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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.
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.
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.
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.
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
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
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:
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:
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 4148 times ]
Fig2_3 minus X, half correct.gif [ 3.38 KiB | Viewed 4144 times ]
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:
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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