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

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Manager
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If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the [#permalink]

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13 Apr 2007, 21:13
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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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14 Apr 2007, 04:51
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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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14 Apr 2007, 05:27
thats right i absolutely agree. M must be greater than 4

Javed.

Cheers!
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14 Apr 2007, 08:12
Thanks vshaunak! My brain goes blank at times and I end up making many questions harder than they should be
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Re: If M = sqrt of 4 + cube root of 4 + quad root of 4, M = ?  [#permalink]

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26 Jan 2014, 21:38
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the [#permalink]

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27 Jan 2014, 00:53
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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 $$\sqrt[4]{4}>1$$ --> $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}=2+(number \ more \ then \ 1)+(number \ more \ then \ 1)>4$$

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] 27 Jan 2014, 00:53
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