\\=\dfrac{3}{\sqrt{17}-\sqrt2}\\\\\\=\dfrac{3(\sqrt{17}+\sqrt{2})}{(\sqrt{17}-\sqrt{2})(\sqrt{17}+\sqrt{2})}\\\\\\=\dfrac{3(\sqrt{17}+\sqrt{2})}{17-2},~~~~~~~~\textup{since}~(a+b)(a-b)=a^2-b^2\\\\\\=\dfrac{3(\sqrt{17}+\sqrt{2})}{15}\\\\=\dfrac{\sqrt{17}+\sqrt{2}}{5}.[/tex]
Thus, the denominator is rationalized and the required fraction to be multiplied is
[tex]\dfrac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}.[/tex]
Option (B) is CORRECT.
