Elements of Algebra

CHAP. VIII. OF ALGEBRA. 363 the square root of the first term, and $p$ is taken such, that the second term must likewise disappear; so that there remains only to compare $p^2x^2$ with the third and fourth term of the formula, namely $cx^2 + dx^3$; for then that equation, being divisible by $x^2$, gives a new value of $x$, which is $x = \frac{p^2-c}{d}$. In the second method, three terms are given to the root; that is to say, if the first term $a = f^2$, we express the root by $f + px + qx^2$; after which, $p$ and $q$ are determined such, that the first three terms of the formula may vanish, which is done in the following manner. Since $f^2+bx+cx^2+dx^3 = f^2+2pfx+2fqx^2+p^2x^2+2pqx^3+q^2x^4$, we must have $b = 2fp$; and, consequently, $p=\frac{b}{2f}$; farther, $c=2fq+p^2$; or $q = \frac{c-p^2}{2f}$; after this, there remains the equation $dx^3 = 2pqx^3+q^2x^4$; and, as it is divisible by $x^3$, we obtain from it $x = \frac{d-2pq}{q^2}$. 118. It may frequently happen, however, even when $a=f^2$, that neither of these methods will give a new value of $x$; as will appear, by considering the formula $f^2 + dx^3$, in which the second and third terms are wanting. For if, according to the first method, we suppose the root to be $f + px$, that is, $f^2+dx^3 = f^2 + 2fpx + p^2x^2$, we shall have $2fp = 0$, and $p = 0$; so that $dx^3 = 0$; and therefore $x = 0$, which is not a new value of $x$. If, according to the second method, we were to make the root $f+px+qx^2$, or $f^2+dx^3 = f^2+2fpx+p^2x^2+2fqx^2+2pqx^3+q^2x^4$, we should find $2fp = 0$, $p^2+2fq = 0$, and $q^2=0$; whence $dx^3=0$, and also $x=0$. 119. In this case, we have no other expedient, than to endeavour to find such a value of $x$, as will make the formula a square; if we succeed, this value will then enable us to find new values, by means of our two methods: and this will apply even to the cases in which the first term is not a square. If, for example, the formula $3+x^3$ must become a square; as this takes place when $x=1$, let $x=1+y$, and we shall thus have $4+3y+3y^2+y^3$, the first term of which is a square.