Elements of Algebra

284 ELEMENTS SECT. IV. 776. Since, therefore, the equation $x^4 - ax^2 - bx - c = 0$, gives the values of the letters $f$, $g$, and $h$, so that $f = \frac{1}{2}a$, $g = \frac{1}{16}a^2 + \frac{1}{4}c$, and $h = \frac{1}{64}b^2$, or $\sqrt{h} = \frac{1}{8}b$, we form from these values the equation of the third degree $z^3 - fz^2 + gz - h = 0$, in order to obtain its roots by the known rule. And if we suppose those roots, 1. $z=p$, 2. $z=q$, 3. $z=r$, one of the roots of our equation of the fourth degree must be, by the supposition, Art. 774, $x = \sqrt{p} + \sqrt{q} + \sqrt{r}$. 777. This method appears at first to furnish only one root of the given equation; but if we consider that every sign $\sqrt{}$ may be taken negatively, as well as positively, we shall immediately perceive that this formula contains all the four roots. Farther, if we chose to admit all the possible changes of the signs, we should have eight different values of $x$, and yet four only can exist. But it is to be observed, that the product of those three terms, or $\sqrt{pqr}$, must be equal to $\sqrt{h} = \frac{1}{8}b$, and that if $\frac{1}{8}b$ be positive, the product of the terms $\sqrt{p}$, $\sqrt{q}$, $\sqrt{r}$, must likewise be positive; so that all the variations that can be admitted are reduced to the four following: 1. $x = \sqrt{p} + \sqrt{q} + \sqrt{r}$, 2. $x = \sqrt{p} - \sqrt{q} - \sqrt{r}$, 3. $x = -\sqrt{p} + \sqrt{q} - \sqrt{r}$, 4. $x = -\sqrt{p} - \sqrt{q} + \sqrt{r}$. In the same manner, when $\frac{1}{8}b$ is negative, we have only the four following values of $x$: 1. $x = \sqrt{p} + \sqrt{q} - \sqrt{r}$, 2. $x = \sqrt{p} - \sqrt{q} + \sqrt{r}$, 3. $x = -\sqrt{p} + \sqrt{q} + \sqrt{r}$, 4. $x = -\sqrt{p} - \sqrt{q} - \sqrt{r}$. This circumstance enables us to determine the four roots in all cases; as may be seen in the following example. 778. Let there be proposed the equation of the fourth degree, $x^4 - 25x^2 + 60x - 36 = 0$, in which the second term is wanting. Now, if we compare this with the general formula, we have $a=25$, $b=-60$, and $c=36$; and after that, $f = \frac{25}{2}$, $g = \frac{625}{16} + 9 = \frac{769}{16}$, and $h = \frac{225}{4}$; by which means our equation of the third degree becomes,