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

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

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22 Oct 2007, 23:02
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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, 04:34, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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22 Oct 2007, 23:04
(C) for me

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

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23 Oct 2007, 10:46
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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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VP
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23 Oct 2007, 11:53
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|x-3|

Don't for get that...

|x-3| = |3-x|

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Director
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23 Oct 2007, 12:02
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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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Manager
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23 Oct 2007, 12:32
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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23 Oct 2007, 14:00
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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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23 Oct 2007, 14:05
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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Director
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23 Oct 2007, 14:30
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.

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

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23 Oct 2007, 14:31
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.

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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Director
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23 Oct 2007, 21:05
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.

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.

thanks.

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Director
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23 Oct 2007, 21:38
1
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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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23 Oct 2007, 22:04
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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Director
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23 Oct 2007, 22:45
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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24 Oct 2007, 01:31
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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24 Oct 2007, 03:27
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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Director
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24 Oct 2007, 06:38
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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24 Oct 2007, 06:55
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 [ 3.54 KiB | Viewed 7723 times ]

Fig2_3 minus X, half correct.gif [ 3.38 KiB | Viewed 7720 times ]

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VP
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24 Oct 2007, 06:59
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]

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19 Feb 2013, 03:04
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, 03:04

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