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In mathematics, a quartic equation is the result of setting
a quartic function to zero. An example of a quartic equation is the
equation
- 2x4 + 4x3 - 26x2 - 28x + 48 =
0;
the general form is
-
A quartic equation always has four solutions (roots). They may be complex and there may be duplicate solutions.
It is known that four is the highest degree of polynomial equation for which
roots can be expressed using a finite number of arithmetic operators and n-th roots.
If a0 = 0, then one of the roots is x = 0, and the other roots can be found
by dividing by x, and solving the resulting cubic
equation,
- a4x3 + a3x2 +
a2x + a1 = 0.
Converting to a depressed quartic
Let
- Ax4 + Bx3 + Cx2 +
Dx + E = 0
be the general quartic equation which it is desired to solve. Divide both sides by A,
-
The first step should be to eliminate the x3 term. To do this, change variables from x to
u, such that
- .
Then
-
Expanding the powers of the binomials produces
-
Collecting the same powers of u yields
-
Now rename the coefficients of u. Let
-
-
-
The resulting equation is
-
which is a depressed quartic equation.
Ferrari's solution
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 equation
- (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:
- (ax + b)2 = a2x2 +
2abx + b2.
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:
- (2ab)2 - 4a2b2 = 0.
Therefore to make the right side of equation (3) into a perfect square, the following equation must be solved:
- β2 - 4(2y + α)(y2 + 2yα + α2
- γ) = 0.
Multiply the binomial with the polynomial,
- β2 - 4(2y3 + 5αy2 + (4α2 -
2γ)y + (α3 - αγ)) = 0
Multiply both sides by −1, then add β2 to both sides, divide both sides by 4, then subtract
β2/4 from both sides,
-
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 solution of equation (5) is
-
therefore the solution of the original nested cubic is
-
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,
so that it can be folded:
-
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. Note: if β equals zero then the ratio β/|β| (see sign function) will be indeterminate, but in such a case equation (1) becomes equivalent to a quadratic
equation so that it is possible to use the quadratic formula to solve for u2 directly.
If two squares are equal, then the sides of the two squares are also equal, viz.
-
Collecting like powers of u produces
-
Equation (8) is a quadratic equation for u. Its
solution is
-
This is the solution of the depressed quartic, therefore the solution of the original quartic equation is
-
Summary
Given the quartic equation
- Ax4 + Bx3 + Cx2 +
Dx + E = 0,
its solution can be found by means of the following calculations:
-
-
-
-
-
-
-
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.
Obtaining alternative solutions
It could happen that the solution obtained through the seven formulae above is complex. It may also be the case that one is only looking for a real solution. Let x1
denote the complex solution. If all the original coefficients A, B, C, D and E are
real -- which should be the case when one desires only real solutions -- then there is another complex solution
x2 which is the complex conjugate of
x1. If the other two roots are denoted as x3 and x4 then the quartic
equation can be expressed as
- (x - x1)(x - x2)(x -
x3)(x - x4) = 0,
but this quartic equation is equivalent to the product of two quadratic equations:
-
and
-
Since
-
then
-
Let
-
- b = [Re(x1)]2 + [Im(x1)]2
so that equation (9) becomes
-
Also let there be (unknown) variables w and v such that equation (10) becomes
-
Multiplying equations (11) and (12) produces
-
Comparing equation (13) to the original quartic equation, it can be seen that
-
-
-
and
-
Therefore
-
-
Equation (12) can be solved for x yielding
-
-
One of these two solutions should be the desired real solution.
See also
Reference
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