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# D01-13

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Intern
Joined: 05 Apr 2017
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31 May 2017, 09:13
Can someone please provide the logic behind not taking the negative root for '√9=3, NOT +3 or -3;'
Is this mentioned anywhere int the OG or any other official material?
How do we know the GMAT doesn't consider the negative root ?

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31 May 2017, 10:20
AllenF wrote:
Can someone please provide the logic behind not taking the negative root for '√9=3, NOT +3 or -3;'
Is this mentioned anywhere int the OG or any other official material?
How do we know the GMAT doesn't consider the negative root ?

This is explained many times. $$\sqrt{}$$ denotes a function. Mathematically the square root function cannot give negative result.

OFFICIAL GUIDE:
$$\sqrt{n}$$ denotes the positive number whose square is n.

Piece of advice, people here, experts/tutors, are here to help students, not to deceive them. So, if you see the same thing repeated many times by experts/tutors you should now that it's true.
_________________

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31 May 2017, 19:33
Bunuel wrote:
CristianJuarez wrote:
stan3544 wrote:
And here we go again. Sometimes i feel as if GMAT has it's own version of math that goes against universal one. I understand that on verbal there can be a whole lot of inerprteations of rules, simply because it is VERBAL, as in not a precise science which is basically a human imagination. But math has to be precise. And now, according to e-GMAT -b^m if m is even --> positive number and you are stating that −2^4=−16 and i made a whole lot of mistakes when practicing quant from both OG and scholaranium that involved sqr.root of a number, because i didn't include the negative value as well. Can comeone give a link or a screenshot of an OG question where when we take sqr root we don't consider the negative value?

The gmat doesn't have its own version of math. The square root is a function, that's why it is always positive. As a function, it can't take a positive and a negative value (if it did, it wouldn't be a function, you can look at the graph to check that out).

When you solve x^2 = 4, applying the square root function and assuming two results for x is incorrect, because you would only get x = 2. The formal and correct method is:
x^2 = 4
x^2 - 4 =0
(x + 2)(x - 2)=0 and from here get both results for x.

That's correct. + 1.

Another way of solving x^2 = 4 would be:

$$x = \sqrt{4}=2$$ or $$x = -\sqrt{4}=-2$$.

Yes, actually I'd like to correct myself, I hope it isn't more confusing.

Bunuel's solution applying square root function comes from the fact that the function f(x)=x^2 allows x to take positive or negative values (or 0):

I don't have 5 posts so I can't include the image , please search for it on google.

Meanwhile, the function $$f(x)= \sqrt{x}$$ doesn't (this is why square root is always positive):

(check image on google)

So, if people want to solve this excercise by applying square root function, you must consider that this is incorrect:

$$x^2=4$$ Apply $$\sqrt{}$$
$$\sqrt{x^2}=\sqrt{4}$$
$$x = \sqrt{4}$$
$$x=2$$ or $$x= -2$$

It is incorrect because $$\sqrt{4}$$ is always 2 and you lost a solution in the process.

So, you have to consider that mathemathically, $$\sqrt{x^2} = |x|$$. This is only because $$f(x)=x^2$$ can take a positive or a negative value for x (or 0 of course). Indeed, this doesn't mean the square root takes a positive or a negative solution, its always positive (or zero)!! (remember, $$|x|$$ is always $$>= 0$$ for any value of $$x$$).

You can check it out like this:
$$\sqrt{4} = \sqrt{2^2}= |2|=2, not -2!!$$ or:
$$\sqrt{4} = \sqrt{(-2)^2}= |-2|=2, not -2!!$$ It's impossible to get to $$-2$$ like that

So the correct process by applying square root function is:
$$x^2=4$$ Apply $$\sqrt{}$$
$$\sqrt{x^2}=\sqrt{4}$$
$$|x| = 2$$
$$x=2$$ or $$x= -2$$

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18 Aug 2017, 13:43
I think this is a high-quality question and I agree with explanation.

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Re D01-13   [#permalink] 18 Aug 2017, 13:43

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# D01-13

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