Finding the real root of the polynomial $2x^3-3x^2+2 $

Question

I want to get exactly roots of this equation...

$2x^3-3x^2+2 = 0$

I try to solve it but can not find the solution. wolframealpha just give me aproximation..

I know the real root is $-1< root <-1/2$.

Answer 1

Here are the three exact solutions, which you can find by cubic root methods such as Cardano's:

$\left\{\frac{1}{2} \left(1-\frac{1}{\sqrt[3]{3-2 \sqrt{2}}}-\sqrt[3]{3-2 \sqrt{2}}\right),\frac{1}{2}+\frac{1}{4} \sqrt[3]{3-2 \sqrt{2}} \left(1-i \sqrt{3}\right)+\frac{1+i \sqrt{3}}{4 \sqrt[3]{3-2 \sqrt{2}}},\frac{1}{2}+\frac{1-i \sqrt{3}}{4 \sqrt[3]{3-2 \sqrt{2}}}+\frac{1}{4} \sqrt[3]{3-2 \sqrt{2}} \left(1+i \sqrt{3}\right)\right\}$

The first root is real, and has a decimal approximation of $-0.677651...$.