- #36
Hurkyl
Staff Emeritus
Science Advisor
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- 26
Benn said: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 provex = 2 and x = -2 are both solutions to x2 = 4
unless you can argue that every step is reversible. (they are in this case)