# Thread: how to solve cubic function.

1. ## how to solve cubic function.

The question says comupte the area by integrating, BUT in order to integrate I have to find the find the intersaction points. How do I find the intersection point of this CUBIC FUNCTION-- X^3+X^2-12. Can anyone help me please?

Well it's area between two curves problem. So In order to do that I have to figure out where the two curves intersect over y- axis.

I set functions y= x^3-6 and y=x^2+6 equal to each other to find the intersection points.

then I got x^3-x^2-12=0

Now I don't know how to solve. I was thinking about quadratic formula, but not sure if that can be used in this case.

Even if it can, Is there any other ways to factor this x^3-x^2-12=0 function to find the x values?
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2. Originally Posted by skboss
The question says comupte the area by integrating, BUT in order to integrate I have to find the find the intersaction points. How do I find the intersection point of this CUBIC FUNCTION-- X^3+X^2-12. Can anyone help me please?
intersection with what?

3. Well it's area between two curves problem. So In order to do that I have to figure out where the two curves intersect over y- axis.

I set functions y= x^3-6 and y=x^2+6 equal to each other to find the intersection points.

then I got x^3-x^2-12=0

Now I don't know how to solve. I was thinking about quadratic formula, but not sure if that can be used in this case.

Even if it can, Is there any other ways to factor this x^3-x^2-12=0 function to find the x values?

4. Originally Posted by skboss
Well it's area between two curves problem. So In order to do that I have to figure out where the two curves intersect over y- axis.

I set functions y= x^3-6 and y=x^2+6 equal to each other to find the intersection points.

then I got x^3-x^2-12=0

Now I don't know how to solve. I was thinking about quadratic formula, but not sure if that can be used in this case.

Even if it can, Is there any other ways to factor this x^3-x^2-12=0 function to find the x values?
the curves $\displaystyle y = x^3 - 6$ and $\displaystyle y = x^2 + 6$ intersect at only a single point , $\displaystyle x \approx 2.676$

solving that cubic equation by hand is not easy.

5. Thanks. Yeah, I figured that by looking at the calculator, but wondering if someone knew how to do it by hand. But thanks anyone.