This is what Ferrari is recognized to have achieved:

Cardano was given a word problem by another mathematician, which equated to this equation:

 

x4 + 6x2 + 36 = 60x (original equation)

Cardano was unable to solve it. He then passed it onto Ferrari who managed to solve it. It is almost a quartic equation (ax4 + bx3 + cx2 + dx + e = 0) except for the fact that it does not have a ‘bx3’ term in it. This makes it a depressed quartic.

 

Ferrari’s Solution to the Equation

 

Reference Structure/Equation/Step

Applying to Cardano’s Equation

Description
2. x4 + px2 + qx + r = 0 x4 + 6x2 – 60x + 36 = 0 Taking 60x from both sides
2. (x2 + p)2 = px2 – qx – r + p2 (x2 + 6)2 = 6x2 + 60x – 36 + 36 Making a perfect square
3. (x2 + p + y)2 = (2y + p)x2 - qx + (y2 + 12y) (x2 + 6 + y)2 = (2y + 6)x2 + 60x + (y2 + 12y) Introducing y into the perfect square. The right-hand side has been made a quadratic equation in x, in the form y = ax2 + bx + c.
4. (-q)2 – 4(p + 2y)(p2 – r + 2py + y2) = 0 3600 – 4(6 + 2y)(12y + y2) = 0 Making a perfect square with y. This is done by making a discriminant zero*.
= –8y3 – 120y2 – 288y – 3600 = 0 Multiplying out the brackets
ax3 + bx2 + cx + d = 0 = -y3 – 15y2 - 36y + 450 = 0 Dividing equation by 8 to simplify. It is now a cubic equation*. This is known as the resolvent cubic of the original quartic.
5. (x2 + 6 + y)2 = (2y + 6)x2 + 60x + (y2 + 12y) (x2 + 6 + 4)2 » (2(4) + 6)x2 + 60x + (42 + 12(4)) Solved the cubic equation and found that y » 4. Now substituting into the equation in step 2.
(x2 + 10)2 » 14x2 + 60x + 64 Simplifying
(x2 + 10)2 » (sqrt.gif (60 bytes)14x + 8)2 Making a perfect square with the right-hand side
6. x2 + 10 » ±(sqrt.gif (60 bytes)14x + 8) Taking the square root of both sides
ax2 + bx + c = 0 x2 ± sqrt.gif (60 bytes)14x +10 ± 8 = 0 In quadratic form (at least enough to substitute into the quadratic formula)
7. –b ± sqrt.gif (60 bytes)b2 – 4ac Solve by using quadratic formula.
x = 3.09557 If you use the positive sign
x = 0.64608 If you use the negative sign