Elements of Algebra

282 ELEMENTS SECT. IV. $x^4 - 6x^3 + 12x^2 - 12x + 4 = 0$, which must be contained in the formula $(x^2 - 3x + p)^2 - (qx + r)^2 = 0$, in the former part of which we have put $-3x$, because $-3$ is half the coefficient, $-6$, of the given equation. This formula being expanded, gives $x^4 - 6x^3 + (2p + 9 - q^2)x^2 - (6p + 2qr)x + p^2 - r^2 = 0$; which, compared with our equation, there will result from that comparison the following equations: 1. $2p + 9 - q^2 = 12$, 2. $6p + 2qr = 12$, 3. $p^2 - r^2 = 4$. The first gives $q^2 = 2p - 3$; the second, $2qr = 12 - 6p$, or $qr = 6 - 3p$; the third, $r^2 = p^2 - 4$. Multiplying $r^2$ by $q^2$, and $p^2 - 4$ by $2p - 3$, we have $q^2r^2 = 2p^3 - 3p^2 - 8p + 12$; and if we square $qr$, and its value, $6 - 3p$, we have $q^2r^2 = 36 - 36p + 9p^2$; so that we have the equation $2p^3 - 3p^2 - 8p + 12 = 9p^2 - 36p + 36$, or $2p^3 - 12p^2 + 28p - 24 = 0$, or $p^3 - 6p^2 + 14p - 12 = 0$, one of the roots of which is $p = 2$; and it follows that $q^2 = 1$, $q = 1$, and $qr - r = 0$. Therefore our equation will be $(x^2 - 3x + 2)^2 = x^2$, and its square root will be $x^2 - 3x + 2 = \pm x$. If we take the upper sign, we have $x^2 = 4x - 2$; and taking the lower sign, we obtain $x^2 = 2x - 2$, whence we derive the four roots $x = 2 \pm \sqrt{2}$, and $x = 1 \pm \sqrt{-1}$. *** CHAP. XV. *Of a new Method of resolving Equations of the Fourth Degree.* 773. The rule of Bombelli, as we have seen, resolves equations of the fourth degree by means of an equation of the third degree; but since the invention of that Rule,