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What is the square root of x^2?

 
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Nov12-12, 12:18 AM   #35
 

What is the square root of x^2?

Quote by tahayassen View Post
Wait a second... Sorry for my repeated lack of understanding, but how did I go from step 2 to step 3 here:

[tex]{ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2[/tex]

If I remember correctly, to take the absolute value of anything, you just ensure that the sign is positive. For example, |-2|=2, |4|=4, |-sqrt(4)|=sqrt(4), and |sqrt(4)|=4.
I think you might mean

[tex]{ x }^{ 2 }=4\\ \sqrt{{ x }^{ 2 }}=\sqrt { 4 } \\ |x| = 2 \\ x=\pm 2[/tex]
 
Nov12-12, 01:51 AM   #36
 
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Quote by Benn View Post
I think you might mean

[tex]{ x }^{ 2 }=4\\ \sqrt{{ x }^{ 2 }}=\sqrt { 4 } \\ |x| = 2 \\ x=\pm 2[/tex]
As a reminder for the OP, while this argument does prove
If x2 = 4, then x = 2 or x = -2
it does not prove
x = 2 and x = -2 are both solutions to x2 = 4
unless you can argue that every step is reversible. (they are in this case)
 
Nov12-12, 08:00 AM   #37
 
Quote by Hurkyl View Post
As a reminder for the OP, while this argument does prove
If x2 = 4, then x = 2 or x = -2
it does not prove
x = 2 and x = -2 are both solutions to x2 = 4
unless you can argue that every step is reversible. (they are in this case)
What do you mean by every step is reversible? Can you give me an example where they are not reversible?
 
Nov12-12, 04:32 PM   #38
 
Mentor
x = -2 ## \Rightarrow## x2 = 4

The steps here are not reversible. If x2 = 4, it does not necessarily imply that x = -2.
 
Nov12-12, 04:39 PM   #39
 
Quote by Hurkyl View Post
As a reminder for the OP, while this argument does prove
If x2 = 4, then x = 2 or x = -2
it does not prove
x = 2 and x = -2 are both solutions to x2 = 4
unless you can argue that every step is reversible. (they are in this case)
Wrong.Is x=2 a solution to x^2=4?-yes and is x=-2 a solution to x^2=4-yes, they are both solutions.Saying x=two values simultaneously is wrong, which I think is what you meant and thinking about this, with everything said generally being correct in this thread about(x^0.5)^2 being= to + or-(x) or abs(x) this means that (x^a)^b doesn't always equal x^(ab), which is an exponentiation law-I suppose there always are exceptions.The steps are irreversible because the inverse has more than one image, which isn't right.
 
Nov12-12, 04:59 PM   #40
 
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Quote by Dalek1099 View Post
Wrong.Is x=2 a solution to x^2=4?-yes and is x=-2 a solution to x^2=4-yes, they are both solutions.Saying x=two values simultaneously is wrong, which I think is what you meant and thinking about this, with everything said generally being correct in this thread about(x^0.5)^2 being= to + or-(x) or abs(x) this means that (x^a)^b doesn't always equal x^(ab), which is an exponentiation law-I suppose there always are exceptions.The steps are irreversible because the inverse has more than one image, which isn't right.
What Hurkyl said was absolutely correct. I think you misinterpreted his post.
 
Nov12-12, 07:51 PM   #41
 
Alright, thanks for the help guys.
 
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