Elements of Algebra

378 ELEMENTS PART II. $64 + 32y = 14p - 14p + p^2y = -56 + 2y$, which gives $y = -4$; that is, the known case again. If we chose to destroy the second terms, we should have $64 = 14p$, and $p = \frac{32}{7}$; and, consequently, dividing the other terms by $y^2$, we should obtain $32 + 8y = -14 + p^2 - 2py$, or $32 + 8y = \frac{338}{49} - \frac{64}{7}y$; whence $y = -\frac{71}{28}$, and $x = -\frac{15}{28}$; that is to say, the same values that we found before. 145. We may proceed, in the same manner, with respect to the general formula, $a + bx + cx^2 + dx^3 + ex^4$, when we know one case, as $x = h$, in which it becomes a square, $k^2$. The constant method is to suppose $x = h + y$: from this, we obtain a formula of as many terms as the other, the first of them being $k^2$. If, after that, we express the root by $k + py + qy^2$; and determine $p$ and $q$ so, that the second and third terms may disappear; the last two, being divisible by $y^3$, will be reduced to a simple equation of the first degree, from which we may easily obtain the value of $y$, and, consequently, that of $x$ also. Still, however, we shall be obliged, as before, to exclude a great number of cases in the application of this method; those, for instance, in which the value found for $x$ is no other than $x = h$, which was given, and in which, consequently, we could not advance one step. Such cases shew either that the formula is impossible in itself, or that we have yet to find some other case in which it becomes a square. 146. And this is the utmost length to which mathematicians have yet advanced, in the resolution of formulæ, that are affected by the sign of the square root. No discovery has hitherto been made for those, in which the quantities under the sign exceed the fourth degree; and when formulæ occur which contain the fifth, or a higher power of $x$, the artifices which we have explained are not sufficient to resolve them, even although a case be given. That the truth of what is now said may be more evident, we shall consider the formula $k^2 + bx + cx^2 + dx^3 + ex^4 + fx^5$, the first term of which is already a square. If, as before, we suppose the root of this formula to be $k + px + qx^2$, and determine $p$ and $q$, so as to make the second and third terms disappear, there will still remain three terms, which,