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 $-1< root <-1/2$.

  • 1
    $\begingroup$ if there is just one real root you can write it using radicals with Cardano's method en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method $\endgroup$
    – Will Jagy
    May 23 '15 at 23:10
  • $\begingroup$ WolframAlpha actually gives $x\cong -0.67755$. $\endgroup$
    – hjhjhj57
    May 23 '15 at 23:11
  • $\begingroup$ There are only three methods which i can state : Cubic root formula, Cardano's method (David's answer) and Newton's method...Can't arrive at anything else. The closest i have got to is $\Rightarrow (x-1)^2 (2x+1) + 1 = 0$ $\endgroup$
    – NeilRoy
    Jun 18 '15 at 5:43

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 $-0.677651...$.


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