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# If x is positive, what is the value of x^(1/2)?

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Re: If x is positive, what is the value of x^(1/2)? [#permalink]
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I agree. And I am considering x to be positive (8). However the question asks for sqrt of x. It doesn't suggest that sqrt of x is positive. hence it can be +- 2(2)^1/2. Am I missing something?
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Re: If x is positive, what is the value of x^(1/2)? [#permalink]
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When the GMAT gives you a square root symbol, it’s referring to one specific value: the positive square root.

So from statement 1: x=2^3=8. Sufficient.
Statement 2: x^2=64
IxI=8; x can be +8 or,-8.
As per question stem x is positive. So only x=8 satisfied.
Hence sufficient

Ans: D

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Re: If x is positive, what is the value of x^(1/2)? [#permalink]
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vhsneha wrote:
If x is positive, what is the value of $$\sqrt{x}$$

(1) $$\sqrt[3]{x}=2$$

(2) x^2=64

Comment: The official answer is D. However, since the question stem doesn't state anything about the sign of x^1/2 (only that x is positive), i am not convinced that there is a unique answer since x^1/2 can be +- 2 {(2)^1/2}

PS: I tried using the math formula buttons. Didnt work for me. Apologize for the format

If x is positive, what is the value of $$\sqrt{x}$$?

(1) $$\sqrt[3]{x}=2$$ --> take to the third power: x = 8 --> $$\sqrt{x}=\sqrt{8}$$. Sufficient.

(2) x^2=64 --> x = 8 or x = -8. Since we are told that x is positive, then x = 8 and $$\sqrt{x}=\sqrt{8}$$. Sufficient.

vhsneha wrote:
Comment: The official answer is D. However, since the question stem doesn't state anything about the sign of x^1/2 (only that x is positive), i am not convinced that there is a unique answer since x^1/2 can be +- 2 {(2)^1/2}

PS: I tried using the math formula buttons. Didnt work for me. Apologize for the format

First of all, we are told that x is positive, so x cannot be $$-2\sqrt{2}$$. Next, the square root cannot give a negative result, that is $$\sqrt{4}=2$$ NOT +2 and -2. (In contrast the equation x^2 = 4 has TWO solutions x = 2 and x = -2).

arosman wrote:
You can't have the square root of a negative number. Irrational numbers are way beyond the scope of GMAT. If a question ask for $$\sqrt{x}$$ you can assume x is positive or zero.

Yes, even roots from negative numbers are not defined for the GMAT ($$\sqrt[even]{negative}$$ is undefined). So, you don't need complex numbers for the GMAT.

GMAT deals with only real numbers: integers (-3, -2, -1, 0, 1, 2, 3, ...), fractions/decimals (3/2, 4/3, 0.7, 17.5, ...) and irrational numbers ($$\sqrt{3}$$, $$\sqrt{2}$$, $$\pi$$, ...).

Check for more below:

2. Properties of Integers

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
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Re: If x is positive, what is the value of x^(1/2)? [#permalink]
1) x^1/3=2
which means x=2^3= 8
therefore x=8 and root x = root 8
(sufficient)

2) x^2=64
which means x=root 64, ie, 8
therefore, x=8 and root x= root 8
(sufficient)
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Re: If x is positive, what is the value of x^(1/2)? [#permalink]
KarishmaB

Madam, I did not get the answer. How could the answer be D? As per statement 2, x^2 = 64 which means x=8 and x^(1/2) could be +/- 2*(2)^(1/2). I would use another simple example to explain this, if x = 4 then square root of x could be +/-2.

Much to my surprise this question is of 5% difficulty level. Please help me with this concept. Thanks a lot in advance Madam!
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Re: If x is positive, what is the value of x^(1/2)? [#permalink]
waytowharton wrote:
KarishmaB

Madam, I did not get the answer. How could the answer be D? As per statement 2, x^2 = 64 which means x=8 and x^(1/2) could be +/- 2*(2)^(1/2). I would use another simple example to explain this, if x = 4 then square root of x could be +/-2.

Much to my surprise this question is of 5% difficulty level. Please help me with this concept. Thanks a lot in advance Madam!

I think you are missing the crucial info from the stem

If x is positive, what is the value of $$\sqrt{x}$$?

So, from x^2=64, when you get that x = 8 or x = -8, you should discard x = -8 because we are explicitly given that x is positive and you'll get only one value: x = 8, thus $$\sqrt{x}=\sqrt{8}=2\sqrt{2}$$.

Does this make sense?
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Re: If x is positive, what is the value of x^(1/2)? [#permalink]
Bunuel wrote:
waytowharton wrote:
KarishmaB

Madam, I did not get the answer. How could the answer be D? As per statement 2, x^2 = 64 which means x=8 and x^(1/2) could be +/- 2*(2)^(1/2). I would use another simple example to explain this, if x = 4 then square root of x could be +/-2.

Much to my surprise this question is of 5% difficulty level. Please help me with this concept. Thanks a lot in advance Madam!

I think you are missing the crucial info from the stem

If x is positive, what is the value of $$\sqrt{x}$$?

