Why Does the Square Root of a Squared Algebraic Term Equal the Term Itself?

[mathdad]

Let A be an a algebraic term.

Why does the sqrt{(A)^2} = A?

In other words, why does the radicand simply come out of the square root?

 

[MarkFL]

That only works if $A$ is non-negative. More precisely, there is the definition:

$$|A|\equiv\sqrt{A^2}$$

Let's now assume that $0\le A$. Then we can think of this along these lines:

$$\sqrt{A^2}=\left(A^2\right)^{\frac{1}{2}}=A^{2\cdot\frac{1}{2}}=A^1=A$$

 

[mathdad]

That only works if $A$ is non-negative. More precisely, there is the definition:

$$|A|\equiv\sqrt{A^2}$$

Let's now assume that $0\le A$. Then we can think of this along these lines:

$$\sqrt{A^2}=\left(A^2\right)^{\frac{1}{2}}=A^{2\cdot\frac{1}{2}}=A^1=A$$

Are you saying that sqrt{(-A)^2} does not equal A?

 

[HallsofIvy]

Let A be an a algebraic term.

Why does the sqrt{(A)^2} = A?

It doesn't. If A= -2 then A^2= 4 and sqrt(4)= 2 not -2.

In other words, why does the radicand simply come out of the square root?

 

[mathdad]

What about sqrt{(x^2 + 1)^2}?

The square root of a quantity squared is the quantity itself.

Does that equal x^2 + 1?

 

[Greg]

Yes, but only because $x^2+1=|x^2+1|$, as $x^2+1$ is always positive.

 

[mathdad]

Yes, but only because $x^2+1=|x^2+1|$, as $x^2+1$ is always positive.

Only because |x^2 + 1| is always positive. Is this what you're saying?

 

[Greg]

Are you familiar with absolute value?

 

[mathdad]

Are you familiar with absolute value?

I am familiar with the basic idea of absolute value. However, in what way is absolute value connected to square roots, particularly my question here?

 

[Greg]

See Mark's post in this topic, http://mathhelpboards.com/pre-calculus-21/sqrt-something-2-a-22221-post100174.html#post100174. (Post #2)

 

[mathdad]

I guess I have a lot more to learn.