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Intern  B
Joined: 07 Jul 2018
Posts: 44
Re: Square Root = Always Positive?  [#permalink]

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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;
$$\sqrt{16} = 2$$, 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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58420
Re: Square Root = Always Positive?  [#permalink]

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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;
$$\sqrt{16} = 2$$, 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:

$$p^2= p^2+2p+1$$;

$$p=-\frac{1}{2}$$.
_________________
Intern  B
Joined: 07 Jul 2018
Posts: 44
Re: Square Root = Always Positive?  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58420
Re: Square Root = Always Positive?  [#permalink]

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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?
_________________
Intern  B
Joined: 07 Jul 2018
Posts: 44
Re: Square Root = Always Positive?  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58420
Re: Square Root = Always Positive?  [#permalink]

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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;
$$\sqrt{16} = 2$$, 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 $$x=\sqrt{9}=3$$.
_________________ Re: Square Root = Always Positive?   [#permalink] 14 Jan 2019, 11:10

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