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Math Expert V
Joined: 02 Sep 2009
Posts: 59561

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stan3544 wrote:
Bunuel 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?

Not sure I understand what you mean but GMAT doe not have its own math. That's simply not correct. The only thing is that GMAT deals with only Real Numbers: Integers, Fractions and Irrational Numbers, so no imaginary numbers. That's why the even roots from negative numbers are not defined for the GMAT.

(-2)^4 = 16 but -2^4 = -16, this is true generally not only for the GMAT.

Hey Bunuel,

Now i am totally confused. Yes it is obvious and universally understood that sqr.root of negative number is not a real number, basically it doesn't exist, why? Because no negative number raised to an even power will ever give us a negative result, yet according to your example -2^4 = -16, so we know that the number -2 raised to the power 4 gives us -16, isn't it a paradox? Now, according to the exponent rules: (-2)^4 = (-2^1)^4 = -2*(-2)*(-2)*(-2) right? Now according to your interpretation -2^4 is not the same, so it basically means -2*2*2*2, which doesn't make sense, because this goes against the whole idea behind exponents, which is: we take a number and multiply by ITSELF n times. So the correct way to write it would be -(2^4) = -16 or -(-2^4) = -16. Now back to the issue at hand: logically 2^2=4 as well as -2*(-2)=4, which is (-2)^2= 4, now i don't see why sqr.root of 4 can only be 2 and not -2 as well.

-2^4 = -(2^4) = -16. It's order of operations thing.

As for the square roots: I tried to explain it several times on previous pages but I'll try this one last time:

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. Even roots have only a positive value on the GMAT.

In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5.

If it's still unclear or confusing then probably it's better to start from basics and brush-up fundamentals or enrol into a class.
Current Student B
Joined: 17 Nov 2016
Posts: 13
Location: United States
Schools: Duke Fuqua
GMAT 1: 720 Q49 V39 GPA: 3.3

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

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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 $$x = -\sqrt{4}=-2$$.
Intern  Joined: 05 Apr 2017
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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 ?
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

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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.
Current Student B
Joined: 17 Nov 2016
Posts: 13
Location: United States
Schools: Duke Fuqua
GMAT 1: 720 Q49 V39 GPA: 3.3

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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):

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 $$x= -2$$
Senior Manager  P
Joined: 17 Mar 2014
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very good question. Definitely eye opener.
Manager  S
Joined: 21 Aug 2017
Posts: 77
Location: United States
Schools: Oxford"20 (A)
GMAT 1: 700 Q43 V42 ### Show Tags

I love this question.

Did anyone come across any similar ones in the forum question banks? I'm not really able to drill down enough on filters to find similar ones with any consistency.
Intern  Joined: 28 Jun 2018
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I think this is a poor-quality question and I don't agree with the explanation. the solution of minus two is correct , you can have the 4th root of (X=-2) it is a 4th root of 16 ,and 16 it is a positive number which can use for the root.
Math Expert V
Joined: 02 Sep 2009
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Eyaltau wrote:
I think this is a poor-quality question and I don't agree with the explanation. the solution of minus two is correct , you can have the 4th root of (X=-2) it is a 4th root of 16 ,and 16 it is a positive number which can use for the root.

You are wrong. Your doubt is addressed MANY times on previous pages:

$$\sqrt{}$$ denotes a function. Mathematically the square root function cannot give negative result. When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

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.

P.S. Generally I would suggest to brush-up fundamental before attempting hard questions.
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Joined: 14 Feb 2017
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Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31 GMAT 6: 600 Q38 V35 GPA: 3
WE: Management Consulting (Consulting)

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Can't we just plug-in the solutions back to the equation to test whether it works? Bunuel - can you please validate this approach?

power of 4 of LHS
$$x^4 = x^3 + 6x^2$$$$x^4 - x^3+6x^2 = 0$$
$$x^2(x^2 - x - 6)= 0$$$$x^2 (x-3)(x+2)=0$$
$$x=3, x=-2$$ and $$x=0$$
Test valid solutions by plugging back in the roots.

$$x=-2$$
$$(-2)^2((-2)^2-(-2)-6)$$$$= 4(4 + 4-6)=0$$
4(2) =8 and is not equal to 0 therefore -2 is not a valid solution

$$x=3$$
$$(3)^2(3^2 - 3-6)$$
$$9(9-3-6)= 9(0) =0$$therefore valid.

Only 3 is a valid solution
Intern  B
Joined: 24 Feb 2014
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Location: United States (GA)
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I think this is a high-quality question and I agree with explanation.
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

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dcummins wrote:
Can't we just plug-in the solutions back to the equation to test whether it works? Bunuel - can you please validate this approach?

power of 4 of LHS
$$x^4 = x^3 + 6x^2$$$$x^4 - x^3+6x^2 = 0$$
$$x^2(x^2 - x - 6)= 0$$$$x^2 (x-3)(x+2)=0$$
$$x=3, x=-2$$ and $$x=0$$
Test valid solutions by plugging back in the roots.

$$x=-2$$
$$(-2)^2((-2)^2-(-2)-6)$$$$= 4(4 + 4-6)=0$$
4(2) =8 and is not equal to 0 therefore -2 is not a valid solution

$$x=3$$
$$(3)^2(3^2 - 3-6)$$
$$9(9-3-6)= 9(0) =0$$therefore valid.

Only 3 is a valid solution

Yes, that would be a valid approach. Solving, then verifying.
Senior Manager  B
Joined: 03 Sep 2018
Posts: 252
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GMAT 1: 710 Q48 V40 GMAT 2: 780 Q50 V49 GMAT 3: 760 Q49 V44 GPA: 4

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Hi Bunuel,

I tried to use vieta's theorem

$$x_1+x_2=\frac{-b}{a}$$

But now I realise that the theorem itself does not guarantee that the solution is not extraneous, right?

Could you confirm that?
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Good luck to you.
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

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SchruteDwight wrote:
Hi Bunuel,

I tried to use vieta's theorem

$$x_1+x_2=\frac{-b}{a}$$

But now I realise that the theorem itself does not guarantee that the solution is not extraneous, right?

Could you confirm that?

Yes because you used it for the quadratics which you got AFTER you squared the whole expression, which (squaring) created an extra, false root.
Senior Manager  B
Joined: 03 Sep 2018
Posts: 252
Location: Netherlands
GMAT 1: 710 Q48 V40 GMAT 2: 780 Q50 V49 GMAT 3: 760 Q49 V44 GPA: 4

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Bunuel wrote:
SchruteDwight wrote:
Hi Bunuel,

I tried to use vieta's theorem

$$x_1+x_2=\frac{-b}{a}$$

But now I realise that the theorem itself does not guarantee that the solution is not extraneous, right?

Could you confirm that?

Yes because you used it for the quadratics which you got AFTER you squared the whole expression, which (squaring) created an extra, false root.

Since squaring is a non-reversible operation. Makes sense, thank you!
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# D01-13

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