- Thread starter
- #1

#### mathmaniac

##### Well-known member

- Mar 4, 2013

- 188

Solve using cardan's method.

\(\displaystyle x^3-13x+12=0\)

\(\displaystyle x=v+u\)

\(\displaystyle 3uv=-p=13\)

\(\displaystyle v^3+u^3=-q=-12\)

\(\displaystyle 27v^6+324v^3-13=0\)

\(\displaystyle v^3=\frac{-324\pm\sqrt{324^2-27*4*13}}{54}\)

Please solve for x.I know I am asking for too much,but seems like I am not able to get the desired answers even though nothing seems wrong with the method.

I would be so grateful towards anybody who helps.

For the moderator: please change * to x,I don't know how to do that.

Thanks

\(\displaystyle x^3-13x+12=0\)

\(\displaystyle x=v+u\)

\(\displaystyle 3uv=-p=13\)

\(\displaystyle v^3+u^3=-q=-12\)

\(\displaystyle 27v^6+324v^3-13=0\)

\(\displaystyle v^3=\frac{-324\pm\sqrt{324^2-27*4*13}}{54}\)

Please solve for x.I know I am asking for too much,but seems like I am not able to get the desired answers even though nothing seems wrong with the method.

I would be so grateful towards anybody who helps.

For the moderator: please change * to x,I don't know how to do that.

Thanks

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