Question
Write the following in ascending order:$5 \sqrt{7}, 7 \sqrt{5}$ and $6 \sqrt{2}$
✓
Answer
$5 \sqrt{7}=\sqrt{5^2 \times 7}=\sqrt{25 \times 7}=\sqrt{175} $
$7 \sqrt{5}=\sqrt{7^2 \times 5}=\sqrt{49 \times 5}=\sqrt{245} $
$6 \sqrt{2}=\sqrt{6^2 \times 2}=\sqrt{36 \times 2}=\sqrt{72}$
Since, $72<175<245$,
we have $\sqrt{72}<\sqrt{175}<\sqrt{245}$.
Hence, $6 \sqrt{2}<5 \sqrt{7}<7 \sqrt{5}$.
$7 \sqrt{5}=\sqrt{7^2 \times 5}=\sqrt{49 \times 5}=\sqrt{245} $
$6 \sqrt{2}=\sqrt{6^2 \times 2}=\sqrt{36 \times 2}=\sqrt{72}$
Since, $72<175<245$,
we have $\sqrt{72}<\sqrt{175}<\sqrt{245}$.
Hence, $6 \sqrt{2}<5 \sqrt{7}<7 \sqrt{5}$.
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