pre cal 11

mixed radicals to entire radical
-a√b =-√(a^2)(b) left out the negative why????
and for cube root you include the -, why is this the case?
pls help

3 answers

In high school, often teachers neglect other than primary roots. Later on in engineering, they have real meaning, and the solutions have meaning. For instance, in electronic engineering cubroot(-8i) has three roots, and all three have meaning.
Here is a magic trick you will learn (out of high school). In the complex plain, 8i is 8@90, and -8i is 8(270)
so the cubrt (-8i)=cubrt(8(270))=2@90; 2@210;2@330 and all those can be converted to the complex plane (0+2i;-1.75- .935i; and one other. The point is all the roots exist, it is just a convenience to ignore some. Sometimes one gets burned doing that. We have a nice theorum that any degree n equation (cubrts are degree3, sqrts are degree2), there are n solutions. Keep that in your back pocket for later life as an engineer or scientist.
It's important to remember that √x = |x|
√9 = 3, NOT ±3
Just because (-3)^2 = 9 does not make -3 = √9

For cube roots, you can take the - sign inside or not, since (for all odd powers)
∛(-x)^3 = ∛[(-1)^3*x^3) = ∛(-1)^3 * ∛x^3 = -1*∛x^3 = -x

But √x^2 ≠ x if x < 0
@Steve So -2√4 = -√16? negative on the outside?