and the equation is on this account said to be algebraically solvable, or more accurately solvable by radicals. Or we may by writing x = -1/2 p + z reduce the equation to z^2 = 1/4(p^2 - 4q), viz. to an equation of the form x^2 = 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 [root 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 = 1/2{p [+-] [root](p^2 - 4q)}, substituting for p, q their given numerical values, we obtain for x an expression of the form x = [alpha] + [beta]i [+-] [root]([gamma] + [delta]i), where [alpha], [beta], [gamma], [delta] 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 [root]([gamma]+[delta]i) 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 [root]([gamma] + [delta]i) has such a value calculable by means of the radical expressions [root]{[root]([gamma]^2 + [delta]^2) [+-] [gamma]} : 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 x^3 + qx - r = 0, and assuming x = a + b, this will be a solution if only 3ab = q and a^3 + b^3 = r, equations which give (a^3 - b^3)^2 = r^2 - (4/27)q^3, a quadric equation solvable by radicals, and giving a^3 - b^3 = [root](r^2 - (4/27)q^3), a 2-valued function of the coefficients: combining this with a^3 + b^3 = r, we have a^3 = 1/2{r + [root](r^2 - (4/27)q^3)}, a 2-valued function: we then have a by means of a cube root, viz. a = [root 3][1/2{r + [root](r^2 - (4/27)q^3)}],