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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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59% (01:37) correct 41% (00:42) wrong based on 565 sessions
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 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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 [#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: 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, 13:00
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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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 [#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
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 [#permalink] New post 24 Oct 2007, 05:55
Fistail wrote:
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


It's not the same debate :)... y = sqrt(2*x) can only be positive or nul. But hobbit and HongHu said that y = the square root of () has another meaning...

I will draw u
> the function : sqrt (9+x^2-6x) in fig 1
> the function : 3-x in fig 2

These 2 draws show that there are not equal :)
Attachments

Fig1_Sqrt__ an hidden Abs.gif
Fig1_Sqrt__ an hidden Abs.gif [ 3.54 KiB | Viewed 3394 times ]

Fig2_3 minus X, half correct.gif
Fig2_3 minus X, half correct.gif [ 3.38 KiB | Viewed 3391 times ]

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 [#permalink] New post 24 Oct 2007, 05:59
Fistail wrote:
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


Fistail,

Not sure if this is your confusion and not trying to complicate things further, but ...
9+x^2-6x = (3-x)^2
So when you plug this back in the original equation, you get:
sqrt(9+x^2-6x) = sqrt((3-x)^2) = |3-x|

This means that
sqrt (9+x^2-6x) = +(3-x) OR -(3-x)

However, if you rearrange 9+x^2-6x to x^2-6x+9, you can get
x^2-6x+9 = (x-3)^2
Plug this back in the equation, you get
sqrt(x^2-6x+9) = sqrt((x-3)^2) = |x-3|

This means there are four solutions to this problem as follows:
sqrt(x^2-6x+9) = +(3-x) AND-(3-x) OR +(x-3) AND -(x-3)
In sum,
sqrt(x^2-6x+9) = |3-x| OR |x-3|

This is true and absolute value properties confirms this because:
|a-b| = |b-a|
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Re: Which of the following is always equal to sqrt (9+x^2-6x)? [#permalink] New post 19 Feb 2013, 02:04
This is true and absolute value properties confirms this because:
|a-b| = |b-a|


Bunuel/KArishma,
Is this always true?
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Re: Which of the following is always equal to sqrt (9+x^2-6x)?   [#permalink] 19 Feb 2013, 02:04

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