Elements of Algebra

348 ELEMENTS PART II. 92. Hitherto, when the given formula contained a second term, we were obliged to expunge it, but the method we have now given may be applied, without taking away that second term, in the following manner. Let $ax^2 + bx + c$ be the given formula, which must be a square, $y^2$, and let us suppose that we already know the case $af^2 + bf + c = g^2$. Now, if we subtract this equation from the first, we shall have $a(x^2 - f^2) + b(x - f) = y^2 - g^2$, which may be expressed by factors in this manner: $(x - f) \times (ax + af + b) = (y - g) \times (y + g)$; and if we multiply both sides by $pq$, we shall have $pq(x - f) \times (ax + af + b) = pq (y - g) \times (y + g)$, which equation may be resolved into these two, 1. $p(x - f) = q(y - g)$, 2. $q(ax + af + b) = p(y + g)$. Now, multiplying the first by $p$, and the second by $q$, and subtracting the first product from the second, we obtain $(aq^2 - p^2)x + (aq^2 + p^2)f + bq^2 = 2gpq$, which gives $x = \frac{2gpq}{aq^2 - p^2} - \frac{(aq^2 + p^2)f}{aq^2 - p^2} - \frac{bq^2}{aq^2 - p^2}$. But the first equation is $p(x - f) = q(y - g) = \dots$ $p(\frac{2gpq}{aq^2 - p^2} - \frac{2afq^2}{aq^2 - p^2} - \frac{bq^2}{aq^2 - p^2})$; so that $y - g = \frac{2gp^2}{aq^2 - p^2} - \frac{2afpq}{aq^2 - p^2} - \frac{bpq}{aq^2 - p^2}$; consequently, $y = g(\frac{aq^2 + p^2}{aq^2 - p^2}) - \frac{2afpq}{aq^2 - p^2} - \frac{bpq}{aq^2 - p^2}$. Now, in order to remove the fractions, let us make, as before, $\frac{aq^2 + p^2}{aq^2 - p^2} = m$, and $\frac{2pq}{aq^2 - p^2} = n$; and we shall have $m + 1 = \frac{2aq^2}{aq^2 - p^2}$, and $\frac{q^2}{aq^2 - p^2} = \frac{m+1}{2a}$; therefore $x = ng - mf - \frac{b(m+1)}{2a}$; and $y = mg - naf - \frac{1}{2}bn$; in which the letters $m$ and $n$ must be such, that, as before, $m^2 = an^2 + 1$. 93. The formulæ which we have obtained for $x$ and $y$, are still mixed with fractions, since some of their terms contain the letter $b$; for which reason they do not answer our