Elements of Algebra

544 ADDITIONS. CHAP. V. and $r = 1$; therefore, $s = \mu = 2$, hence $y' = \frac{r}{s} = \frac{1}{2}$; but we know that we must multiply the value of $y'$ by 2; so that we shall have $y' = 1$; wherefore, tracing back the steps, we obtain $\frac{q'}{p} = 1$; whence $q' = p$; and the equation $-12p^2 = (-12q + 41q')^2 - 13q'^2$ will give $(-12q + 41p)^2 = p^2$; that is, $-12q + 41p = p$; so that $12q = 40p$; therefore, $y = \frac{q}{p} = \frac{40}{12} = \frac{10}{3}$; but as we must divide the value of $y$ by 2, we shall have $y = \frac{5}{3}$; which will be the root of the given formula, $7 + 15x + 13x^2$; so that making $7 + 15x + 13x^2 = \frac{25}{9}$, we shall find, by resolving the equation, that $26x + 15 = \pm \frac{7}{3}$; whence, $x = -\frac{19}{39}$, or $= -\frac{2}{3}$. We might have also taken $-12q + 41p = -p$, and should have had $y = \frac{q}{p} = \frac{42}{12} = \frac{7}{2}$; and, dividing by 2, $y = \frac{7}{4}$; then making $7 + 15x + 13x^2 = (\frac{7}{4})^2$, we shall find $26x + 15 = \pm \frac{35}{8}$; whence, $x = -\frac{85}{208}$, or $= -\frac{3}{2}$. If we wished to have other values of $x$, we should only have to seek other solutions of the equation $r^2 = 3s^2 + s^2$, which is resolvible in general by the methods that are known; but when we know a single value of $x$, we may immediately deduce from it all the other satisfactory values, by the method explained in Chap. IV. of the preceding Treatise. 57. Scholium. Suppose, in general, that the quantity $a + bx + cx^2$ becomes equal to a square $g^2$, when $x = f$, so that we have $a + bf + cf^2 = g^2$; then $a = g^2 - bf - cf^2$; substituting this value in the given formula, it will become $g^2 + b(x - f) + c(x^2 - f^2)$. Now, let us take $g + m(x - f)$ for the root of this quantity, ($m$ being an indeterminate number), and we shall have the equation, $g^2 + b(x - f) + c(x^2 - f^2) =$ $g^2 + 2mg(x - f) + m^2(x - f)^2$; that is, expunging $g^2$ on both sides, and then dividing by $x - f$, we have $b + c(x + f) = 2mg + m^2(x - f)$; whence we find $x = \frac{fm^2 - 2gm + b + cf}{m^2 - c}$. And it is evident,