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\cl { \bf A BRIEF HISTORY OF THE SOLUTIONS }
\cl {\bf OF THE CUBIC AND QUARTIC EQUATIONS}
\noi
Throughout the 15th and and early 16th centuries, many mathematicians worked on
the problem of giving an algebraic solution to the cubic equation (i.e. similar
to the quadratic formula for quadratic equations). Such a solution should
involve only the operations of $+,-,\times,\div$ and extracting roots. Pacioli noted
in 1494 that no such solution yet existed for the general cubic. 
\noi
del Ferro, in the early 1500's, gave a general
solution to cubics of the form $x^3+cx=d$.
\noi
It was common in Italian academia at the
time for public challenges to occur. These challenges were often problem
solving competitions where the winner would receive money and/or a
professorship. Thus, to give himself an advantage in these competitions, del
Ferro kept his solution secret. However, before he died, he did pass his work
along to Fiore and della Nave (although it wasn't published).
\noi
Tartaglia had boasted of solving cubics of
the form $x^3 +bx^2=d$, and in 1535, Fiore challenged Tartaglia. Tartaglia won,
and in the process discovered the solution to the form $x^3+cx=d$.
\noi
Cardano was working on what would become the {\it Ars Magna}, and asked
Tartaglia for permission to publish his solutions in his text (with full credit
to Tartaglia). However, Tartaglia said no. He finally relented and gave Cardano
his solutions (in verse!), but only with the promise from Cardano not to
publish them (of course, Tartaglia planned to publish his results himself).
\noi
Cardano kept his word to Tartaglia, but began working on the solution to the
cubic himself (along with his assistant Ferrari). He visited della Nave and was
granted access to del Ferro's papers. Here he discovered del Ferro's solutions,
and since this work preceeded Tartaglia's, Cardano no longer felt obligated to
Tartaglia to conceal the results. In 1545, Cardano published the {\it Ars
Magna} which contained these results.
\noi
Ferrari himself discovered the solution to the general quartic, and this work
was also included in the {\it Ars Magna}.
\noi
Tartaglia was furious that his results were published by Cardano, and
challenged Ferrari to a contest. Ferrari won, and to this day the solution to
the cubic is known as Cardano's formula.
\noi
Finally, in the early $19^{th}$ century it was shown that no algebraic solution
exists for the general $n^{th}$ degree polynomial equation for all $n\geq 5$.
\next
\cl  {{\bf Tartaglia's Solution to the Cubic Equation}  (as in the {\it Ars Magna)}}
\noi
Any cubic equation of the form $$y^3+Ay^2+By+C=0$$ 
can  be converted to a cubic equation of the form 
$$X^3+cX=d$$ by using the substitution $y=X-\frac{A}{3}$. 
Thus in order to solve {\it any} cubic equation, it suffices to solve ones of the form $X^3+cX=d$. Cardano describes the following method to do this in the {\it Ars Magna}.
\noi
\cl { { How to solve $X^3+cX=d$}}
\begin{questions}
\question
Find $u$ and $v$ so that 
\begin{parts}
\part $u-v=d$
\part $uv=\left( \frac{c}{3} \right)^3$
\end{parts}
\question
The solution to these equations for $u$ and $v$ are given by
\begin{parts}
\part $u=\sqrt{\left( \frac{d}{2} \right)^2+\left( \frac{c}{3} \right)^3}+\frac{d}{2}$
\part $v=\sqrt{\left( \frac{d}{2} \right)^2+\left( \frac{c}{3} \right)^3}-\frac{d}{2}$
\end{parts}
\question
The solution to $X^3+cX=d$ is then given by $X=\root 3 \of u -\root 3 \of v$
\question
The other two solutions, not given in the {\it Ars Magna}, are
$$X=-\frac{1}{2}\left( \root 3 \of u -\root 3 \of v \right) \pm \frac{\sqrt{-3}}{2}
\left( \root 3 \of u +\root 3 \of v \right)$$
\question
Use the above method to solve $X^3+6X=20$. Note that $c=6$ and $d=20$.
\noi
Set $u=\sqrt{\left( \frac{20}{2} \right)^2+\left( \frac{6}{3} \right)^3}+\frac{20}{2}$ and
$v=\sqrt{\left( \frac{20}{2} \right)^2+\left( \frac{6}{3} \right)^3}-\frac{20}{2}$
so that 
$$u=\sqrt{108}+10 \ \ \ \ \ v=\sqrt{108}-10$$
The solution given in the {\it Ars Magna} is then $$X=\root 3 \of {\sqrt{108}+10} - \root 3 \of {\sqrt{108}-10}=2$$
The other two solutions would be given by 
$$X=-1 \pm \frac{\sqrt{-3}}{2} \left( \root 3 \of {\sqrt{108}+10} + \root 3 \of {\sqrt{108}-10}\right)=-1\pm3i$$

\end{questions}

\cl {{\bf Ferrari's Solution to the Quartic Equation} (as in the {\it Ars Magna})} 

\noi
Any quartic equation 
$$y^4+Ay^3+By^2+Cy+D=0$$
can be converted to a quartic of the form
$$X^4+aX^2+bX+c=0 \eqno(1)$$
by using the substitution $y=X-\frac{A}{4}$. Thus to solve any quartic equation it suffices to solve ones of the form (1)  (i.e. with no cube term).

