Elements of Algebra

CHAP. II. ADDITIONS. 509 has two real irrational roots, they must both be reduced to continued fractions. This operation may be performed with the greatest ease by a method which we have elsewhere explained, and which it may be proper to repeat here, since it is naturally deduced from the formulæ of Art. 25, and likewise contains all the principles necessary for the complete and general solution of the problem proposed. Let us, therefore, denote the root which is to be thrown into a continued fraction by $a$, which we shall suppose to be always positive; at the same time, let $b$ be the other root, and we shall evidently have $a + b = -\frac{B}{A}$, and $ab = \frac{C}{A}$; whence $a - b = \frac{\sqrt{(B^2 - 4AC)}}{A}$; or, for the sake of abridgment, making $B^2 - 4AC = E$, $a - b = \frac{\sqrt{E}}{A}$, where the radical $\sqrt{E}$ may be positive, or negative: it will be positive, when the root $a$ is the greater of the two, and negative, when that root is the less; therefore $a = \frac{-B + \sqrt{E}}{2A}$, $b = \frac{-B - \sqrt{E}}{2A}$. Now, if we preserve the denominations of Art. 25, we shall only have to substitute for $a$ the preceding value, and the difficulty will only consist in determining the integer, approximate values, $\mu'$, $\mu''$, $\mu'''$, &c. To facilitate these determinations, I multiply the numerator and the denominator of the fractions, $\frac{p^0 - aq^0}{aq' - p'}$, $\frac{aq' - p'}{p'' - aq''}$, $\frac{p'' - aq''}{aq''' - p'''}$, &c. respectively by $A(bq' - p')$, $A(p'' - bq'')$, $A(bq''' - p''')$, &c. and as we have $A(p^0 - aq^0) \times (p^0 - bq^0) = A$ $A(aq' - p') \times (bq' - p') = Ap'^2 - A(a+b)p'q' + Aabq'^2 = Ap'^2 + Bp'q' + Cq'^2$, $A(p'' - aq'') \times (p'' - bq'') = Ap''^2 - A(a+b)p''q'' + Aabq''^2 = Ap''^2 + Bp''q'' + Cq''^2$, &c. $A(p^0 - aq^0) \times (bq' - p') = -\mu A - \frac{1}{2}B - \frac{1}{2}\sqrt{E}$,