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

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Re: Which of the following is always equal to sqrt (9+x^2-6x)? [#permalink] New post 19 Feb 2013, 04:09
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Sachin9 wrote:
This is true and absolute value properties confirms this because:
|a-b| = |b-a|


Bunuel/KArishma,
Is this always true?


Yes, since both |a-b| and |b-a| represent the distance between a and b on the number line.

COMPLETE SOLUTION:

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

\(\sqrt{9+x^2-6x}=\sqrt{(3-x)^2}=|3-x|\).

Answer: C.
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Re: Which of the following is always equal to sqrt (9+x^2-6x)? [#permalink] New post 19 Feb 2013, 05:16
Bunuel wrote:
Sachin9 wrote:
This is true and absolute value properties confirms this because:
|a-b| = |b-a|


Bunuel/KArishma,
Is this always true?


Yes, since both |a-b| and |b-a| represent the distance between a and b on the number line.

COMPLETE SOLUTION:

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

\(\sqrt{9+x^2-6x}=\sqrt{(3-x)^2}=|3-x|\).

Answer: C.


amazing, I dont know why I fail/forget to consider |a-b| as the distance between a and b. I repeatedly commit this mistake..
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 20 Feb 2013, 20:31
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


Hi, can anyone explain me how to go abt this...the above answers have got a
lil confusing for me.

ill go one option at a time:

A. x-3

when we square x-3 it give x^2+9-6x
if we squareroot x^2+9-6x then we get the same exp...so for me this seems
to be an answer

B. 3+x

when we square 3+x it gives 9+6x+x^2
this is all positive unlike 9+x^2-6x given in the question, so not an answer

C. |3-x|

given sqrt{9+x^2-6x}

solving sqrt(3-x)

|3-x|=sqrt(3-x)

this too seems to be a possible answer

D. |3+x|

this will give 9+x^2+6x which is not equal to the equation given hence not
an answer

E. 3-x

=9+x^2-6x

this too seems fine

can anyone please clarify where im going wrong.... in all the options...


Thanks
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 21 Feb 2013, 02:28
Expert's post
mehasingh 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


Hi, can anyone explain me how to go abt this...the above answers have got a
lil confusing for me.

ill go one option at a time:

A. x-3


Notice that the square root function cannot give negative result: \(\sqrt{{some \ expression}}\geq{0}\).

So, \(\sqrt{9+x^2-6x}=\sqrt{(3-x)^2}\geq{0}\).

Now, in option A we have x-3, which can be negative if x<3, so A cannot be the correct answer.

Hope it's clear.
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 24 Mar 2013, 10:11
sqrt (9+x^2-6x) = sqrt( (3-x)^2 )
= |3-x|

sqrt (9+x^2-6x) = sqrt( (x-3)^2 )
= |x-3|

but we have only |3-x| as option, so "C"
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 13 Apr 2013, 05:51
From the property |X| = sqrt( X^2 )

| 3 - X | = sqrt ( (3 - X)^2 )
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 13 Apr 2013, 10:46
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


x^2 - 6x+9 = (3-x)^2 thus\(\sqrt{9+x^2-6x}\) = x-3 or 3-x this is equivel to /3-x/
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 15 Jun 2013, 08:57
I originally said (x-3) was the right answer.

In essence, this is a "square root of a square" problem, is it not? In that case, isn't the result always a positive number?
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 15 Jun 2013, 09:02
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WholeLottaLove wrote:
I originally said (x-3) was the right answer.

In essence, this is a "square root of a square" problem, is it not? In that case, isn't the result always a positive number?


Whenever you have an expression in the form \(\sqrt{x^2}\) it becomes \(|x|\).
So in this case \(\sqrt{(x-3)^2}=|x-3|\)

For example if \(\sqrt{x^2}=3\) x could be 3 and \(\sqrt{3^2}=3\)
but could also be -3 as \(\sqrt{(-3)^2}=3\).

That's why we need the abs value \(x=|3|\)
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Re: [#permalink] New post 15 Jun 2013, 12:19
bkk145 wrote:
The answer is indeed
|x-3|

Don't for get that...

|x-3| = |3-x|

C is the answer.


excellent, thank you for the help.
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 01 Jul 2013, 10:13
Which of the following is always equal to √(9+x^2-6x)?

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

√(9+x^2-6x)
√(x^2 - 6x + 9)
√(x - 3)*(x - 3)
√(x - 3)^2
|x - 3| (Square root of a square...)

Lets choose two values for x: 6, -6
|6-3| = 3
|-6-3| = 9
3,9

Let's plug 6, -6 into the answer choices:

C.) |3 - x|
|3 - 6| = 3
|3- (-6)| = 9
3,9

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



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] New post 07 Apr 2015, 09:36
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
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 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 \(x=\sqrt{9}=3\).

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About \(\sqrt{x^2}=|x|\):

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

Hope it helps.
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 07 Apr 2015, 10:51
Thanks Bunuel
That really helps :)
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Re: Which of the following is always equal to sqrt(9+x^2-6x)?   [#permalink] 07 Apr 2015, 10:51

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