question on the imaginary unit and the real number system.

ochocki

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Feb 7, 2005
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So I know that there are no real number solutions to a number like the square root of -16. so can someone please explain to me why a cube root or a fifth root of -16 results in a real number solution? It seems that when the index of a radical is odd and the radicand is negative the solution is real, however, with an even index there are no real number solutions. Why is this?
 
At a very very basic level, one can graph x<SUP>3</SUP>+16.
The graph will cross the x-axis at some negative value.
That will be the real cube root of -16.

At higher level we can use calculus to get an approx. value.
 
How would you graph x^3 + 16, could you just show me how to set it up please?
 
First, you cannot graph these expressions. An equation is required so that the axes can be defined.

Can you graph y = x^3? That's a good place to start.
 
ochocki said:
How would you graph x^3 + 16, could you just show me how to set it up please?
Surely you have a calulator?
 
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