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

Guest Oct 2, 2017
 #1
avatar+91071 
+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 !!!

 

 

cool cool cool

CPhill  Oct 2, 2017
edited by CPhill  Oct 2, 2017
 #2
avatar+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)

Pyramid  Oct 2, 2017

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