Is There More Than One Solution for Cube Root Equations?

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The discussion centers on the restrictions of radical equations involving cubic functions. Unlike quadratic equations, which yield both positive and negative roots, cube roots only produce one real root and two complex conjugate roots. The equation inside the radical must be greater than or equal to zero to determine valid inputs. The conversation also references external resources for further clarification on cubic equations and roots. Ultimately, understanding the nature of cube roots is crucial for solving these types of equations effectively.
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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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