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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

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31 Oct 2014, 10:30

Hello from the GMAT Club BumpBot!

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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

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26 Dec 2014, 11:21

Answer to this question is +/- (x-3) both,but we need to define the range of x because Domain y is always positive i.e.above x-axis. So, for x>3 y=(x-3) and for x<3, y=(3-x) It can be re-written as |x-3| or |3 - x| ............. which is in agreement with property of modulus ie |x-a|=|a-x|

Ideally one must always remember that

Sqrt(x-a)^2 = |x-a|

and, |x-a|= x-a , if x>a = -(x-a) , if x<a.

Hope, it makes sense .

GK_Gmat wrote:

Which of the following is always equal to \(\sqrt{9+x^2-6x}\)?

A. x - 3 B. 3 + x C. |3 - x| D. |3 + x| E. 3 - x

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"Arise, Awake and Stop not till the goal is reached"

Bunuel Can you please help here? According to my understanding there are two cases in the GMAT: Case 1 :

sqrt[(-5)*(-5)] = -5

Case 2 :

sqrt[(5)*(5)] = 5

Is my understanding correct?

Thanks in advance Ankit

No, it's totally wrong.

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 \(x=\sqrt{9}=3\).

Again, the point here is that since square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples: If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

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21 Apr 2016, 05:38

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

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25 Jul 2016, 02:39

1

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GK_Gmat wrote:

Which of the following is always equal to \(\sqrt{9+x^2-6x}\)?

A. x - 3 B. 3 + x C. |3 - x| D. |3 + x| E. 3 - x

There should be no confusion regarding this question. SQUAREROOT and MODULUS have the same property regarding the polarity of a number. THEY BOTH YIELDS ONLY POSITIVE OUTPUTS.

Q:- what is the surest way to make any value positive. ? A:- Take the modulus of the value |x|

So we know \(\sqrt{9+x^2-6x}\) will gives us ONLY POSITIVE VALUE, Then we should make sure that the option also matches this property. Therefore only |3 - x| and |3+x| are the one that will always give positive value BUT |3+x| is not a root or solution of \(\sqrt{9+x^2-6x}\) So the only option left is |3-x|

ANSWER IS C
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Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.

Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

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07 Jul 2017, 11:09

After factorization I got sq root x- 3 ^2 => (x-3) ^2 so x = 3 I tried this algebraically - keeping in mind 'always true'. I would be glad to hear any comments on the same 1) x - 3 = 0 can be 3 => 3 - 3 = or x = 3 but I wasn't sure if this will hold always 2) x = -3 (not correct) 3) |3-x|: 3 - x =0; so x=3 and -3+x=0 so x = 3 (here both negative or positive sign; value remains the same) 4) |3+x| = 0 will not hold 5) 3 - x= 0; so -x = - 3 or x = 3

Now I know, that C will ALWAYS be true. But Why is A, E wrong? They are giving us 3 also Can anyone explain the logic for A, E to be wrong?

Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

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07 Jul 2017, 21:14

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Madhavi1990 wrote:

After factorization I got sq root x- 3 ^2 => (x-3) ^2 so x = 3 I tried this algebraically - keeping in mind 'always true'. I would be glad to hear any comments on the same 1) x - 3 = 0 can be 3 => 3 - 3 = or x = 3 but I wasn't sure if this will hold always 2) x = -3 (not correct) 3) |3-x|: 3 - x =0; so x=3 and -3+x=0 so x = 3 (here both negative or positive sign; value remains the same) 4) |3+x| = 0 will not hold 5) 3 - x= 0; so -x = - 3 or x = 3

Now I know, that C will ALWAYS be true. But Why is A, E wrong? They are giving us 3 also Can anyone explain the logic for A, E to be wrong?

You can't equate it to 0 and compute value of x as 3. Anyways A and E are subset of the solution. In case of C it is considering both the solutions which equate to square root expression given in question stem.

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