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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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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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