# What is the square root of x^2?

by tahayassen
Tags: root, square
P: 273
 Quote by Hurkyl As a reminder for the OP, while this argument does proveIf x2 = 4, then x = 2 or x = -2it does not provex = 2 and x = -2 are both solutions to x2 = 4unless 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?
 Mentor P: 21,261 x = -2 ## \Rightarrow## x2 = 4 The steps here are not reversible. If x2 = 4, it does not necessarily imply that x = -2.
P: 17
 Quote by Hurkyl As a reminder for the OP, while this argument does proveIf x2 = 4, then x = 2 or x = -2it does not provex = 2 and x = -2 are both solutions to x2 = 4unless 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.
Mentor
P: 18,244
 Quote by Dalek1099 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.
 P: 273 Alright, thanks for the help guys.

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