\noi
To solve (1), first rewrite as 
$$X^4+aX^2=-bX-c$$
Add $aX^2+a^2$ to both sides in order to get a perfect square
$$X^4+2aX^2+a^2=aX^2-bX+a^2-c$$
The LHS is a perfect square, so
$$(X^2+a)^2=aX^2-bX+a^2-c \eqno(2)$$
Now add $z$ (to be determined later) {\it inside} the square on the LHS  in order to try and make the RHS a perfect square also. Expanding the LHS would give
$$(X^2+a+z)^2=X^4+2aX^2+a^2+2X^2z+2az+z^2$$
Thus adding $z$ to equation (2) as above would actually result in adding $2X^2z+2az+z^2$. We thus add this to the RHS of (2) to balance the equation 
$$(X^2+a+z)^2=(aX^2-bX+a^2-c)+2X^2z+2az+z^2$$
Grouping like terms on the RHS gives
$$(X^2+a+z)^2=(2z+a)X^2-bX+(a^2-c+2az+z^2) \eqno(3)$$
Now we can try to choose $z$ so that the RHS of (3) is a perfect square.
For this to be the case, notice that the RHS must be of the form
$$((\sqrt{2z+a} )X \pm \sqrt{a^2-c+2az+z^2})^2$$
Expanding this shows that the RHS of (3) must be of the form
$$(2z+a)X^2\pm(2\sqrt{2z+a}\sqrt{a^2-c+2az+z^2})X+(a^2-c+2az+z^2)$$
Thus $-b$, which is the coefficient on $X$ in equation (3), must be equal to the coefficient on $X$ in the above expression
$$-b=\pm  2\sqrt{2z+a}\sqrt{a^2-c+2az+z^2} \eqno(4)$$
Squaring both sides gives
$$b^2=4(2z+a)(a^2-c+2az+z^2)$$
Regrouping this gives 
$$4(2z+a)(a^2-c+2az+z^2)-b^2=0 \eqno(5)$$
Equation (4) is a cubic equation in $z$. Solving it for $z$ would then force the RHS of equation (3) to be a perfect square. The methods given in {\it Ars Magna} to solve cubic equations would have been used to solve this equation for $z$. So assuming we solve for $z$, we then can write equation (3) as
$$(X^2+a+z)^2=((\sqrt{2z+a} )X \pm \sqrt{a^2-c+2az+z^2})^2 \eqno(6)$$
To determine whether to use $\pm$ in equation (6), we can go back and check equation (4).
Since both sides are now perfect squares, we can square root both sides to generate two quadratic equations in $X$
$$(X^2+a+z)=\pm ((\sqrt{2z+a} )X \pm \sqrt{a^2-c+2az+z^2}) \eqno(7)$$
Using the quadratic formula on these equations gives the four solutions to the original quartic equation.
\noi
\cl {Example given in {\it Ars Magna}}
Solve $x^4-12x+3=0$, for which $a=0$, $b=-12$, and $c=3$. We first examine the cubic equation in $z$ given in (5).
$$0=4(2z+0)(0^2-3+2\cdot0\cdot z +z^2)-(-12)^2$$
This simplifies to 
$$0=8z^2-24z-144$$
Tartaglia's method gives the solution $z=3$. Checking equation (4), we see that $-b=12$ so that the $\pm$ in equation (6) is a $+$. We then apply this to equation (7) to solve the original quartic
$$(X^2+0+3)=\pm ((\sqrt{2\cdot 3+0} )X +\sqrt{0^2-3+2\cdot 0 \cdot 3+3^2})$$
This simplifies to 
$$X^2+3=\pm (\sqrt 6 X +\sqrt 6)$$
Running these two equations through the quadratic formula gives the four solutions 
$$X=\frac{\sqrt 6 \pm \sqrt{4\sqrt 6 -6}}{2} $$
$$X=\frac{-\sqrt 6 \pm \sqrt{-4\sqrt 6 -6}}{2} $$
However, the latter two solutions are not reported in the {\it Ars Magna} as they are imaginary!

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