Editing Quartic function (section)

====Solving with algebraic geometry====
There is an alternative solution using algebraic geometry<ref>{{Citation|last = Faucette|first = William M.|journal = [[American Mathematical Monthly]]|pages = 51–57|title = A Geometric Interpretation of the Solution of the General Quartic Polynomial|volume = 103|year = 1996|issue = 1|doi = 10.2307/2975214|jstor = 2975214|mr = 1369151}}</ref> In brief, one interprets the roots as the intersection of two quadratic curves, then finds the three [[degenerate conic|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 {{math|''x''<sup>4</sup> + ''px''<sup>2</sup> + ''qx'' + ''r'' {{=}} 0}} may also be expressed as the {{mvar|x}} coordinates of the intersections of the two quadratic equations {{math|''y''<sup>2</sup> + ''py'' + ''qx'' + ''r'' {{=}} 0}} and {{math|''y'' − ''x''<sup>2</sup> {{=}} 0}} i.e., using the substitution {{math|''y'' {{=}} ''x''<sup>2</sup>}} that two quadratics intersect in four points is an instance of [[Bézout's theorem]]. Explicitly, the four points are {{math|''P<sub>i</sub>'' ≔ (''x<sub>i</sub>'', ''x<sub>i</sub>''<sup>2</sup>)}} for the four roots {{math|''x<sub>i</sub>''}} of the quartic.

These four points are not collinear because they lie on the irreducible quadratic {{math|''y'' {{=}} ''x''<sup>2</sup>}} and thus there is a 1-parameter family of quadratics (a [[pencil of curves]]) passing through these points. Writing the projectivization of the two quadratics as [[quadratic form]]s in three variables:
:<math>\begin{align}
F_1(X,Y,Z) &:= Y^2 + pYZ + qXZ + rZ^2,\\
F_2(X,Y,Z) &:= YZ - X^2
\end{align}</math>
the pencil is given by the forms {{math|''λF''<sub>1</sub> + ''μF''<sub>2</sub>}} for any point {{math|[''λ'', ''μ'']}} in the projective line — in other words, where {{math|''λ''}} and {{math|''μ''}} are not both zero, and multiplying a quadratic form by a constant does not change its quadratic curve of zeros.

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 <math>\textstyle{\binom{4}{2}}</math>&nbsp;=&nbsp;{{math|6}} different ways. Denote these {{math|''Q''<sub>1</sub> {{=}} ''L''<sub>12</sub> + ''L''<sub>34</sub>}}, {{math|''Q''<sub>2</sub> {{=}} ''L''<sub>13</sub> + ''L''<sub>24</sub>}}, and  {{math|''Q''<sub>3</sub> {{=}} ''L''<sub>14</sub> + ''L''<sub>23</sub>}}. Given any two of these, their intersection has exactly the four points.

The reducible quadratics, in turn, may be determined by expressing the quadratic form {{math|''λF''<sub>1</sub> + ''μF''<sub>2</sub>}} as a {{math|3×3}}&nbsp;matrix: reducible quadratics correspond to this matrix being singular, which is equivalent to its determinant being zero, and the determinant is a homogeneous degree three polynomial in {{math|''λ''}} and {{math|''μ''}} and corresponds to the resolvent cubic.