Elements of Algebra

CHAP. XV. OF ALGEBRA. 285 $z^3 - \frac{25}{2}z^2 + \frac{769}{16}z - \frac{225}{4} = 0$. First, to remove the fractions, let us make $z = \frac{u}{4}$; and we shall have $\frac{u^3}{64} - \frac{25u^2}{32} + \frac{769u}{64} - \frac{225}{4} = 0$, and multiplying by the greatest denominator, we obtain $u^3 - 50u^2 + 769u - 3600 = 0$. We have now to determine the three roots of this equation; which are all three found to be positive; one of them being $u = 9$: then dividing the equation by $u - 9$, we find the new equation $u^2 - 41u + 400 = 0$, or $u^2 = 41u - 400$, which gives $u = \frac{41}{2} \pm \sqrt{(\frac{1681}{4} - \frac{1600}{4})} = \frac{41 \pm 9}{2}$; so that the three roots are $u = 9$, $u = 16$, and $u = 25$. Consequently, as $z = \frac{u}{4}$ the roots are 1. $z = \frac{9}{4}$, 2. $z = 4$, 3. $z = \frac{25}{4}$. These, therefore, are the values of the letters $p$, $q$, and $r$; that is to say, $p = \frac{9}{4}$, $q = 4$, and $r = \frac{25}{4}$. Now, if we consider that $\sqrt{pqr} = \sqrt{h} = -\frac{15}{2}$, and that therefore this value $= \frac{1}{2}b$ is negative, we must, agreeably to what has been said with regard to the signs of the roots $\sqrt{p}$, $\sqrt{q}$, and $\sqrt{r}$, take all those three roots negatively, or take only one of them negatively; and consequently, as $\sqrt{p} = \frac{3}{2}$, $\sqrt{q} = 2$, and $\sqrt{r} = \frac{5}{2}$, the four roots of the given equation are found to be: 1. $x = \frac{3}{2} + 2 - \frac{5}{2} = 1$, 2. $x = \frac{3}{2} - 2 + \frac{5}{2} = 2$, 3. $x = -\frac{3}{2} + 2 + \frac{5}{2} = 3$, 4. $x = -\frac{3}{2} - 2 - \frac{5}{2} = -6$. From these roots are formed the four factors, $(x-1) \times (x-2) \times (x-3) \times (x+6) = 0$. The first two, multiplied together, give $x^2 - 3x + 2$; the product of the last two is $x^2 + 3x - 18$; again multiplying these two products together, we obtain exactly the equation proposed. 779. It remains now to shew how an equation of the fourth degree, in which the second term is found, may be transformed into another, in which that term is wanting: for which we shall give the following Rule *. * An investigation of this rule may be seen in Maclaurin's Algebra, Part II. chap. 3.