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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]

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19 Feb 2013, 05:09
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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

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

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

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19 Feb 2013, 06:16
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

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

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]

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20 Feb 2013, 21:31
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

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

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]

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21 Feb 2013, 03:28
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]

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24 Mar 2013, 11:11
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]

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13 Apr 2013, 06:51
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]

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13 Apr 2013, 11:46
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]

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15 Jun 2013, 09:57
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]

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15 Jun 2013, 10:02
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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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15 Jun 2013, 13:19
bkk145 wrote:
|x-3|

Don't for get that...

|x-3| = |3-x|

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]

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01 Jul 2013, 11:13
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]

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31 Oct 2014, 10:30
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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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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

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07 Apr 2015, 10:36
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?

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

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07 Apr 2015, 10:48
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ankittiss wrote:
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?

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 $$x=\sqrt{9}=3$$.

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59

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]

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07 Apr 2015, 11:51
Thanks Bunuel
That really helps
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

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21 Apr 2016, 05:38
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] 21 Apr 2016, 05:38

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