So, from x^2=64, when you get that x = 8 or x = -8, you should discard x = -8 because we are explicitly given that x is positive and you'll get only one value: x = 8, thus $$\sqrt{x}=\sqrt{8}=2\sqrt{2}$$.

Does this make sense?

Thanks Bunuel for your reply. I did not miss the condition that x is positive. But i would like to highlight that we are not given that x^(1/2) will be positive. It could be both positive and negative. Since square of neg is also positice. For example, If we are given x=4 then square root of x is +2 and -2.

Please do let me know flaw in my reasoning. Thanks in advance!

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If x is positive, what is the value of x^(1/2)? [#permalink]
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waytowharton wrote:
Bunuel wrote:
waytowharton wrote:
KarishmaB

Madam, I did not get the answer. How could the answer be D? As per statement 2, x^2 = 64 which means x=8 and x^(1/2) could be +/- 2*(2)^(1/2). I would use another simple example to explain this, if x = 4 then square root of x could be +/-2.

Much to my surprise this question is of 5% difficulty level. Please help me with this concept. Thanks a lot in advance Madam!

I think you are missing the crucial info from the stem

If x is positive, what is the value of $$\sqrt{x}$$?

So, from x^2=64, when you get that x = 8 or x = -8, you should discard x = -8 because we are explicitly given that x is positive and you'll get only one value: x = 8, thus $$\sqrt{x}=\sqrt{8}=2\sqrt{2}$$.

Does this make sense?

Thanks Bunuel for your reply. I did not miss the condition that x is positive. But i would like to highlight that we are not given that x^(1/2) will be positive. It could be both positive and negative. Since square of neg is also positice. For example, If we are given x=4 then square root of x is +2 and -2.

Please do let me know flaw in my reasoning. Thanks in advance!

Posted from my mobile device

So, you are saying that $$\sqrt{8}$$ could be $$2\sqrt{2}$$ or $$-2\sqrt{2}$$. That's not true.

$$\sqrt{...}$$ 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 (and generally in math) 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;

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 $$x=\sqrt{9}=3$$.

Thus, $$\sqrt{8}=2\sqrt{2}$$ only.

Hope it helps.
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Re: If x is positive, what is the value of x^(1/2)? [#permalink]
Thanks a lot Bunuel! This is really very helpful.
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If x is positive, what is the value of x^(1/2)? [#permalink]
Bunuel wrote:
vhsneha wrote:
If x is positive, what is the value of $$\sqrt{x}$$

(1) $$\sqrt[3]{x}=2$$

(2) x^2=64

Comment: The official answer is D. However, since the question stem doesn't state anything about the sign of x^1/2 (only that x is positive), i am not convinced that there is a unique answer since x^1/2 can be +- 2 {(2)^1/2}

PS: I tried using the math formula buttons. Didnt work for me. Apologize for the format

If x is positive, what is the value of $$\sqrt{x}$$?

(1) $$\sqrt[3]{x}=2$$ --> take to the third power: x = 8 --> $$\sqrt{x}=\sqrt{8}$$. Sufficient.

(2) x^2=64 --> x = 8 or x = -8. Since we are told that x is positive, then x = 8 and $$\sqrt{x}=\sqrt{8}$$. Sufficient.

vhsneha wrote:
Comment: The official answer is D. However, since the question stem doesn't state anything about the sign of x^1/2 (only that x is positive), i am not convinced that there is a unique answer since x^1/2 can be +- 2 {(2)^1/2}

PS: I tried using the math formula buttons. Didnt work for me. Apologize for the format

First of all, we are told that x is positive, so x cannot be $$-2\sqrt{2}$$. Next, the square root cannot give a negative result, that is $$\sqrt{4}=2$$ NOT +2 and -2. (In contrast the equation x^2 = 4 has TWO solutions x = 2 and x = -2).

arosman wrote:
You can't have the square root of a negative number. Irrational numbers are way beyond the scope of GMAT. If a question ask for $$\sqrt{x}$$ you can assume x is positive or zero.

Yes, even roots from negative numbers are not defined for the GMAT ($$\sqrt[even]{negative}$$ is undefined). So, you don't need complex numbers for the GMAT.

GMAT deals with only real numbers: integers (-3, -2, -1, 0, 1, 2, 3, ...), fractions/decimals (3/2, 4/3, 0.7, 17.5, ...) and irrational numbers ($$\sqrt{3}$$, $$\sqrt{2}$$, $$\pi$$, ...).

Check for more below:

2. Properties of Integers

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hey Bunuel, I still can't wrap my head around this question.

Although i DO agree that only x can be determined to be +8 based on the stem, how can we presume that the sqrt of +8 cannot be negative?

For example, given that x = +4, then sqrt of +4 can be +/-2.

multiplying -2*-2 will return the positive x value +4.

In the case of the question - why would this not apply..?

it could effectively be written as:

$$[(-\sqrt{8})^2]^(1/3)$$

EDIT: i believe the confusing part about this questions is that the question is not asking you to TAKE the square root of x, but rather assume that the expression provided is already $$\sqrt{x}$$, implying that the number will always be positive.
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