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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]
31 Oct 2014, 09:30

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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]
26 Dec 2014, 10:21

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"

Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]
07 Apr 2015, 09:48

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ankittiss wrote:

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 \(\sqrt{x^2}=|x|\).

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