Quadratic and Cubic formulas

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 14 November 2011

Quadratic formula

To be added before the page is put up.

Cubic formula

We wish to solve $$x^3+a_2{x}^2{a}_{1}x+a_0=0.$$ Put $y=x-a_2/3.$ Then $$y^3+\left(\frac{a_2^2}{3}-\frac{2a_2^2}{3}+a_1\right)y+\left(\frac{a_2^3}{9}-\frac{a_2^3}{27}-\frac{a_1{a}_2}{3}+a_0\right)=0,$$ and so we may assume that our original equation was of the form $$x^3+px+q=0.$$ Let $x=u-v.$ Then $$u^3-3u{v}^2+3u^{2}v-v^3+pu-pv+q=0,for\; all\; u,v \; such\; that\; x=u-v,$$ implies that $$u^3-v^3+q=0and(3uv-p)(u-v)=0.$$ So $$v=\frac{p}{3u}qand{u}^3-{\left(\frac{p}{3u}\right)}^3+q=0.$$ Thus $27u^6-p^3+27q{u}^3=0$ and so $$u^3=\frac{-27q\pm \sqrt{{27}^2{q}^2+4\cdot 27p^3}}{2\cdot 27}and{v}^3=u^3+q.$$ So $$u=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q}{2}{\left(p,3\right)}^3}},v=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q}{2}{\left(p,3\right)}^3}},$$ and $$x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q}{2}{\left(p,3\right)}^3}}-\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q}{2}{\left(p,3\right)}^3}}.$$ This is the solution of $x^3+px+q=0$ by radicals.

Notes and References

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References

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