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# how would i order these set of numbers from least to greatest?

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1. $${3\sqrt{6}, 10, 7\sqrt{2}}$$

2.$${-2\sqrt{3}, -4, -3\sqrt{2}, -2\sqrt{\frac{7}{2}}}$$

3.$${\sqrt[3]{21}, 3 \sqrt [3]{2}, 2.8, 2 \sqrt[3]{5}}$$

Oct 2, 2017

#1
+95884
+1

They are trying to be tricky, here  !!!

Remember that    a√b  is just  √ [a^2 * b ]

And    a  =  √a^2

And    a∛ b     is just  ∛  [ a^3 * b ]

So

3√ 6 ,  10  ,  7√2    can be written as

√54 , √100 ,  √ 98          see  ???.......you can arrange these for yourself

Similarly

-2√3 ,  -4 , -3√2,  -2√ [7/2]       becomes

- √12, -√16, -√18  and  -√ [ (4 * 7) / 2 ]  =  - √14

Again you can arrange these....just remember that, for instance,  -√13 > -√25   !!!!

Lastly

∛21 , 3∛2 , 2.8 , 2∛5    become

∛21, ∛54, ∛21.952  and  ∛40

And you can handle this one, too !!!

Oct 2, 2017
edited by CPhill  Oct 2, 2017
#2
+271
+1

I guess you could do this sort of quesion by raising the numbers to a power until they are whole numbers and then working it out.

Raising the first one by squaring it gives

$$6*9,100,49*2=54,100,98$$

therefore the result would be

$$3\sqrt{6},7\sqrt{2},10$$

for the second one squaring it would be

$$-4*3,-16,-9*2,-4*\frac{7}{2}=-12,-16,-18,-14$$

so it is

$$-3\sqrt{2},-4,-2\sqrt{\frac{7}{2}},-2\sqrt{3}$$

(note that I changed to negative because of that)

for the third one you need to cube it

$$21,27*2,21.952,8*5=21,54,21.952,40$$

Then it becomes

$$\sqrt[3]{21},2.8,2\sqrt[3]{5},3\sqrt[3]{2}$$

(Sorry for errors, I kind of rushed it)

Oct 2, 2017