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If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the

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If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the [#permalink] New post 13 Apr 2007, 20:13
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Question Stats:

66% (01:54) correct 34% (00:48) wrong based on 54 sessions
If M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}, then the value of M is:

A. Less than 3
B. Equal to 3
C. Between 3 and 4
D. Equal to 4
E. Greater than 4

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-m-root-4-cube-root-4-fourth-root-4-then-the-93340.html
[Reveal] Spoiler: OA
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 [#permalink] New post 14 Apr 2007, 03:51
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M = sqrt of 4 + cube root of 4 + quad root of 4
No need to calculate ...
Sqrt of 4 = 2
Cube root of 4 should be > 1
Quad root of 4 should be > 1
Hence M should be > 4

The rule is nth root of any positive integer >1, will be always be > 1
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 [#permalink] New post 14 Apr 2007, 04:27
thats right i absolutely agree. M must be greater than 4


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 [#permalink] New post 14 Apr 2007, 07:12
Thanks vshaunak! My brain goes blank at times and I end up making many questions harder than they should be :x
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Re: If M = sqrt of 4 + cube root of 4 + quad root of 4, M = ?  [#permalink] New post 26 Jan 2014, 20:38
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the [#permalink] New post 26 Jan 2014, 23:53
Expert's post
If M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}, then the value of M is:

A. Less than 3
B. Equal to 3
C. Between 3 and 4
D. Equal to 4
E. Greater than 4

Here is a little trick: any positive integer root from a number more than 1 will be more than 1.

For instance: \sqrt[1000]{2}>1.

Hence \sqrt[3]{4}>1 and \sqrt[4]{4}>1 --> M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}=2+(number \ more \ then \ 1)+(number \ more \ then \ 1)>4

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-m-root-4-cube-root-4-fourth-root-4-then-the-93340.html
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the   [#permalink] 26 Jan 2014, 23:53
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