Elements of Algebra

CHAP. XV. \qquad OF ALGEBRA. \qquad 287 $y^4 = x^4 + 8x^3 + 24x^2 + 32x + 16$ $-8y^3 = -8x^3 - 48x^2 - 96x - 64$ $14y^2 = 14x^2 + 56x + 56$ $4y = 4x + 8$ $-8 = -8$ *** $x^4 + 0 - 10x^2 - 4x + 8 = 0$. This equation being compared with our general formula, gives $a = 10$, $b = 4$, $c = -8$; whence we conclude, that $f = 5$, $g = \frac{1}{2}$, $h = \frac{1}{4}$; and $\sqrt{h} = \frac{1}{2}$; that the product $\sqrt{pqr}$ will be positive; and that it is from the equation of the third degree, $z^3 - 5z^2 + \frac{1}{2}z - \frac{1}{4} = 0$, that we are to seek for the three roots $p, q, r$. (Art. 774.) 782. Let us first remove the fractions from this equation, by making $z = \frac{u}{2}$, and we shall thus have, after multiplying by 8, the equation $u^3 - 10u^2 + 17u - 2 = 0$, in which all the roots are positive. Now, the divisors of the last term are 1 and 2; if we try $u=1$, we find $1 - 10 + 17 - 2 = 6$; so that the equation is not reduced to nothing; but trying $u=2$, we find $8 - 40 + 34 - 2 = 0$, which answers to the equation, and shews that $u=2$ is one of the roots. The two others will be found by dividing by $u-2$, as usual; then the quotient $u^2 - 8u + 1 = 0$ will give $u^2 = 8u - 1$, and $u = 4 \pm \sqrt{15}$. And since $z = \frac{1}{2}u$, the three roots of the equation of the third degree are, 1, $z = p = 1$, 2, $z = q = \frac{4+\sqrt{15}}{2}$, 3, $z = r = \frac{4-\sqrt{15}}{2}$. 783. Having therefore determined $p, q, r$, we have also their square roots; namely, $\sqrt{p} = 1$, $\sqrt{q} = \frac{\sqrt{(8+2\sqrt{15})}}{2}$*, and $\sqrt{r} = \frac{\sqrt{(8-2\sqrt{15})}}{2}$. * This expression for the square root of $q$ is obtained by multiplying the numerator and denominator of $\frac{4+\sqrt{15}}{2}$ by 2, and extracting the root of the latter, in order to remove the surd: Thus, $\frac{4+\sqrt{15}}{2} \times 2 = \frac{8+2\sqrt{15}}{4}$; and $\frac{\sqrt{(8+2\sqrt{15})}}{\sqrt{4}} \ldots$ $= \frac{\sqrt{(8+2\sqrt{15})}}{2}$.