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# D01-13

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Intern
Joined: 07 Dec 2015
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17 Mar 2016, 08:21
I think this the explanation isn't clear enough, please elaborate.
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17 Mar 2016, 08:24
I think this the explanation isn't clear enough, please elaborate. i dont understand why -2 is not a solution .
for instance squrt of 4 has 2 soln i.e 2 and -2. please explain?
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17 Mar 2016, 10:27
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I think this the explanation isn't clear enough, please elaborate. i dont understand why -2 is not a solution .
for instance squrt of 4 has 2 soln i.e 2 and -2. please explain?

That's not true.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, 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.
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26 Mar 2016, 16:24
Hi Bunuel, would the following be a valid approach to this question or just luck?

1.) Re-arrange and factor out $$x^2: x^2(x^2−x−6)=0$$ --> So one solution is 0 -> $$x^2$$
2.) Use discriminant for the left equation $$(x^2−x−6)=0$$: $$b^2 - 4ac$$ --> 1-(4*1*(-6)) -> 25 -> Since 25 is positive the left equation has two solutions
3.) 1.) + 2.) = 3 Solutions
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27 Mar 2016, 02:47
Hi Bunuel, would the following be a valid approach to this question or just luck?

1.) Re-arrange and factor out $$x^2: x^2(x^2−x−6)=0$$ --> So one solution is 0 -> $$x^2$$
2.) Use discriminant for the left equation $$(x^2−x−6)=0$$: $$b^2 - 4ac$$ --> 1-(4*1*(-6)) -> 25 -> Since 25 is positive the left equation has two solutions
3.) 1.) + 2.) = 3 Solutions

You should find the actual roots and find out whether all of them are valid. Only 2 of them are valid for the equation at hand. Please refer to the discussion on previous 2 pages.
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Joined: 17 Mar 2016
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19 Apr 2016, 13:07
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 19 Jul 2016
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20 Jul 2016, 01:24
I am still not able to convince my self that -2 is not the root of the given equation. Because as per my knowledge this should have been the one solution. Now, the dilemma is what i will do in similar questions.
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Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
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07 Aug 2016, 02:49
I think this is a high-quality question and I agree with explanation.
Intern
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21 Sep 2016, 10:38
how did the following question become x*4-x*3-6x*2=0?
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22 Sep 2016, 00:59
manchitkapoor wrote:
how did the following question become x*4-x*3-6x*2=0?

It does not.

Take $$x=\sqrt[4]{x^3+6x^2}$$ to the 4th power to get $$x^4=x^3+6x^2$$, which when re-arranged becomes $$x^4-x^3-6x^2=0$$. It's x to the 4th power, minus x cubed, minus 6 times x squared, equal 0.
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GMAT 1: 520 Q43 V27
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25 Nov 2016, 08:17
This is a nice question. I forgot that 0 is a multiple of 9.
Manager
Joined: 14 Oct 2012
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01 Apr 2017, 09:38
Bunuel wrote:
manchitkapoor wrote:
how did the following question become x*4-x*3-6x*2=0?

It does not.

Take $$x=\sqrt[4]{x^3+6x^2}$$ to the 4th power to get $$x^4=x^3+6x^2$$, which when re-arranged becomes $$x^4-x^3-6x^2=0$$. It's x to the 4th power, minus x cubed, minus 6 times x squared, equal 0.

I think this should clarifies everybody's confusion regarding the answer, i hope everyone understands how we got 0,3,(-2)

$$x=\sqrt[4]{x^3+6x^2}$$
x=0
$$0=\sqrt[4]{0^3+6*0^2}$$
$$0=\sqrt[4]{0}$$
$$0=0$$

$$x=\sqrt[4]{x^3+6x^2}$$
x=3
$$3=\sqrt[4]{3^3+6*3^2}$$
$$3=\sqrt[4]{27+6*9}$$
$$3=\sqrt[4]{27+54}$$
$$3=\sqrt[4]{81}$$
$$3=3$$

$$x=\sqrt[4]{x^3+6x^2}$$
x=(-2)
$$(-2)=\sqrt[4]{(-2)^3+6*(-2)^2}$$
$$(-2)=\sqrt[4]{-8+6*4}$$
$$(-2)=\sqrt[4]{-8+24}$$
$$(-2)=\sqrt[4]{16}$$
$$(-2) =| 2$$
Thus solution x = (-2) isn't applicable.
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Joined: 26 Sep 2016
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12 Apr 2017, 04:49
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?
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12 Apr 2017, 04:58
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.
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12 Apr 2017, 05:58
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.
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12 Apr 2017, 06:05
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 $$\sqrt[4]{x}$$, 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.
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12 Apr 2017, 07:37
Thanks for the explanation

Maybe i wasn't able to explain clearly what i was confused about, but i found the answer on the Magoosh website, boldfaced is the explanation i needed: The equation x^2 = 16 has two solutions, x = +4 and x = -4, because 4^2 = 16 and (-4)^2 = 16, and the GMAT will impale you for only remembering one of those two. At the same time, sqrt{16} has only one output: sqrt{16} = +4 only. When you yourself undo a square by taking a square root, that’s a process that results in two possibilities, but when you see this symbol as such, printed as part of the problem, it means find the positive square root only.

Hope this clears all the doubts for everyone.
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01 May 2017, 09:38
Bunuel wrote:
Official Solution:

If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is:

A. $$-2$$
B. $$0$$
C. $$1$$
D. $$3$$
E. $$5$$

Take the given expression to the 4th power: $$x^4=x^3+6x^2$$;

Re-arrange and factor out x^2: $$x^2(x^2-x-6)=0$$;

Factorize: $$x^2(x-3)(x+2)=0$$;

So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.

The sum of all possible solutions for x is 0+3=3.

Thank you for this valuable info. Because in engineering we used to consider root(4) = + / - 2 . Now, I know that GMAT only considers positive values for even roots. . Ahh i screwed up one question in Main GMAT because of my ignorance in regards to this knowledge.
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30 May 2017, 23:18
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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.
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30 May 2017, 23:21
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 $$x = -\sqrt{4}=-2$$.
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