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Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]
 22 Oct 2007, 23:02
Question Stats:
52% (01:28) correct
47% (00:39) wrong based on 22 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
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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Re: Which of the following is always equal to sqrt (9+x^2-6x)? [#permalink]
19 Feb 2013, 05:09
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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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(C) for me
sqrt (9+x^2-6x)
= sqrt( (3-x)^2 )
= |3-x|
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I get x-3?
sqrt(9+x^2-6x)
can be written as
sqrt(x^2-6x+9)
sqrt(x-3)^2
x-3
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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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what's wrong with fresinha12's explanation ?
why cant you re-arrange the equation?
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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]
23 Oct 2007, 12:02
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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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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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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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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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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Fig wrote: (C) for me sqrt (9+x^2-6x) = sqrt( (3-x)^2 ) = |3-x| Perfect !
P.S - smilies are back !
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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^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.
please correct me and also clearify your reasoning..
thanks.
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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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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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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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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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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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