M^3 - M -1 =0

how to solve for M...?

i take approximation for M=1.33, am i right?

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- Oct 29th 2009, 05:42 AMkinproblems of Roots of polynomials
M^3 - M -1 =0

how to solve for M...?

i take approximation for M=1.33, am i right? - Oct 29th 2009, 06:13 AMproscientia
There are many numerical methods you can use for finding approximations to the real root of the equation. Use the intermediate-value theorem to narrow it down first; in this case, the real root is between $\displaystyle 1$ and $\displaystyle 1.5.$ Then you can apply, for example, Newton’s method to the polynomial.

For an analytic solution, you can use Cardano’s method, which in this case yields an exact solution of $\displaystyle \sqrt[3]{\frac{1+\sqrt{\frac{23}{27}}}2}+\sqrt[3]{\frac{1-\sqrt{\frac{23}{27}}}2}$ for the real root (which is approximately $\displaystyle 1.3247).$