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555-605 Level|   Algebra|      
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Bunuel
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The numbers a, b, and c are all non-zero integers. Is a > 0? 02 Mar 2015, 08:15
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D
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The numbers a, b, and c are all non-zero integers. Is a > 0?

(1) a = b^2
(2) \(a=\sqrt{c}\)


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D
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The numbers a, b, and c are all non-zero integers. Is a > 0? 02 Mar 2015, 13:09
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(1) Sufficient. A square of a number is always positive. So, a>0.
(2) Sufficient. A square root of a number will give a positive answer. So, a>0.

Answer D
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The numbers a, b, and c are all non-zero integers. Is a > 0? 02 Mar 2015, 21:34
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viktorija
(1) Sufficient. A square of a number is always positive. So, a>0.
(2) Sufficient. A square root of a number will give a positive answer. So, a>0.

Answer D

Note that the square of a number or a square root can be 0 too (if the number is 0). Here, you can given that a, b and c are non zero integers. That is why you know that the square root or square will not be 0.
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The numbers a, b, and c are all non-zero integers. Is a > 0? 03 Mar 2015, 08:23
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viktorija
(1) Sufficient. A square of a number is always positive. So, a>0.
(2) Sufficient. A square root of a number will give a positive answer. So, a>0.

Answer D

Note that the square of a number or a square root can be 0 too (if the number is 0). Here, you can given that a, b and c are non zero integers. That is why you know that the square root or square will not be 0.

Good catch. If the problem didn't state non-zero integers, the answer would be (E).

As per the problem:

St(1) tells us that that a is positive because the square of any integer, (number for that matter) other than zero is positive.

St (2) tell us that a is positive because you cannot take the square root of a negative number. This is a property you should know.

It's also important to note the impact the statement "non-zero integeres" in the problem. If this wasn't specified, the answer would be E.

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The numbers a, b, and c are all non-zero integers. Is a > 0? 04 Mar 2015, 08:38
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[1] : Sufficient: Square of negative or positive number is always positive. Since b is non-zero integer, hence a>0.
[2] : Sufficient: Only square root of positive number produces real number. Square root of negative number is complex. Since a and c is non-zero integer(real), so square root of c will always be positive. i.e. a>0 .

Answer D.
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The numbers a, b, and c are all non-zero integers. Is a > 0? 08 Mar 2015, 16:06
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The numbers a, b, and c are all non-zero integers. Is a > 0?

(1) a = b^2
(2) \(a=\sqrt{c}\)


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MAGOOSH OFFICIAL SOLUTION:

All that is given in the prompt is that a, b, and c are non-zero integers.

Statement #1: the result of square anything is always positive, so whether b is negative or positive, a must be positive. This statement, by itself, is sufficient.

Statement #2: since the square root symbol is printed as part of the problem, the output of the sqrt{c} must be positive. We know for a fact that a must be positive. Again, this statement, by itself, is sufficient.

Both statements are sufficient. Answer = D.
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The numbers a, b, and c are all non-zero integers. Is a > 0? 23 Apr 2018, 18:33
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Bunuel chetan2u VeritasPrepKarishma amanvermagmat niks18


Quote:
The numbers a, b, and c are all non-zero integers. Is a > 0?

(1) a = b^2
(2) \(a=\sqrt{c}\)


Is this approach correct for St 2?

\(\sqrt{4}\) = 2
\(\sqrt{4}\) = -2

My both conditions of all three numbers as non-zero integers are satisfied and
I am still getting two values for a.

I did not understand this part of OE:
since the square root symbol is printed as part of the problem, the output of the sqrt{c} must be positive.
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The numbers a, b, and c are all non-zero integers. Is a > 0? 23 Apr 2018, 21:17
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Bunuel chetan2u VeritasPrepKarishma amanvermagmat niks18


Quote:
The numbers a, b, and c are all non-zero integers. Is a > 0?

(1) a = b^2
(2) \(a=\sqrt{c}\)


Is this approach correct for St 2?

\(\sqrt{4}\) = 2
\(\sqrt{4}\) = -2

My both conditions of all three numbers as non-zero integers are satisfied and
I am still getting two values for a.

I did not understand this part of OE:
since the square root symbol is printed as part of the problem, the output of the sqrt{c} must be positive.

\(\sqrt{4}=2\) ONLY, not +2/-2.

I tried to explain this to you in this topic: https://gmatclub.com/forum/how-many-bit ... l#p2029387

\(\sqrt{...}\) 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;

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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The numbers a, b, and c are all non-zero integers. Is a > 0? 21 Mar 2022, 03:52
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