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.
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\(M=4^{1/2} + 4^{1/3} + 4^{1/4}\)

Now we know that \(4^{1/2} = 2\)

We also know that \(4^{1/4} = \sqrt{2} \approx 1.414 > 1\)

And finally \(4^{1/3} > 4^{1/4} \Rightarrow 4^{1/3}>1\)

So combining all three together \(M > 2+1+1 \Rightarrow M > 4\)
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Welcome to GMAT Club. Below might help to clear your doubts.


1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

2. Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, \(\sqrt{25}=+5\) and \(-\sqrt{25}=-5\). Even roots have only non-negative value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

For more check Number Theory chapter of Math Book: math-number-theory-88376.html
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Factoring out is not correct:
4^(1/2)*4^(2/3) does not equal to 4^(1/3).
4^(1/2)*4^(1/2) does not equal to 4^(1/4).

\(a^n*a^m=a^{n+m}\) not \(a^{nm}\).

Hope it helps.
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√4
√4 = 2

∛4
∛1 = 1
∛8 = 2
So, ∛4 is BETWEEN 1 and 2.
In other words, ∛4 = 1.something

∜4
∜1 = 1
∜16 = 2
So, ∜ is BETWEEN 1 and 2.
In other words, ∜ = 1.something

So, √4 + ∛4 + ∜4 = 2 + 1.something + 1.something
= more than 4
= E

Cheers,
Brent
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\(4^{\frac{1}{4}}=(2^2)^{\frac{1}{4}}=2^{\frac{2}{4}}=2^{\frac{1}{2}}=\sqrt{2}\)
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We are given that M = √4 + ^3√4 + ^4√4. We need to determine the approximate value of M.

Since √4 = 2, we need to determine the value of 2 + ^3√4 + ^4√4

Let’s determine the approximate value of ^3√4. To find this value, we need to find the perfect cube roots just below and just above the cube root of 4.

^3√1 < ^3√4 < ^3√8

1 < ^3√4 < 2

Let’s next determine the approximate value of ^4√4. To find this value, we need to find the perfect fourth roots just below and just above the fourth root of 4.

^4√1 < ^4√4 < ^4√16

1 < ^4√4 < 2

Since both ^3√4 and ^4√4 are greater than 1, so √4 + ^3√4 + ^4√4 > 2 + 1 + 1, and thus, √4 + ^3√4 + ^4√4 > 4.

Answer: E
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The square root function CANNOT produce negative result. EVER.

\(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;
Similarly \(\sqrt{\frac{1}{16}} = \frac{1}{4}\), NOT +1/4 or -1/4.


Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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Similar but harder question to practice: https://gmatclub.com/forum/new-tough-an ... l#p1029227
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