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# M26-25

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Math Expert
Joined: 02 Sep 2009
Posts: 44650

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10 Dec 2016, 05:58
cuhmoon wrote:
Doubt for this q:

$$\sqrt{-y * |y|}$$

= $$\sqrt{y^2}$$

and Square root of Y has to be the positive root.. Thus = y

Thus the answer must be -1-y.

I do understand that $$\sqrt{x^2}$$= |x| ... and if X is -ve then mod x = -x, but we don't have to simplify$$\sqrt{y^2}$$ to mod y.
We can just take the positive root.. since per GMAT anything under the square root sign is the positive root ?

What am I missing here?

I guess my q is on the GMAT, square root of numbers are ONLY positive..

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Joined: 11 Oct 2016
Posts: 88
Location: India
GMAT 1: 610 Q47 V28

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25 Mar 2017, 20:46
Very Good question
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Joined: 09 Jun 2016
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05 Dec 2017, 23:09
Bunuel wrote:
blitheclyde wrote:
Hi,

Having trouble wrapping my head around why $$\sqrt{x^2}$$ is equal to |x|

I looked at the post on this page, but did not see any explanation.

For example, how is the equation $$\sqrt{4}$$ = {-2, 2} any different from $$\sqrt{2^2}$$ = ? Is it because the solution of $$\sqrt{x^2}$$ must be equal to the solution of $$(\sqrt{x})^2$$? if so, why?

Thanks in advance for any help!

(with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "$$\sqrt{25}$$ = 5" as though this is given, but my understanding is that $$\sqrt{25}$$ could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion)

$$\sqrt{4}=2$$ ONLY, not 2 or -2. The square root function cannot give negative result. Check here: http://gmatclub.com/forum/m26-184464.html#p1729559

Hi Bunuel, certain application of the sqrt function seem to be contradictory. If sqrt function can't give negative numbers, why do we substitute both -ve and +ve values of $$\sqrt{expression}$$ while solving algebraic equations.
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Joined: 02 Sep 2009
Posts: 44650

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05 Dec 2017, 23:22
abhishek911 wrote:
Bunuel wrote:
blitheclyde wrote:
Hi,

Having trouble wrapping my head around why $$\sqrt{x^2}$$ is equal to |x|

I looked at the post on this page, but did not see any explanation.

For example, how is the equation $$\sqrt{4}$$ = {-2, 2} any different from $$\sqrt{2^2}$$ = ? Is it because the solution of $$\sqrt{x^2}$$ must be equal to the solution of $$(\sqrt{x})^2$$? if so, why?

Thanks in advance for any help!

(with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "$$\sqrt{25}$$ = 5" as though this is given, but my understanding is that $$\sqrt{25}$$ could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion)

$$\sqrt{4}=2$$ ONLY, not 2 or -2. The square root function cannot give negative result. Check here: http://gmatclub.com/forum/m26-184464.html#p1729559

Hi Bunuel, certain application of the sqrt function seem to be contradictory. If sqrt function can't give negative numbers, why do we substitute both -ve and +ve values of $$\sqrt{expression}$$ while solving algebraic equations.

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: M26-25   [#permalink] 05 Dec 2017, 23:22

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# M26-25

Moderators: chetan2u, Bunuel

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