Last update: 30 June 2012

We wish to solve $ax2+bx+c =0, when a≠0.$ Divide by $a$ to get $x2+a1x +a0 = 0, where a1 =ba and a2 =ca.$ Since ${\left(x+\frac{1}{2}{a}_{1}\right)}^{2}={x}^{2}+{a}_{1}x+\frac{1}{4}{a}_{1}^{2},$ $0=x2+a1x+a0 = ( x+12 a1 )2 - 14 a12 +a0.$ So $( x+12 a1 )2 = 14 a12-a0 .$ So $x+12 a1 = ± 14 a12 -a0 and x= -12 a1 ± 14 a12 -a0 .$ Replacing ${a}_{1}=\frac{b}{a}$ and ${a}_{2}=\frac{c}{a}$ gives $x = -12 ba ± 14 b2a2 - ca = -b2a ± b2-4ac 4a2 = -b2a ± 12a b2-4ac .$ So the two solutions for $x$ are $x= -b+ b2-4ac 2a and x = -b- b2-4ac 2a .$

## Cubic formula

We wish to solve $x3+ a2x2+ a1x+a0 =0.$ Put $y=x-\frac{1}{3}{a}_{2}$. Then $y3 + ( a223 - 2a223 +a1)y +( a239 - a2327 - a1a23 +a0 ) =0,$ and so we may assume that our original equation was of the form $x3 +px+q=0.$ Let $x=u-v.$ Then $u3- 3uv2+ 3u2v- v3+ pu-pv+ q=0, for all u,v such that x=u-v,$ implies that $u3-v3+q =0 and (3uv-p) (u-v)=0.$ So $v= p3uq and u3- (p3u) 3 +q=0.$ Thus $27{u}^{6}-{p}^{3}+27q{u}^{3}=0$ and so $u3= -27q± 272q2 +4⋅27p3 2⋅27 and v3= u3+q.$ So $u= -q2+ (q2)2 + (p3)3 3 , v= q2+ (q2)2 + (p3)3 3 ,$ and $x= -q2+ (q2)2 + (p3)3 3 - q2+ (q2)2 + (p3)3 3 .$ This is the solution of ${x}^{3}+px+q=0$ by radicals.

## Notes and References

There are solutions by radicals of quadratic, cubic and fourth degree polynomials. Abel and Galois proved that the general solutions of a degree 5 polynomial in $x$ is not given by radicals. This page gives derivations of the solutions by radicals for quadratic and cubic polynomials.

The presentation of the qudratic formula was handwritten in 2012 to fill out this page which already had the derivation of the cubic formula. The derivation of the cubic formula is from the "Group cohomology" page of Work2004/BookNewalg/PartV.pdf (page 20). No reference for the source of this derivation is in the original notes.

## References

None known at this time.