Elements of Algebra

CHAP. XV. OF ALGEBRA. 283 another method has been discovered of performing the same resolution: and, as it is altogether different from the first, it deserves to be separately explained*. 774. We will suppose that the root of an equation of the fourth degree has the form, $x = \sqrt p + \sqrt q + \sqrt r$, in which the letters $p, q, r$, express the roots of an equation of the third degree, such as, $z^3 - fz^2 + gz - h = 0$; so that $p+q+r = f$; $pq + pr + qr = g$; and $pqr = h$. This being laid down, we square the assumed formula, $x = \sqrt p + \sqrt q + \sqrt r$, and we obtain $x^2 = p+q+r+2\sqrt{pq} + 2\sqrt{pr} + 2\sqrt{qr}$; and, since $p+q+r = f$, we have $x^2 - f = 2\sqrt{pq} + 2\sqrt{pr} + 2\sqrt{qr}$. We again take the squares, and find $x^4 - 2fx^2 + f^2 = 4pq + 4pr + 4qr + 8\sqrt{p^2qr} + 8\sqrt{pq^2r} + 8\sqrt{pqr^2}$. Now, $4pq + 4pr + 4qr = 4g$; so that the equation becomes $x^4 - 2fx^2 + f^2 - 4g = 8\sqrt{pqr} \times (\sqrt p + \sqrt q + \sqrt r)$; but $\sqrt p + \sqrt q + \sqrt r = x$, and $pqr = h$, or $\sqrt{pqr} = \sqrt h$; wherefore we arrive at this equation of the fourth degree, $x^4 - 2fx^2 - 8x\sqrt h + f^2 - 4g = 0$, one of the roots of which is $x = \sqrt p + \sqrt q + \sqrt r$; and in which $p, q,$ and $r$, are the roots of the equation of the third degree, $z^3 - fz^2 + gz - h = 0$. 775. The equation of the fourth degree, at which we have arrived, may be considered as general, although the second term $x^3y$ is wanting; for we shall afterwards shew, that every complete equation may be transformed into another, from which the second term has been taken away. Let there be proposed the equation $x^4 - ax^2 - bx - c = 0$, in order to determine one of its roots. We will first compare it with the formula, in order to obtain the values of $f, g,$ and $h$; and we shall have, 1. $2f = a$, and, consequently, $f = \frac{a}{2}$; 2. $8\sqrt h = b$, so that $h = \frac{b^2}{64}$; 3. $f^2 - 4g = -c$, or $\frac{a^2}{4} - 4g + c = 0$, or $\frac{1}{4}a^2 + c = 4g$; consequently, $g = \frac{1}{16}a^2 + \frac{1}{4}c$. * This method was the invention of Euler himself. He has explained it in the sixteenth volume of the Ancient Commentaries of Petersburg. F. T.