Question
Find the rationalizing factor of $\sqrt{27}$
✓
Answer
$\sqrt{27}=\sqrt{9 \times 3}=3 \sqrt{3} \quad \therefore 3 \sqrt{3} \times \sqrt{3}=3 \times 3=9$ is a rational number.
$\therefore \sqrt{3}$ is the rationalizing factor of $\sqrt{27}$.
Note that, $\sqrt{27}=3 \sqrt{3}$ means $3 \sqrt{3} \times 3 \sqrt{3}=9 \times 3=27$.
Hence $3 \sqrt{3}$ is also a rationalizing factor of $\sqrt{27}$. In the same way $4 \sqrt{3}, 7 \sqrt{3}, \ldots$ are also the rationalizing factors of $\sqrt{27}$. Out of all these $\sqrt{3}$ is the simplest rationalizing factor.
$\therefore \sqrt{3}$ is the rationalizing factor of $\sqrt{27}$.
Note that, $\sqrt{27}=3 \sqrt{3}$ means $3 \sqrt{3} \times 3 \sqrt{3}=9 \times 3=27$.
Hence $3 \sqrt{3}$ is also a rationalizing factor of $\sqrt{27}$. In the same way $4 \sqrt{3}, 7 \sqrt{3}, \ldots$ are also the rationalizing factors of $\sqrt{27}$. Out of all these $\sqrt{3}$ is the simplest rationalizing factor.
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