Elements of Algebra

368 ELEMENTS PART II. $b + cx + dx^2 = 2fp + p^2x^2$, an equation of the second degree, which gives $x = \frac{p^2 - c + \sqrt{p^4 - 2cp^2 + 8dfp + c^2 - 4bd}}{2d}$ So that the question is now reduced to finding such values of $p$, as will make the formula $p^4 - 2cp^2 + 8dfp + c^2 - 4bd$ become a square. But as it is the fourth power of the required number $p$ which occurs here, this case belongs to the following chapter. *** CHAP. IX. *Of the Method of rendering Rational the incommensurable Formula $\sqrt{a + bx + cx^2 + dx^3 + ex^4}$.* 128. We are now come to formulæ, in which the indeterminate number, $x$, rises to the fourth power; and this must be the limit of our researches on quantities affected by the sign of the square root; since the subject has not yet been prosecuted far enough to enable us to transform into squares any formulæ, in which higher powers of $x$ are found. Our new formula furnishes three cases: the first, when the first term, $a$, is a square; the second, when the last term, $ex^4$, is a square; and the third, when both the first term and the last are squares. We shall consider each of these cases separately. 129. 1st. Resolution of the formula $\sqrt{f^2 + bx + cx^2 + dx^3 + ex^4}$. As the first term of this is a square, we might, by the first method, suppose the root to be $f + px$, and determine $p$ in such a manner, that the first two terms would disappear, and the others be divisible by $x^2$; but we should not fail still to find $x^2$ in the equation, and the determination of $x$ would depend on a new radical sign. We shall therefore have recourse to the second method; and represent the root by $f + px + qx^2$; and then determine $p$ and $q$, so as to remove the first three terms, and then dividing by $x^3$, we shall arrive at a simple equation of the first degree, which will give $x$ without any radical signs.