Elements of Algebra

Now, squaring the first of those equations, we have $p^2 + 2pq + q^2 = m^2$; and if from this we subtract the second, taken four times, or $4pq = 4n - 8$, there remains $p^2 - 2pq + q^2 = m^2 - 4n + 8$; and taking the square root, we find $p - q = \sqrt{(m^2 - 4n + 8)}$; also, $p + q = m$; we shall therefore have, by addition, $2p = m + \sqrt{(m^2 - 4n + 8)}$, or $p = \frac{m + \sqrt{(m^2 - 4n + 8)}}{2}$; and, by subtraction, $2q = m - \sqrt{(m^2 - 4n + 8)}$, or $q = \frac{m - \sqrt{(m^2 - 4n + 8)}}{2}$. Having therefore found $p$ and $q$, we have only to suppose each factor $= 0$, in order to determine the value of $x$. The first gives $x^2 + pax + a^2 = 0$, or $x^2 = -pax - a^2$, whence we obtain $x = -\frac{pa}{2} \pm \sqrt{(\frac{p^2a^2}{4} - a^2)}$, or $x = -\frac{pa}{2} \pm \frac{1}{2}a\sqrt{(p^2 - 4)}$. The second factor gives $x = -\frac{qa}{2} \pm \frac{1}{2}a\sqrt{(q^2 - 4)}$; and these are the four roots of the given equation. 762. To render this more clear, let there be given the equation $x^4 - 4x^3 - 3x^2 - 4x + 1 = 0$. We have here $a = 1$, $m = -4$, $n = -3$; consequently, $m^2 - 4n + 8 = 36$, and the square root of this quantity is $= 6$; therefore $p = \frac{-4+6}{2} = 1$, and $q = \frac{-4-6}{2} = -5$; whence result the four roots, 1st and 2d, $x = -\frac{1}{2} \pm \frac{1}{2}\sqrt{-3} = \frac{-1 \pm \sqrt{(-3)}}{2}$; and 3d and 4th, $x = \frac{5}{2} \pm \frac{1}{2}\sqrt{21} = \frac{5 \pm \sqrt{21}}{2}$; that is, the four roots of the given equation are: 1. $x = \frac{-1+\sqrt{-3}}{2}$, 2. $x = \frac{-1-\sqrt{-3}}{2}$, 3. $x = \frac{5+\sqrt{21}}{2}$, 4. $x = \frac{5-\sqrt{21}}{2}$. The first two of these roots are imaginary, or impossible; but the last two are possible; since we may express $\sqrt{21}$ to any degree of exactness, by means of decimal fractions. In fact, $21$ being the same with $21.00000000$, we have only to extract the square root, which gives $\sqrt{21} = 4.5825$.