Is (square root of x)^2 an identity?

Question

True for all nonnegative values of x^2? Please explain

Answer 1

Alexis is not correct. Several people seem to be misreading the question as √(x²), even though the questioner made it very explicit that the expression is (√x)². All √x means is "one of the two numbers which, when squared, equals x (specifically, the positive one, assuming either number is real)." Thus, (√x)²=x by definition. This is in contrast to √(x²) -- although x is one of the two numbers which, when squared, equals x², so is -x, and it may be the case that -x is the one selected by the square root function. For real x, this happens precisely when x<0, so √(x²) becomes the function {-x when x<0, x when x≥0}, which is precisely |x| (note: this does not hold for complex x in general. There, the function is {x when 0≤arg(x)<π, -x when π≤arg(x)<2π}, assuming the usual choice of principal value for square roots. Note that this is very different from the modulus function for complex numbers).

To recap:

(√x)² = x
√(x²) = {x when 0≤arg(x)<π, -x when π≤arg(x)<2π}, which for real x simplifies to |x|.

Answer 2

Alexis is correct

sqrt(x^2) = |x|

Since you're speaking in the domain of nonnegative values of x, then it is an identity, because

|x| = {x if x >= 0
{-x if x < 0

So |x| = x.
sqrt(x^2) = |x|, and |x| = x, means
sqrt(x^2) = x

Answer 3

Square root of x^2 is x.

Answer 4

sqr root x means x^1/2....so (x^1/2)^2...the 2 n 2 cancells and x^1 is left...therefore the answer is x

Answer 5

Let x be <0, then (sq[x])^2 < 0
Let x be >0, then (sq[x])^2 > 0
I guess you might call it an identity, kinda

Answer 6

(√x)^2=x is an identity.

Answer 7

Its equal to the absolute value of x.

Answer 8

no it's not.