Elements of Algebra

CHAP. XIV. OF ALGEBRA. 279 in order that we may obtain the equation proposed. By squaring, and ordering this new equation, we shall have $x^4 + ax^3 + \frac{1}{4}a^2x^2 + apx + p^2 + 2px^2 - 2qrx - r^2 - q^2x^2$. Now, the first two terms are already the same here as in the given equation; the third term requires us to make $\frac{1}{4}a^2 + 2p - q^2 = b$, which gives $q^2 = \frac{1}{4}a^2 + 2p - b$; the fourth term shews that we must make $ap - 2qr = c$, or $2qr = ap - c$; and, lastly, we have for the last term $p^2 - r^2 = d$, or $r^2 = p^2 - d$. We have therefore three equations which will give the values of $p, q$, and $r$. 767. The easiest method of deriving those values from them is the following: if we take the first equation four times, we shall have $4q^2 = a^2 + 8p - 4b$; which equation, multiplied by the last, $r^2 = p^2 - d$, gives $4q^2r^2 = 8p^3 + (a^2 - 4b)p^2 - 8dp - d(a^2 - 4b)$. Farther, if we square the second equation, we have $4q^2r^2 = a^2p^2 - 2acp + c^2$. So that we have two values of $4q^2r^2$, which, being made equal, will furnish the equation $8p^3 + (a^2 - 4b)p^2 - 8dp - d(a^2 - 4b) = a^2p^2 - 2acp + c^2$; or, bringing all the terms to one side, and arranging, $8p^3 - 4bp^2 + (2ac - 8d)p - a^2d + 4bd - c^2 = 0$, an equation of the third degree, which will always give the value of $p$ by the rules already explained. 768. Having therefore determined three values of $p$ by the given quantities $a, b, c, d$, when it was required to find only one of those values, we shall also have the values of the two other letters $q$ and $r$; for the first equation will give $q = \sqrt{(\frac{1}{4}a^2 + 2p - b)}$, and the second gives $r = \frac{ap-c}{2q}$. Now, these three values being determined for each given case, the four roots of the proposed equation may be found in the following manner: This equation having been reduced to the form $(x^2 + \frac{1}{2}ax + p)^2 - (qx + r)^2 = 0$, we shall have $(x^2 + \frac{1}{2}ax + p)^2 = (qx + r)^2$, and, extracting the root, $x^2 + \frac{1}{2}ax + p = qx + r$, or $x^2 + \frac{1}{2}ax + p = -qx - r$. The first equation gives $x^2 = (q - \frac{1}{2}a)x - p + r$, from which we may find two roots; and the second equation, to which we may give the form $x^2 = -(q + \frac{1}{2}a)x - p - r$, will furnish the two other roots.