Ferrari’s solution of a biquadratic

Ferrari’s solution of the Biquadratic. 

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

The second term can be removed by the substitution s=2t+cEquation E

and the equation D becomes 4t^{3}-It+J=0Equation 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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s