Is There More Than One Solution for Cube Root Equations?

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SUMMARY

The discussion clarifies that when dealing with cube root equations, there is only one real cube root for any non-zero real number, unlike quadratic equations which yield both positive and negative roots. The cubic function's behavior indicates that while there is one real root, there are also two complex conjugate roots. This distinction is crucial for understanding the restrictions of radical equations involving cubic functions.

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  • Understanding of cubic functions and their properties
  • Knowledge of radical equations and their restrictions
  • Familiarity with complex numbers and conjugates
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  • Explore the concept of complex conjugates and their applications
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Mathematicians, students studying algebra, educators teaching cubic functions, and anyone interested in the properties of radical equations.

Coco12
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Lets say I was trying to figure out the restrictions of a radical equation and the function inside the radical was a cubic function. I know you have to make the equation inside greater than or equal to 0.
In the case of a quadratic equation, you have to square root it once you bring everything to one side of the equality, giving you a positive and negative answer, is this the same for cube rooting it when figuring out the restrictions? You will have a positive and negative cube root?
 
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Coco12 said:
Lets say I was trying to figure out the restrictions of a radical equation and the function inside the radical was a cubic function. I know you have to make the equation inside greater than or equal to 0.
In the case of a quadratic equation, you have to square root it once you bring everything to one side of the equality, giving you a positive and negative answer, is this the same for cube rooting it when figuring out the restrictions? You will have a positive and negative cube root?
No. (-x)^3= -x^3 so there are not "positive and negative cube roots" of the same number. There will be one real cube root of a real number (other than 0) and two complex conjugate (non-real) roots.
 
HallsofIvy said:
No. (-x)^3= -x^3 so there are not "positive and negative cube roots" of the same number. There will be one real cube root of a real number (other than 0) and two complex conjugate (non-real) roots.

Ok thank you
 

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