Question

to rationalize the denominator of \\(\frac{2sqrt{10}}{3sqrt{11}}\\), you should multiply the expression by which fraction?\\(\frac{sqrt{10}}{sqrt{10}}\\)\\(\frac{sqrt{11}}{sqrt{11}}\\)\\(\frac{2-sqrt{10}}{2-sqrt{10}}\\)\\(\frac{3-sqrt{11}}{3-sqrt{11}}\\)

Explanation:

Step1: Recall Denominator Rationalization

To rationalize a denominator with a square root (like \( \sqrt{a} \)), we multiply the numerator and denominator by \( \sqrt{a} \) to eliminate the square root from the denominator. Here, the denominator is \( 3\sqrt{11} \), so we need to multiply by \( \frac{\sqrt{11}}{\sqrt{11}} \) (since multiplying by 1 doesn't change the value, and \( \frac{\sqrt{11}}{\sqrt{11}} = 1 \) when \( \sqrt{11}
eq 0 \)).

Step2: Analyze the Options

• Option 1: \( \frac{\sqrt{10}}{\sqrt{10}} \) – Denominator has \( \sqrt{10} \), not related to \( \sqrt{11} \) in the denominator. Eliminate.
• Option 2: \( \frac{\sqrt{11}}{\sqrt{11}} \) – Multiplying by this will make the denominator \( 3\sqrt{11} \times \sqrt{11} = 3 \times 11 = 33 \), which is rational. This works.
• Option 3: \( \frac{3 - \sqrt{11}}{3 - \sqrt{11}} \) – This is for rationalizing a denominator with a binomial (like \( a - \sqrt{b} \)), but our denominator is a monomial with \( \sqrt{11} \), not a binomial. Eliminate.
• Option 4: \( \frac{2 - \sqrt{10}}{2 - \sqrt{10}} \) – Denominator and numerator have \( \sqrt{10} \) and 2, not related to \( \sqrt{11} \). Eliminate.

Answer:

The correct option is the one with \( \frac{\sqrt{11}}{\sqrt{11}} \) (the middle option among the four, or the second option if we list them as first: \( \frac{\sqrt{10}}{\sqrt{10}} \), second: \( \frac{\sqrt{11}}{\sqrt{11}} \), third: \( \frac{3 - \sqrt{11}}{3 - \sqrt{11}} \), fourth: \( \frac{2 - \sqrt{10}}{2 - \sqrt{10}} \)). So the answer is the option with \( \frac{\sqrt{11}}{\sqrt{11}} \).