# Ferrari’s solution of a biquadratic

Writing the equation $u=ax^{4}+4bx^{3}+6cx^{2}+4ax+c=0$Equation A

we assume that $au=(ax^{2}+2bx+s)^{2}-(2mx+n)^{2}$Equation B

Expanding and equating coefficients, we have

$2m^{2}=as+2b^{2}-3ac$ and $mn=bs-ad$ and $n^{2}=s^{2}-ae$ ….Equation C

Eliminating m, n,

$(s^{2}-ae)(as+2b^{2}-3ac)=2(bs-ad)^{2}$ which reduces to

$s^{2}-3cs^{2}+(4bd-ae)s+(3ace-2ad^{2}-2eb^{2})=0$Equation D

The second term can be removed by the substitution $s=2t+c$Equation E

and the equation D becomes $4t^{3}-It+J=0$Equation F, which is the “reducing cubic”.

Equations C become

$m^{2}=at+b^{2}-ac=at-H$

$mn=2bt+bc-ad$

$n^{2}=(2t+c)^{2}-ae$

Call the above three equations as G.

Thus, if $t_{1}$ is a root of F and

$m_{1}=\sqrt{at_{1}+b^{2}-ac}$ and $n_{1}=\frac{(2bt_{1}+bc-ad)}{m_{1}}$, the equation

$u=0$ can be put in the form

$(ax^{2}+2bx+c+2t_{1})^{2}-(2m_{1}x+n_{1})^{2}=0$,

and its roots are the roots of the quadratics

$ax^{2}+2b^{x}+c+2t_{1}=\pm (2m_{1}x+n_{1})$.

It should be noted that the three roots of F correspond to the three ways of expressing u as the product of two quadratic factors.

Homework: Solve the equation $u=x^{4}+3x^{3}+x^{2}-2=0$.

Hope you enjoyed it…

More later,

Nalin Pithwa

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