Elements of Algebra

CHAP. II. ADDITIONS. 521 We shall now give the ambiguous sign $\pm$ to B, as well as to $\sqrt{E}$, and shall then take these signs such, that $a = \frac{\pm 59 \pm \sqrt{79}}{9}$ may be a positive quantity (Art. 39); whence we see, that we must always take the upper sign for the number 59; and, that for the radical $\sqrt{79}$, we may either take the upper, or the under. So that we shall always make $q^0 = -\frac{1}{2}u$, and $\sqrt{E}$ may be taken, successively, plus and minus. First, therefore, if $\frac{1}{2}\sqrt{E} = \sqrt{79}$ with the positive sign, we shall make (Art. 33), the following calculation: $q^{vii} = 9 \times 1 - 5 = 4$, $q^{vi} = -6 \times 2 + 7 = -5$, $q^v = 5 \times 3 - 8 = 7$, $q^{iv} = -3 \times 5 + 7 = -8$, $q''' = 10 \times 1 - 3 = 7$, $q'' = -7 \times 1 + 4 = -3$, $q' = 9 \times 7 - 59 = 4$, $q^0 = -59$. $p^{viii} = \frac{16-79}{9} = -7$, $p^{vii} = \frac{25-79}{-6} = 9$, $p^{vi} = \frac{49-79}{5} = -6$, $p^v = \frac{64-79}{-3} = 5$, $p^{iv} = \frac{49-79}{10} = -3$, $p''' = \frac{9-79}{-7} = 10$, $p'' = \frac{16-79}{9} = -7$, $p' = 9$. $q^{viii} = \frac{-4-\sqrt{79}}{-7} = 1$, $q^{vii} = \frac{5+\sqrt{79}}{9} = 1$, $q^{vi} = \frac{-7-\sqrt{79}}{-6} = 2$, $q^v = \frac{8+\sqrt{79}}{5} = 3$, $q^{iv} = \frac{-7-\sqrt{79}}{-3} = 5$, $q''' = \frac{3+\sqrt{79}}{10} = 1$, $q'' = \frac{-4-\sqrt{79}}{-7} = 1$, $q' = \frac{59+\sqrt{79}}{9} = 7$. \ldots \ldots \ldots Here I stop, because I perceive that $q^{viii} = q''$, and