Write in ascending order- 6 sqrt-5- 7 sqrt-3- and 8 sqrt-2-

Question

Write in ascending order- 6 sqrt-5- 7 sqrt-3-  and  8 sqrt-2-

Toppr Admin · Accepted Answer

- 6 -sqrt-5-sqrt-6-2- -times 5-sqrt-180- -We know that- - 128-lt-147-lt-180 - - -therefore -sqrt-128-lt-sqrt-147-lt-sqrt-180- - -Rightarrow 8 -sqrt-2-lt-7 -sqrt-3-lt-6 -sqrt-5- -

Write in ascending order:
$$ 6 \sqrt{5}, 7 \sqrt{3} $$ and $$ 8 \sqrt{2} $$

$$ 6 \sqrt{5}=\sqrt{6^{2} \times 5}=\sqrt{180} $$
We know that, $$ 128<147<180 $$
$$ \therefore \sqrt{128}<\sqrt{147}<\sqrt{180} $$
$$ \Rightarrow 8 \sqrt{2}<7 \sqrt{3}<6 \sqrt{5} $$

Write in ascending order:$$ 6 \sqrt{5}, 7 \sqrt{3} $$ and $$ 8 \sqrt{2} $$

$$ 6 \sqrt{5}=\sqrt{6^{2} \times 5}=\sqrt{180} $$We know that, $$ 128<147<180 $$ $$ \therefore \sqrt{128}<\sqrt{147}<\sqrt{180} $$$$ \Rightarrow 8 \sqrt{2}<7 \sqrt{3}<6 \sqrt{5} $$

Write in ascending order:
$$ 6 \sqrt{5}, 7 \sqrt{3} $$ and $$ 8 \sqrt{2} $$

$$ 6 \sqrt{5}=\sqrt{6^{2} \times 5}=\sqrt{180} $$
We know that, $$ 128<147<180 $$
$$ \therefore \sqrt{128}<\sqrt{147}<\sqrt{180} $$
$$ \Rightarrow 8 \sqrt{2}<7 \sqrt{3}<6 \sqrt{5} $$