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Square Root = Always Positive?

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Re: Square Root = Always Positive?  [#permalink]

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New post 14 Jan 2019, 05:40
Bunuel wrote:
MartinSC wrote:
I was counting on that, but what about this one:

If (x-1)^2=400, which of the following could be the value of ?
(A) 15
(B) 14
(C) –24
(D) –25
(E) –26

I thought "16" applying the "always positive" mantra, but the explanation goes as follows:

"Work the problem by taking the square root of both sides and solving for x.

Thus,

(x-1)^2=400
x-1= +/-20 (wait, I was not expecting to see the negative root!!)

So x= -19 or x=21 and
x-5=-24 or x-5=16" (which is not even an option).

Seeing that 16 was not among the choices, I could have picked (C)-24, which would have been correct, but it goes against the theory of the positive root...so I was left scratching my head.

Anyone has an idea?

Thks!


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 positive 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 \(x=\sqrt{9}=3\).





\(p^2 = (q+1)^2\)

Now according to you, this should give 2 solutions.
But If I square root this, then it should give only 1 solution.

I don't get it, it confuses me even more, what is the answer to this equation? The official answer says it has only 1 solution, Now how is that?
What would this give?
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Re: Square Root = Always Positive?  [#permalink]

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New post 14 Jan 2019, 05:51
Akshit03 wrote:
Bunuel wrote:
MartinSC wrote:
I was counting on that, but what about this one:

If (x-1)^2=400, which of the following could be the value of ?
(A) 15
(B) 14
(C) –24
(D) –25
(E) –26

I thought "16" applying the "always positive" mantra, but the explanation goes as follows:

"Work the problem by taking the square root of both sides and solving for x.

Thus,

(x-1)^2=400
x-1= +/-20 (wait, I was not expecting to see the negative root!!)

So x= -19 or x=21 and
x-5=-24 or x-5=16" (which is not even an option).

Seeing that 16 was not among the choices, I could have picked (C)-24, which would have been correct, but it goes against the theory of the positive root...so I was left scratching my head.

Anyone has an idea?

Thks!


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 positive 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 \(x=\sqrt{9}=3\).





\(p^2 = (q+1)^2\)

Now according to you, this should give 2 solutions.
But If I square root this, then it should give only 1 solution.

I don't get it, it confuses me even more, what is the answer to this equation? The official answer says it has only 1 solution, Now how is that?
What would this give?


I think you messed up something there... If it's \(p^2 = (q+1)^2\), then this equation with TWO unknowns p and q will have infinitely many solutions.

If it'st \(p^2=(p+1)^2\), then:

\(p^2= p^2+2p+1\);

\(p=-\frac{1}{2}\).
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Re: Square Root = Always Positive?  [#permalink]

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New post 14 Jan 2019, 10:03
Bunuel


I'm sorry for not mentioning that I'm not looking for absolute value.
This was a DS question I was stuck on due to this concept that I still cannot understand.

I only need to know whether p would be (q+1) OR 2 values{(q+1),-(q+1)}
Now taking this into consideration and what I said previously, Can you correct my logic/approach/concept? I have spent 1 hour on this only.
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Re: Square Root = Always Positive?  [#permalink]

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New post 14 Jan 2019, 10:20
Akshit03 wrote:
Bunuel


I'm sorry for not mentioning that I'm not looking for absolute value.
This was a DS question I was stuck on due to this concept that I still cannot understand.

I only need to know whether p would be (q+1) OR 2 values{(q+1),-(q+1)}
Now taking this into consideration and what I said previously, Can you correct my logic/approach/concept? I have spent 1 hour on this only.


\(x^2 = positive \ number\) ALWAYS has TWO solutions: \(x = \sqrt{positive \ number}\) and \(x = -\sqrt{positive \ number}\).

For example, x^2 = 9 --> \(x = \sqrt{9}=3\) or \(x = -\sqrt{9}=-3\).

\(p^2=(q+1)^2\) means that \(|p|=|q+1|\), so p=q+1 or p=-(q+1)

P.S. What DS question are you talking about?
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Re: Square Root = Always Positive?  [#permalink]

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New post 14 Jan 2019, 10:42
Bunuel wrote:
Akshit03 wrote:
Bunuel


I'm sorry for not mentioning that I'm not looking for absolute value.
This was a DS question I was stuck on due to this concept that I still cannot understand.

I only need to know whether p would be (q+1) OR 2 values{(q+1),-(q+1)}
Now taking this into consideration and what I said previously, Can you correct my logic/approach/concept? I have spent 1 hour on this only.


\(x^2 = positive \ number\) ALWAYS has TWO solutions: \(x = \sqrt{positive \ number}\) and \(x = -\sqrt{positive \ number}\).

For example, x^2 = 9 --> \(x = \sqrt{9}=3\) or \(x = -\sqrt{9}=-3\).

\(p^2=(q+1)^2\) means that \(|p|=|q+1|\), so p=q+1 or p=-(q+1)

P.S. What DS question are you talking about?



Q: If p and q are positive integers, is the greatest common factor of p and q greater than 1?
1: \(p^2 = pq + p\)
2: \(p^2 = q^2 + 2q + 1\)

So you mean to say, if GMAT gives you square root, we take only positive.
But when it gives square like \(x^2 = 9\) , we take both values?
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Re: Square Root = Always Positive?  [#permalink]

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New post 14 Jan 2019, 11:10
Akshit03 wrote:
Bunuel wrote:
Akshit03 wrote:
Bunuel


I'm sorry for not mentioning that I'm not looking for absolute value.
This was a DS question I was stuck on due to this concept that I still cannot understand.

I only need to know whether p would be (q+1) OR 2 values{(q+1),-(q+1)}
Now taking this into consideration and what I said previously, Can you correct my logic/approach/concept? I have spent 1 hour on this only.


\(x^2 = positive \ number\) ALWAYS has TWO solutions: \(x = \sqrt{positive \ number}\) and \(x = -\sqrt{positive \ number}\).

For example, x^2 = 9 --> \(x = \sqrt{9}=3\) or \(x = -\sqrt{9}=-3\).

\(p^2=(q+1)^2\) means that \(|p|=|q+1|\), so p=q+1 or p=-(q+1)

P.S. What DS question are you talking about?



Q: If p and q are positive integers, is the greatest common factor of p and q greater than 1?
1: \(p^2 = pq + p\)
2: \(p^2 = q^2 + 2q + 1\)

So you mean to say, if GMAT gives you square root, we take only positive.
But when it gives square like \(x^2 = 9\) , we take both values?


Notice that in the question we are told that p and q are POSITIVE. So, here from \(p^2=(q+1)^2\) it follows that \(p=q+1\) only.

Finally, to answer your question, check the discussion on page 1. For example, this one: https://gmatclub.com/forum/square-root- ... l#p1437877

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 positive 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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Re: Square Root = Always Positive?   [#permalink] 14 Jan 2019, 11:10

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