Arranging $3\sqrt{9}, 4\sqrt{20}, 6\sqrt{25}$ in ascending order

Question

How to arrange $3\sqrt{9}, 4\sqrt{20}, 6\sqrt{25}$ in ascending order?

Welcome to Math.SE: In order to get the best possible answers, it is helpful to state what your thoughts and attempts on the problem are; this will prevent people from telling you things you already know, and help them give their answers at the right level. — projectilemotion, Jul 09 '17 at 11:13
The problem is really easy if you are allowed to use a calculator. You should include what resources you are allowed to use. — projectilemotion, Jul 09 '17 at 11:16

score 5 · Answer 1 · answered Jul 09 '17 at 11:18

$3\sqrt{9}=\sqrt{9}\sqrt{9}=\sqrt{81}$

$4\sqrt{20}=\sqrt{16}\sqrt{20}=\sqrt{320}$

$6\sqrt{25}=\sqrt{36}\sqrt{25}=\sqrt{900}$

Now, $81<320<900$, Then $\sqrt{81}<\sqrt{320}<\sqrt{900}$

Hence, $3\sqrt{9}<4\sqrt{20}<6\sqrt{25}\space\space\space\blacksquare$

score 0 · Answer 2 · answered Jul 09 '17 at 11:13

0

HINT: use that $$3\sqrt{9}=9,4\sqrt{20}=8\sqrt{5},6\sqrt{25}=30$$

answered Jul 09 '17 at 11:13

3

And note the obvious inequality $2<\sqrt{5}<3$ – projectilemotion Jul 09 '17 at 11:17

2 Answers2