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Director
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Which of the following is always equal to sqrt(9+x^26x)? [#permalink]
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22 Oct 2007, 23:02
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Which of the following is always equal to \(\sqrt{9+x^26x}\)? A. x  3 B. 3 + x C. 3  x D. 3 + x E. 3  x
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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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(C) for me
sqrt (9+x^26x)
= sqrt( (3x)^2 )
= 3x



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Dont forget that square root produces a positive and negative result.
I end up with +/ (x3). In absolute value terms, this would be x3 .... I dont see this as an answer choice though :S



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The answer is indeed
x3
Don't for get that...
x3 = 3x
C is the answer.



Director
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Re: PS: Square Root [#permalink]
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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^26x)?
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
(3x) and (x3) 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) = l3xl
so C should be it.



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So in other words sq rt of (x3)^2 is +ve or ve x3? 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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pmenon wrote: Dont forget that square root produces a positive and negative result.
I end up with +/ (x3). In absolute value terms, this would be x3 .... 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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Fistail wrote: GK_Gmat wrote: Which of the following is always equal to sqrt (9+x^26x)?
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(3x) and (x3) 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) = l3xl so C should be it.
Sorry as well... the reasonning is wrong
We have no clues on the sign neither of 3x nor of x3.... So we cannot say sqrt( "Something" ) = Unknown sign expression



Director
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Fig wrote: Fistail wrote: GK_Gmat wrote: Which of the following is always equal to sqrt (9+x^26x)?
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(3x) and (x3) 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) = l3xl so C should be it. Sorry as well... the reasonning is wrong We have no clues on the sign neither of 3x nor of x3.... 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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Fistail wrote: Fig wrote: Fistail wrote: GK_Gmat wrote: Which of the following is always equal to sqrt (9+x^26x)?
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(3x) and (x3) 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) = l3xl so C should be it. Sorry as well... the reasonning is wrong We have no clues on the sign neither of 3x nor of x3.... 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...



Director
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Fig wrote: Fistail wrote: Fig wrote: Fistail wrote: GK_Gmat wrote: Which of the following is always equal to sqrt (9+x^26x)?
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(3x) and (x3) 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) = l3xl so C should be it. Sorry as well... the reasonning is wrong We have no clues on the sign neither of 3x nor of x3.... 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 3x but it can be x3 as well because we can rewrite "sqrt (9 + x^2  6x)" as "sqrt (x^2  6x + 9)", which is x3.
please correct me and also clearify your reasoning..
thanks.



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Fig wrote: (C) for me sqrt (9+x^26x) = sqrt( (3x)^2 )= 3x
Fig can you pls. explain:
sqrt (9 + x^2  6x) = sqrt (x3)^2
From this how do you arrive at :
= sqrt( (3x)^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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GK_Gmat wrote: Fig wrote: (C) for me sqrt (9+x^26x) = sqrt( (3x)^2 )= 3xFig can you pls. explain: sqrt (9 + x^2  6x) = sqrt (x3)^2 From this how do you arrive at : = sqrt( (3x)^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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GK_Gmat wrote: KillerSquirrel wrote: GK_Gmat wrote: Fig wrote: (C) for me sqrt (9+x^26x) = sqrt( (3x)^2 )= 3xFig can you pls. explain: sqrt (9 + x^2  6x) = sqrt (x3)^2 From this how do you arrive at : = sqrt( (3x)^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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Fistail wrote: GK_Gmat wrote: KillerSquirrel wrote: GK_Gmat wrote: Fig wrote: (C) for me sqrt (9+x^26x) = sqrt( (3x)^2 )= 3xFig can you pls. explain: sqrt (9 + x^2  6x) = sqrt (x3)^2 From this how do you arrive at : = sqrt( (3x)^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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Fistail wrote: GK_Gmat wrote: KillerSquirrel wrote: GK_Gmat wrote: Fig wrote: (C) for me sqrt (9+x^26x) = sqrt( (3x)^2 )= 3xFig can you pls. explain: sqrt (9 + x^2  6x) = sqrt (x3)^2 From this how do you arrive at : = sqrt( (3x)^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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Fig wrote: Fistail wrote: GK_Gmat wrote: KillerSquirrel wrote: GK_Gmat wrote: Fig wrote: (C) for me sqrt (9+x^26x) = sqrt( (3x)^2 )= 3xFig can you pls. explain: sqrt (9 + x^2  6x) = sqrt (x3)^2 From this how do you arrive at : = sqrt( (3x)^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^26x) = 3x



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Fistail wrote: Fig wrote: Fistail wrote: GK_Gmat wrote: KillerSquirrel wrote: GK_Gmat wrote: Fig wrote: (C) for me sqrt (9+x^26x) = sqrt( (3x)^2 )= 3xFig can you pls. explain: sqrt (9 + x^2  6x) = sqrt (x3)^2 From this how do you arrive at : = sqrt( (3x)^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=20for me, sqrt (9+x^26x) = 3x
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^26x) in fig 1
> the function : 3x in fig 2
These 2 draws show that there are not equal
Attachments
Fig1_Sqrt__ an hidden Abs.gif [ 3.54 KiB  Viewed 6688 times ]
Fig2_3 minus X, half correct.gif [ 3.38 KiB  Viewed 6685 times ]



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Fistail wrote: Fig wrote: Fistail wrote: GK_Gmat wrote: KillerSquirrel wrote: GK_Gmat wrote: Fig wrote: (C) for me sqrt (9+x^26x) = sqrt( (3x)^2 )= 3xFig can you pls. explain: sqrt (9 + x^2  6x) = sqrt (x3)^2 From this how do you arrive at : = sqrt( (3x)^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=20for me, sqrt (9+x^26x) = 3x
Fistail,
Not sure if this is your confusion and not trying to complicate things further, but ...
9+x^26x = (3x)^2
So when you plug this back in the original equation, you get:
sqrt(9+x^26x) = sqrt((3x)^2) = 3x
This means that
sqrt (9+x^26x) = +(3x) OR (3x)
However, if you rearrange 9+x^26x to x^26x+9, you can get
x^26x+9 = (x3)^2
Plug this back in the equation, you get
sqrt(x^26x+9) = sqrt((x3)^2) = x3
This means there are four solutions to this problem as follows:
sqrt(x^26x+9) = +(3x) AND(3x) OR +(x3) AND (x3)
In sum,
sqrt(x^26x+9) = 3x OR x3
This is true and absolute value properties confirms this because:
ab = ba



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Re: Which of the following is always equal to sqrt (9+x^26x)? [#permalink]
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19 Feb 2013, 03:04
This is true and absolute value properties confirms this because: ab = ba Bunuel/KArishma, Is this always true?
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Re: Which of the following is always equal to sqrt (9+x^26x)?
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