If M = 4^(1/2) + 4^(1/3) + 4^(1/4), then the value of M is: : Problem Solving (PS)

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26 Apr 2010, 22:52

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zz0vlb

Find the value of M (see attachment). I want to see other approaches to this problem.

Source: GMAT Prep

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.

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21 Apr 2012, 13:14

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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!

Bunuel

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21 Apr 2012, 13:18

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immune

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!

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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10 Oct 2013, 22:14

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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

Bunuel

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11 Oct 2013, 05:13

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somalwar

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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07 Dec 2015, 10:23

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Quote:

If M = √4 + ∛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

√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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17 Feb 2018, 14:30

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shrouded1

udaymathapati

Please explain the answer.

Attachment:

Image2.JPG

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

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 \(M > 2+1+1 \Rightarrow M > 4\)

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

Bunuel

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17 Feb 2018, 15:12

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dave13

shrouded1

udaymathapati

Please explain the answer.

Attachment:

Image2.JPG

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

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 \(M > 2+1+1 \Rightarrow M > 4\)

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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08 Mar 2018, 16:19

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zz0vlb

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

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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12 Mar 2018, 16:35

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zz0vlb

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

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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Bunuel

immune

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!

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: https://gmatclub.com/forum/math-number-theory-88376.html

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.

Bunuel

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11 Jun 2019, 02:39

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Diwabag

Bunuel

immune

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!

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: https://gmatclub.com/forum/math-number-theory-88376.html

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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zz0vlb

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

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)

Bunuel

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09 Oct 2019, 02:26

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zz0vlb

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

Similar but harder question to practice: https://gmatclub.com/forum/new-tough-an ... l#p1029227

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Given that \(M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}\) and we need to find the value of M

\(M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}\)
=> M = 2 + \(\sqrt[3]{4} + 2^(2/4)\) = 2 + \(\sqrt[3]{4} + \sqrt[2]{2}\)

Now, \(\sqrt[3]{4}\) will be between 1 and 2 as \(1^3\) = 1 and \(2^3\) = 8
\(\sqrt[2]{2}\) = 1.414

=> M = 2 + 1.414 + number between 1 and 2
=> M > 2 + 1.414 + 1
=> M > 4.414

So, Answer will be E
Hope it helps!

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If M = 4^(1/2) + 4^(1/3) + 4^(1/4), then the value of M is:

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