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

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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sqrt(4) = 2
sqrt(sqrt(4)) = 1.414 approx
hence sqrt(4) + sqrt(sqrt(4)) = 3.414
cuberoot(4) > 1 atleast
hence answer is M>4.

I was wondering why we are not considering the negative roots of 4 in this case. For instance, sq root of 4 would be 2 and -2...same for the 4th root... Any rule / trick that I might be missing here?

Look forward to the answer.

Cheers!
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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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Did in this way:

4 ^1/2 + 4 ^ 1/3 + 4 ^ 1/4

= 4 ^1/2 ( 1 + 4 ^2/3 + 4 ^1/2)

= 2 (1 + 2 + 4 ^2/3)

The above value is obviously above 6, so answer = E

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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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.



If M=sqrt(4)+cuberoot(4) +sqrt(sqrt(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


We know that sqrt(4)=2.
Since 1^3=1 < (cuberoot(4))^3=4 < 2^3=8, 1<cuberoot(4)<2.
Similarly 1^4=4 < (sqrt(sqrt(4)))^4=4 <2^4=16 implies that 1<sqrt(sqrt(4))<2.
So 2+1+1<M<2+2+2.

M is between 4 and 6.

The answer is, therefore, E.
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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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How can it \(4^{1/4}\) be \(\sqrt{2}\) what function does exponent 1/4 have

\(\sqrt{2}\) without 1/4 exponent equals aprox 1.414

\(4^{\frac{1}{4}}=(2^2)^{\frac{1}{4}}=2^{\frac{2}{4}}=2^{\frac{1}{2}}=\sqrt{2}\)
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Main idea:Approximate RHS by taking the cube root.

Details:

We have M= 4^ (1/3) + 4^ (1/3)+ 4^ (1/3) which is greater than 4

Hence E.
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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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Hi Bunuel ,

In some data sufficiency questions, the answer becomes insufficient because the square root produces a plus and a minus value. I've been roasted many times for making this careless mistake. But in this question, I got it wrong because I considered -2 as a secondary answer in \(\sqrt{4}\)

Can you please help me understand why that is?

Thank you.

Thanks.

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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Even if you're not sure about the concept Bunuel describes, you can prove E easily:

The third term, \(\sqrt[4]{4} = 4^{1/4} = 2^{2/4} = 2^{1/2}\) or \(\sqrt[]{2} ≈ 1.4\)
So the second term has to be greater than 1.4, and thus M has to be greater than 4 (2 + number bigger than 1.4 + 1.4)

Similar but harder question to practice: https://gmatclub.com/forum/new-tough-an ... l#p1029227
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Can we not just add the powers here?
Since the numerator is the same, adding 1/2 + 1/3 + 1/4 is a positive number and 4 raised to any positive number must be greater than 4 itself.

Bunuel:
Can't you still factor this problem out by factoring out the square root of 4 here?

= 4^(1/2) * ( 1 + 4 + 4^(1/2))
= 4^(1/2) * (5+2)
= 4^(1/2) * 7
= 14, which is > 4 so therefore the answer is E.