# Radical Equations (and Extraneous Roots?)

• April 19th 2013, 06:24 PM
qanda
Hello. I am having trouble understanding the logic behind the property that states radical expressions with even-number indices cannot be equal to a negative value.

Property: nth-root(xn) = |x|, if n is an even number.

Let's take sqrt(25) for example: sqrt(25) = 5 is a valid statement, but sqrt(25) = -5 is an invalid statement, even though both 52 and (-5)2 equal 25.

Why?

Thanks!
• April 19th 2013, 07:17 PM
Gusbob
Re: Radical Equations (and Extraneous Roots?)
The principal root of 5 is defined to be $\sqrt{25}=5$ and more generally, $\sqrt{x^2}=|x|$. In this sense, the square root function does not actually give you any number which squares to 25, but a specific one.

To clarify, 5 and -5 are both square roots of 25, but the $\sqrt$ sign refers only to the principal root.
• April 19th 2013, 07:34 PM
qanda
Re: Radical Equations (and Extraneous Roots?)
Thanks, Gusbob!

So that I understand the syntax: evaluating -√25 would be to find the negative principal root of 25? Or simply its negative square root?
• April 19th 2013, 09:47 PM
Gusbob
Re: Radical Equations (and Extraneous Roots?)
For the square root case, yes you're evaluating the negative square root.

Without going into higher level maths, I'll just let you know that every real number (even negative ones!) has exactly n number of nth roots in the complex plane. You fix one of them to be a principal root, and the others can be given in terms of this principal root by some complex formula. For the square root case, this complex formula reduces to multiplication by negative 1.