and the equation is on this account said to be algebraically solvable, or more accurately solvable by radicals. Or we may by writing x = −½ p + z reduce the equation to z² = ¼ (p² − 4q), viz. to an equation of the form x² = a; and in virtue of its being thus reducible we say that the original equation is solvable by radicals. And the question for an equation of any higher order, say of the order n, is, can we by means of radicals (that is, by aid of the sign m√( ) or ( )1/m, using as many as we please of such signs and with any values of m) find an n-valued function (or any function) of the coefficients which substituted for x in the equation shall satisfy it identically?
It will be observed that the coefficients p, q ... are not explicitly considered as numbers, but even if they do denote numbers, the question whether a numerical equation admits of solution by radicals is wholly unconnected with the before-mentioned theorem of the existence of the n roots of such an equation. It does not even follow that in the case of a numerical equation solvable by radicals the algebraical solution gives the numerical solution, but this requires explanation. Consider first a numerical quadric equation with imaginary coefficients. In the formula x = ½ {p ± √(p² − 4q) }, substituting for p, q their given numerical values, we obtain for x an expression of the form x = α + βi ± √(γ + δi), where α, β, γ, δ are real numbers. This expression substituted for x in the quadric equation would satisfy it identically, and it is thus an algebraical solution; but there is no obvious a priori reason why √(γ + δi) should have a value = c + di, where c and d are real numbers calculable by the extraction of a root or roots of real numbers; however the case is (what there was no a priori right to expect) that √(γ + δi) has such a value calculable by means of the radical expressions √{√(γ² + δ²) ± γ}; and hence the algebraical solution of a numerical quadric equation does in every case give the numerical solution. The case of a numerical cubic equation will be considered presently.
17. A cubic equation can be solved by radicals.
Taking for greater simplicity the cubic in the reduced form x³ + qx − r = 0, and assuming x = a + b, this will be a solution if only 3ab = q and a³ + b³ = r, equations which give (a³ − b³)² = r² − 4⁄27 q³, a quadric equation solvable by radicals, and giving a³ − b³ = √(r² − 4⁄27 q³), a 2-valued function of the coefficients: combining this with a³ + b³ = r, we have a³ = ½ {r + √(r² − 4⁄27 q³) }, a 2-valued function: we then have a by means of a cube root, viz.
a = 3√[½ {r + √(r² − 4⁄27 q³) }],
a 6-valued function of the coefficients; but then, writing q = b/3a, we have, as may be shown, a + b a 3-valued function of the coefficients; and x = a + b is the required solution by radicals. It would have been wrong to complete the solution by writing
b = 3√[½ {r − √(r² − 4⁄27 q³) } ],
for then a + b would have been given as a 9-valued function having only 3 of its values roots, and the other 6 values being irrelevant. Observe that in this last process we make no use of the equation 3ab = q, in its original form, but use only the derived equation 27a³b³ = q³, implied in, but not implying, the original form.
An interesting variation of the solution is to write x = ab(a + b), giving a³b³ (a³ + b³) = r and 3a³b³ = q, or say a³ + b³ = 3r/q, a³b³ = 1⁄3 q; and consequently
| a³ = | 3⁄2 | {r + √(r² − 4⁄27 q³) }, b³ = | 3⁄2 | {r − √(r² − 4⁄27 q³) }, |
| q | q |