The discriminant, whose vanishing is the condition that f may possess two equal roots, has the expression j 2 - 6 i 3; it is nine times the discriminant of the cubic **resolvent** k 3 - 2 ik- 3j, and has also the expression 4(1, t') 6 .

For, since -2t 2 =0 3 -21f 2, 6,-3j(-f) 3, he compares the right-hand side with cubic **resolvent** k 3 -21X 2 k - j 2.

Of f=0, :and notices that they become identical on substituting 0 for k, and -f for X; hence, if k1, k2, k 3 be the roots of the **resolvent** -21 2 = (o + k if) (A + k 2f)(o + k 3f); and now, if all the roots of f be different, so also are those of the **resolvent**, since the latter, and f, have practically the same discriminant; consequently each of the three factors, of -21 2, must be perfect squares and taking the square root 1 t = -' (1)ï¿½x4; and it can be shown that 0, x, 1P are the three conjugate quadratic factors of t above mentioned.