1. \({3\sqrt{6}, 10, 7\sqrt{2}}\)

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

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

Guest Oct 2, 2017

#1**+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 !!!

CPhill Oct 2, 2017

#2**+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)

Pyramid Oct 2, 2017