to rationalize the denominator of $\frac{5 - sqrt{7}}{9-sqrt{14}}$, you should multiply the expression by…

to rationalize the denominator of $\frac{5 - sqrt{7}}{9-sqrt{14}}$, you should multiply the expression by which fraction?\n$\frac{5+sqrt{7}}{9 - sqrt{14}}$\n$\frac{9-sqrt{14}}{9 - sqrt{14}}$\n$\frac{9+sqrt{14}}{9+sqrt{14}}$\n$\frac{sqrt{14}}{sqrt{14}}$

to rationalize the denominator of $\frac{5 - sqrt{7}}{9-sqrt{14}}$, you should multiply the expression by which fraction?\n$\frac{5+sqrt{7}}{9 - sqrt{14}}$\n$\frac{9-sqrt{14}}{9 - sqrt{14}}$\n$\frac{9+sqrt{14}}{9+sqrt{14}}$\n$\frac{sqrt{14}}{sqrt{14}}$

Answer

Explanation:

Step1: Recall rationalizing rule

To rationalize the denominator of a fraction with the form $a - \sqrt{b}$ in the denominator, we multiply by its conjugate $a+\sqrt{b}$. Here, the denominator is $9-\sqrt{14}$, and its conjugate is $9 + \sqrt{14}$. We multiply the fraction $\frac{5-\sqrt{7}}{9-\sqrt{14}}$ by $\frac{9+\sqrt{14}}{9+\sqrt{14}}$ because multiplying a fraction by $\frac{c}{c}$ (where $c\neq0$) is equivalent to multiplying by 1 and does not change the value of the fraction.

Answer:

C. $\frac{9+\sqrt{14}}{9+\sqrt{14}}$