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=20for 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|
close
