Elements of Algebra

548 ADDITIONS. CHAP. VI. equality will be resolved by the method already explained; and knowing $u$, we shall likewise have $x = \frac{u^2-c}{d}$. If the double equality were $ax^2 + bx = \square$, $cx^2 + dx = \square$, we should only have to make $x = \frac{1}{x'}$, and then multiplying both formulæ by the square $x'^2$, we should get these two equalities, $a + bx' = \square$, and $c + dx' = \square$, which are similar to the preceding. Thus, we may resolve, in general, all the double equalities, in which the unknown quantity does not exceed the first degree, and those in which the unknown quantity is found in all the terms, provided it does not exceed the second degree; but it is not the same when we have equalities of this form, $a + bx + cx^2 = \square$, $\alpha + \beta x + \gamma x^2 = \square$. If we resolve the first of these equalities by our method, and call $f$ the value of $x$, which makes $a + bx + cx^2 = g^2$, we shall have, in general (Art. 57.), $x = \frac{fm^2 - 2gm + b + cf}{m^2 - c}$; wherefore, substituting this expression of $x$ in the other formula, $\alpha + \beta x + \gamma x^2$, and then multiplying it by $(m^2-c)^2$, we shall have to resolve the equality, $\alpha(m^2-c)^2 + \beta(m^2-c) \times (fm^2 - 2gm + b + cf) + \gamma(fm^2 - 2gm + b + cf)^2 = \square$; in which, the unknown quantity, $m$, rises to the fourth degree. Now, we have not yet any general rule for resolving such equalities; and all we can do is to find successively different solutions, when we already know one. (See Chap. IX.) 63. If we had the triple equality, $\left. \begin{array}{r} ax+by \\ cx+dy \\ hx+ky \end{array} \right\} = \square$, we must make $ax + by = t^2$, $cx + dy = u^2$, and $hx + ky = s^2$,