cabbghd2:
OK. FINE.
The link at the bottom also has:
Obtaining alternative solutions the hard way
Ferrari's solution
Otherwise, the depressed quartic can be solved by means of a method discovered by Ferrari. Once the depressed quartic has been obtained, the next step is to add the valid identity
(u2 + α)2 − u4 − 2αu2 = α2
to equation (1), yielding
The effect has been to fold up the u4 term into a perfect square: (u2 + α)2. The second term, α u2 did not disappear, but its sign has changed and it has been moved to the right side.
The next step is to insert a variable y into the perfect square on the left side of equation (2), and a corresponding 2y into the coefficient of u2 in the right side. To accomplish these insertions, the following valid formulas will be added to equation (2),
and
0 = (α + 2y)u2 − 2yu2 − αu2
These two formulas, added together, produce
which added to equation (2) produces
(u2 + α + y)2 + βu + γ = (α + 2y)u2 + (2yα + y2 + α2).
This is equivalent to
The objective now is to choose a value for y such that the
right side of equation (3) becomes a perfect square. This can be done
by letting the discriminant of the quadratic function become zero. To
explain this, first expand a perfect square so that it equals a
quadratic function:
(su + t)2 = (s2)u2 + (2st)u + (t2).
The quadratic function on the right side has three coefficients. It
can be verified that squaring the second coefficient and then
subtracting four times the product of the first and third coefficients
yields zero:
(2st)2 − 4(s2)(t2) = 0.
Therefore to make the right side of equation (3) into a perfect square, the following equation must be solved:
( − β)2 − 4(2y + α)(y2 + 2yα + α2 − γ) = 0.
Multiply the binomial with the polynomial,
β2 − 4(2y3 + 5αy2 + (4α2 − 2γ)y + (α3 − αγ)) = 0
Divide both sides by −4, and move the −β2/4 to the right,
This is a cubic equation for y. Divide both sides by 2,
Conversion of the nested cubic into a depressed cubic
Equation (4) is a cubic equation nested within the quartic equation.
It must be solved in order to solve the quartic. To solve the cubic,
first transform it into a depressed cubic by means of the substitution
Equation (4) becomes
Expand the powers of the binomials,
Distribute, collect like powers of v, and cancel out the pair of v2 terms,
This is a depressed cubic equation.
Relabel its coefficients,
The depressed cubic now is
Solving the nested depressed cubic
The solutions (any solution will do, so pick any of the three complex roots) of equation (5) are
let
![U=\sqrt[3]{{Q\over 2}\pm \sqrt{{Q^{2}\over 4}+{P^{3}\over 27}}}](http://upload.wikimedia.org/math/3/7/f/37f11fad89252a07480c5681cc59c0e6.png)
(taken from
Cubic equation)
therefore the solution of the original nested cubic is
Remember 1:
Remember 2:
![\lim_{P\to 0}{P \over \sqrt[3]{{Q\over 2} + \sqrt{{Q^{2}\over 4}+{P^{3}\over 27}}}}=0](http://upload.wikimedia.org/math/9/8/f/98f330f9f936654f29670a874144a315.png)
Folding the second perfect square
With the value for y given by equation (6), it is now known that the right side of equation (3) is a perfect square of the form
This is correct for both signs of square root, as long as the same
sign is taken for both square roots. A ± is redundant, as it would be
absorbed by another ± a few equations further down this page.
so that it can be folded:
.
Note: If
β ≠ 0 then
α + 2
y ≠ 0. If
β = 0 then this would be a biquadratic equation, which we solved earlier.
Therefore equation (3) becomes
.
Equation (7) has a pair of folded perfect squares, one on each side of the equation. The two perfect squares balance each other.
If two squares are equal, then the sides of the two squares are also equal, as shown by:
.
Collecting like powers of u produces
.
Note: The subscript
s of
and
is to note that they are dependent.
Equation (8) is a quadratic equation for u. Its solution is
Simplifying, one gets
This is the solution of the depressed quartic, therefore the solutions of the original quartic equation are
Remember: The two
come from the same place in equation (7'), and should both have the same sign, while the sign of
is independent.
Summary of Ferrari's method
Given the quartic equation
Ax4 + Bx3 + Cx2 + Dx + E = 0,
its solution can be found by means of the following calculations:
if
β = 0 solve
u4 + αu2 + γ = 0 and substitute
finding the roots
.
, (either sign of the square root will do)
![U = \sqrt[3]{R}](http://upload.wikimedia.org/math/9/0/0/90034f31688ee818ff7c9ef96f161e26.png)
, (there are 3 complex roots, any one of them will do)
The two ±
s must have the same sign, the ±
t is independent. To get all roots, find x for ±
s,±
t
= +,+ and for +,− and for −,+ and for −,−. Double roots will be given
twice, triple roots 3 times and quadruple roots would be given 4 times
(although then
β = 0, which is a special case). The order of the roots depends on which cubic root
U one chose. (see note for (8) vis-à-vis (8'))
Quod Erat Faciendum. There are other methods of solving the quartic equations, perhaps more optimal. Ferrari was the first to discover one of these labyrinthine solutions. The equation which he solved was
x4 + 6x2 − 60x + 36 = 0
which was already in depressed form. It has a pair of solutions which can be found with the set of formulas shown above.
http://en.wikipedia.org/wiki/Quartic_equation
Obtaining alternative solutions the hard way
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