Quartic function

[2] The solution of the quartic was published together with that of the cubic by Ferrari's mentor Gerolamo Cardano in the book Ars Magna.

The notes left by Évariste Galois prior to dying in a duel in 1832 later led to an elegant complete theory of the roots of polynomials, of which this theorem was one result.

[4] Each coordinate of the intersection points of two conic sections is a solution of a quartic equation.

Here are examples of other geometric problems whose solution involves solving a quartic equation.

To calculate its location relative to a triangulated surface, the position of a horizontal torus on the z-axis must be found where it is tangent to a fixed line, and this requires the solution of a general quartic equation to be calculated.

[7][8][9] Finding the distance of closest approach of two ellipses involves solving a quartic equation.

[10] Intersections between spheres, cylinders, or other quadrics can be found using quartic equations.

The four roots x1, x2, x3, and x4 for the general quartic equation with a ≠ 0 are given in the following formula, which is deduced from the one in the section on Ferrari's method by back changing the variables (see § Converting to a depressed quartic) and using the formulas for the quadratic and cubic equations.

Since x2 − xz + m = 0, the quartic equation P(x) = 0 may be solved by applying the quadratic formula twice.

After regrouping the coefficients of the power of y on the right-hand side, this gives the equation which is equivalent to the original equation, whichever value is given to m. As the value of m may be arbitrarily chosen, we will choose it in order to complete the square on the right-hand side.

This was not a problem at the time of Ferrari, when one solved only explicitly given equations with numeric coefficients.

For a general formula that is always true, one thus needs to choose a root of the cubic equation such that m ≠ 0.

Therefore, the solutions of the original quartic equation are A comparison with the general formula above shows that √2m = 2S.

Descartes[14] introduced in 1637 the method of finding the roots of a quartic polynomial by factoring it into two quadratic ones.

Since the coefficient of y3 is 0, we get s = −u, and: One can now eliminate both t and v by doing the following: If we set U = u2, then solving this equation becomes finding the roots of the resolvent cubic which is done elsewhere.

In order to determine the right sign of the square roots, one simply chooses some square root for each of the numbers α, β, and γ and uses them to compute the numbers r1, r2, r3, and r4 from the previous equalities.

Since α, β, and γ are the roots of (2), it is a consequence of Vieta's formulas that their product is equal to q2 and therefore that √α√β√γ = ±q.

This argument suggests another way of choosing the square roots: Of course, this will make no sense if α or β is equal to 0, but 0 is a root of (2) only when q = 0, that is, only when we are dealing with a biquadratic equation, in which case there is a much simpler approach.

The symmetric group S4 on four elements has the Klein four-group as a normal subgroup.

All these different expressions may be deduced from one of them by simply changing the numbering of the xi.

These expressions are unnecessarily complicated, involving the cubic roots of unity, which can be avoided as follows.

There is an alternative solution using algebraic geometry[18] In brief, one interprets the roots as the intersection of two quadratic curves, then finds the three reducible quadratic curves (pairs of lines) that pass through these points (this corresponds to the resolvent cubic, the pairs of lines being the Lagrange resolvents), and then use these linear equations to solve the quadratic.

The four roots of the depressed quartic x4 + px2 + qx + r = 0 may also be expressed as the x coordinates of the intersections of the two quadratic equations y2 + py + qx + r = 0 and y − x2 = 0 i.e., using the substitution y = x2 that two quadratics intersect in four points is an instance of Bézout's theorem.

This pencil contains three reducible quadratics, each corresponding to a pair of lines, each passing through two of the four points, which can be done

Graph of a polynomial of degree 4, with 3 critical points and four real roots (crossings of the x axis) (and thus no complex roots). If one or the other of the local minima were above the x axis, or if the local maximum were below it, or if there were no local maximum and one minimum below the x axis, there would only be two real roots (and two complex roots). If all three local extrema were above the x axis, or if there were no local maximum and one minimum above the x axis, there would be no real root (and four complex roots). The same reasoning applies in reverse to polynomial with a negative quartic coefficient.
Solution of written out in full. This formula is too unwieldy for general use; hence other methods, or simpler formulas for special cases, are generally used.