Elements of Algebra

CHAP. XIII. OF ALGEBRA. 277 Since, therefore, $\sqrt{21} = 4 \cdot 5825$, the third root is very nearly $x = 4 \cdot 7912$, and the fourth, $x = 0 \cdot 2087$. It would have been easy to have determined these roots with still more precision: for we observe that the fourth root is very nearly $\frac{2}{10}$, or $\frac{1}{5}$, which value will answer the equation with sufficient exactness. In fact, if we make $x = \frac{1}{5}$, we find $\frac{1}{625} - \frac{4}{125} - \frac{6}{25} + \frac{4}{5} + 1 = \frac{31}{625}$. We ought however to have obtained 0, but the difference is evidently not great. 763. The second case in which such a resolution takes place, is the same as the first with regard to the coefficients, but differs from it in the signs, for we shall suppose that the second and the fourth terms have different signs; such, for example, as the equation $x^4 + max^3 + na^2x^2 - ma^3x + a^4 = 0$, which may be represented by the product, $(x^2 + pax - a^2) \times (x^2 + qax - a^2) = 0$. For the actual multiplication of these factors gives $x^4 + (p+q)ax^3 + (pq-2)a^2x^2 - (p+q)a^3x + a^4$, a quantity equal to that which was given, if we suppose, in the first place, $p+q=m$, and in the second place, $pq-2=n$, or $pq=n+2$; because in this manner the fourth terms become equal of themselves. If now we square the first equation, as before, (Art. 761.) we shall have $p^2 + 2pq + q^2 = m^2$; and if from this we subtract the second, taken four times, or $4pq = 4n+8$, there will remain $p^2 - 2pq + q^2 = m^2 - 4n - 8$; the square root of which is $p-q = \sqrt{(m^2 - 4n - 8)}$, and thence, by adding $p+q=m$, we obtain $p = \frac{m+\sqrt{(m^2-4n-8)}}{2}$; and, by subtracting $p+q$, \ldots $q = \frac{m-\sqrt{(m^2-4n-8)}}{2}$. Having therefore found $p$ and $q$, we shall obtain from the first factor (as in Art. 761.) the two roots $x = -\frac{1}{2}pa \pm \frac{1}{2}a\sqrt{(p^2+4)}$, and from the second factor the two roots $x = -\frac{1}{2}qa \pm \frac{1}{2}a\sqrt{(q^2+4)}$; that is, we have the four roots of the equation proposed. 764. Let there be given the equation $x^4 - 3 \cdot 2x^3 + 3 \cdot 8x + 16 = 0$. Here we have $a=2$, $m=-3$, and $n=0$; so that $\sqrt{(m^2-4n-8)} = 1$, $=p-q$; and, consequently, $p = \frac{-3+1}{2} = -1$, and $q = \frac{-3-1}{2} = -2$. Therefore the first two roots are $x = 1 \pm \sqrt{5}$, and the