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

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Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink] New post 22 Oct 2007, 22:02
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Feb 2013, 03:34, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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 [#permalink] New post 23 Oct 2007, 10:53
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The answer is indeed
|x-3|

Don't for get that...

|x-3| = |3-x|

C is the answer.
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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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 [#permalink] New post 23 Oct 2007, 13:00
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pmenon wrote:
Dont forget that square root produces a positive and negative result.

I end up with +/- (x-3). In absolute value terms, this would be |x-3| .... I dont see this as an answer choice though :-S


Sorry, but this statment is wrong... Sqrt( "Something" ) >= 0... It's never negative. In addition, "Somtehing" must be positive or 0.

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

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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 [#permalink] New post 22 Oct 2007, 22:04
(C) for me :)

sqrt (9+x^2-6x)
= sqrt( (3-x)^2 )
= |3-x|
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 [#permalink] New post 23 Oct 2007, 09:46
Dont forget that square root produces a positive and negative result.

I end up with +/- (x-3). In absolute value terms, this would be |x-3| .... I dont see this as an answer choice though :-S
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Re: PS: Square Root [#permalink] New post 23 Oct 2007, 11:02
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

Pls. explain.


C. sqrt (9 + x^2 - 6x) can be written as = sqrt (x^2 - 6x + 9)
sqrt (9 + x^2 - 6x) = 3 - x
sqrt (x^2 - 6x + 9) = x - 3

(3-x) and (x-3) are not same but if we multiply either of them by -ve, we get the other. Therefore

sqrt (9 + x^2 - 6x) = sqrt (x^2 - 6x + 9) = l3-xl
so C should be it.
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 [#permalink] New post 23 Oct 2007, 11:32
So in other words sq rt of (x-3)^2 is +ve or -ve x-3? Sorry, but i don't really get that. I thought when we have a square of a sq rt, the answer is just what's under the sq rt, without any power.

It's not sq rt of 9 to be +-3, it's sq rt of 9^2, which is 9.
Please correct me if i am wrong.
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 [#permalink] New post 23 Oct 2007, 13:05
Fistail wrote:
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

Pls. explain.


C. sqrt (9 + x^2 - 6x) can be written as = sqrt (x^2 - 6x + 9)
sqrt (9 + x^2 - 6x) = 3 - x
sqrt (x^2 - 6x + 9) = x - 3


(3-x) and (x-3) are not same but if we multiply either of them by -ve, we get the other. Therefore

sqrt (9 + x^2 - 6x) = sqrt (x^2 - 6x + 9) = l3-xl
so C should be it.


Sorry as well... the reasonning is wrong :)

We have no clues on the sign neither of 3-x nor of x-3.... So we cannot say sqrt( "Something" ) = Unknown sign expression :)
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 [#permalink] New post 23 Oct 2007, 13:30
Fig wrote:
Fistail wrote:
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

Pls. explain.


C. sqrt (9 + x^2 - 6x) can be written as = sqrt (x^2 - 6x + 9)
sqrt (9 + x^2 - 6x) = 3 - x
sqrt (x^2 - 6x + 9) = x - 3


(3-x) and (x-3) are not same but if we multiply either of them by -ve, we get the other. Therefore

sqrt (9 + x^2 - 6x) = sqrt (x^2 - 6x + 9) = l3-xl
so C should be it.


Sorry as well... the reasonning is wrong :)

We have no clues on the sign neither of 3-x nor of x-3.... So we cannot say sqrt( "Something" ) = Unknown sign expression :)


i did not get your point. :roll:

sqrt (9 + x^2 - 6x) and sqrt (x^2 - 6x + 9), both, are +ves.
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 [#permalink] New post 23 Oct 2007, 13:31
Fistail wrote:
Fig wrote:
Fistail wrote:
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

Pls. explain.


C. sqrt (9 + x^2 - 6x) can be written as = sqrt (x^2 - 6x + 9)
sqrt (9 + x^2 - 6x) = 3 - x
sqrt (x^2 - 6x + 9) = x - 3


(3-x) and (x-3) are not same but if we multiply either of them by -ve, we get the other. Therefore

sqrt (9 + x^2 - 6x) = sqrt (x^2 - 6x + 9) = l3-xl
so C should be it.


