Elements of Algebra

CHAP. XIV. OF ALGEBRA. 281 $q = \sqrt{(25 + 14 - 35)} = 2$, and $r = \frac{-70+50}{4} = -5$, whence result the two equations of the second degree, 1. $x^2 = 7x - 12$, 2. $x^2 = 3x - 2$; the first gives $x = \frac{7}{2} \pm \sqrt{\frac{1}{4}}$, or $x = \frac{7 \pm 1}{2}$, so that $x = 4$, and $x = 3$; the second furnishes the root $x = \frac{3}{2} \pm \sqrt{\frac{1}{4}} = \frac{3 \pm 1}{2}$, and, consequently, $x = 2$, and $x = 1$; therefore, by this second supposition, the same four roots are found as by the first. Lastly, the same roots are found, by the third value of $p$, $= \frac{11}{2}$; for, in this case, we have $q = \sqrt{(25 + 11 - 35)} = 1$, and $r = \frac{-55+50}{2} = -\frac{5}{2}$; so that the two equations of the second degree become, 1. $x^2 = 6x - 8$, 2. $x^2 = 4x - 3$. Whence we obtain from the first, $x = 3 \pm \sqrt{1}$, that is to say, $x = 4$, and $x = 2$; and from the second, $x = 2 \pm \sqrt{1}$, that is to say, $x = 3$, and $x = 1$, which are the same roots that we originally obtained. 771. Let there now be proposed the equation $x^4 - 16x - 12 = 0$, in which $a = 0$, $b = 0$, $c = -16$, $d = -12$; and our equation of the third degree will be $8p^3 + 96p - 256 = 0$, or $p^3 + 12p - 32 = 0$, and we may make this equation still more simple, by writing $p = 2t$; for we have then $8t^3 + 24t - 32 = 0$, or $t^3 + 3t - 4 = 0$. The divisors of the last term are 1, 2, 4; whence one of the roots is found to be $t = 1$; therefore $p = 2$, $q = \sqrt{4} = 2$, and $r = \frac{16}{4} = 4$. Consequently, the two equations of the second degree are $x^2 = 2x + 2$, and $x^2 = -2x - 6$; which give the roots $x = 1 \pm \sqrt{3}$, and $x = -1 \pm \sqrt{-5}$. 772. We shall endeavour to render this resolution still more familiar, by a repetition of it in the following example. Suppose there were given the equation