Sorry as well... the reasonning is wrong :)

We have no clues on the sign neither of 3-x nor of x-3.... So we cannot say sqrt( "Something" ) = Unknown sign expression :)


i did not get your point. :roll:

sqrt (9 + x^2 - 6x) and sqrt (x^2 - 6x + 9), both, are +ves.


Yes... Both left sides in bold are ok... but the right sides are not :)... They could be negative...
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 [#permalink] New post 23 Oct 2007, 20:05
Fig wrote:
Fistail wrote:
Fig wrote:
Fistail wrote:
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

Pls. explain.


C. sqrt (9 + x^2 - 6x) can be written as = sqrt (x^2 - 6x + 9)
sqrt (9 + x^2 - 6x) = 3 - x
sqrt (x^2 - 6x + 9) = x - 3


(3-x) and (x-3) are not same but if we multiply either of them by -ve, we get the other. Therefore

sqrt (9 + x^2 - 6x) = sqrt (x^2 - 6x + 9) = l3-xl
so C should be it.


Sorry as well... the reasonning is wrong :)

We have no clues on the sign neither of 3-x nor of x-3.... So we cannot say sqrt( "Something" ) = Unknown sign expression :)


i did not get your point. :roll:

sqrt (9 + x^2 - 6x) and sqrt (x^2 - 6x + 9), both, are +ves.


Yes... Both left sides in bold are ok... but the right sides are not :)... They could be negative...


but Fig i am still not clear on whether you mean "sqrt (9 + x^2 - 6x)" is only +ve or -ve too.

I say it is only +ve because "sqrt (9 + x^2 - 6x)" is a factor of "(9 + x^2 - 6x)" and the other fasctor is "-sqrt(9 + x^2 - 6x)". so

"sqrt (9 + x^2 - 6x)" can only be 3-x but it can be x-3 as well because we can re-write "sqrt (9 + x^2 - 6x)" as "sqrt (x^2 - 6x + 9)", which is x-3.

please correct me and also clearify your reasoning..

thanks.
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 [#permalink] New post 23 Oct 2007, 20:38
Fig wrote:
(C) for me :)

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


Fig can you pls. explain:

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

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.
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 [#permalink] New post 23 Oct 2007, 21:04
GK_Gmat wrote:
Fig wrote:
(C) for me :)

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


Fig can you pls. explain:

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

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.


by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

:)
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 [#permalink] New post 23 Oct 2007, 21:45
GK_Gmat wrote:
KillerSquirrel wrote:
GK_Gmat wrote:
Fig wrote:
(C) for me :)

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


Fig can you pls. explain:

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

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.


by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

:)


Whew! Finally it makes sense. Thanks KS.


wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.
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 [#permalink] New post 24 Oct 2007, 00:31
Fistail wrote:
GK_Gmat wrote:
KillerSquirrel wrote:
GK_Gmat wrote:
Fig wrote:
(C) for me :)

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


Fig can you pls. explain:

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

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.


by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

:)


Whew! Finally it makes sense. Thanks KS.


wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.


the unknown x in sqrt(x^2) can be negative - the outcome will be positive ---> sqrt(-2^2) = 2 ---> x=-2

:)
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 [#permalink] New post 24 Oct 2007, 02:27
Fistail wrote:
GK_Gmat wrote:
KillerSquirrel wrote:
GK_Gmat wrote:
Fig wrote:
(C) for me :)

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


Fig can you pls. explain:

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

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.


by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

:)


Whew! Finally it makes sense. Thanks KS.


wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.


No... :)... In bold, we can only say : sqrt(x^2) = |x| :)

KillerSquirrel gave an exemple that proves it as well :)
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 [#permalink] New post 24 Oct 2007, 05:38
Fig wrote:
Fistail wrote:
GK_Gmat wrote:
KillerSquirrel wrote:
GK_Gmat wrote:
Fig wrote:
(C) for me :)

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


Fig can you pls. explain:

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

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.


by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

:)


Whew! Finally it makes sense. Thanks KS.


wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.


No... :)... In bold, we can only say : sqrt(x^2) = |x| :)

KillerSquirrel gave an exemple that proves it as well :)


Fig, I am still not convinced. I am in line with what Honghu and hobbit said here:

http://www.gmatclub.com/forum/t40319?po ... c&start=20


for me, sqrt (9+x^2-6x) = 3-x
  [#permalink] 24 Oct 2007, 05:38

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