Elements of Algebra

364 ELEMENTS PART II square. If, therefore, we suppose, according to the first method, the root to be $2 + py$, we shall have $4 + 3y + 3y^2 + y^3 = 4 + 4py + p^2y^2$. In order that the second term may disappear, we must make $4p = 3$; and, consequently, $p = \frac{3}{4}$; whence $3 + y = p^2$, and $y = p^2 - 3 = \frac{9}{16} - \frac{48}{16} = \frac{-39}{16}$; therefore $x = \frac{-23}{16}$, which is a new value of $x$. If, again, according to the second method, we represent the root by $2 + py + qy^2$, we shall have $4+3y+3y^2+y^3 = 4+4py+4qy^2+p^2y^2+2pqy^3+q^2y^4$, from which the second term will be removed, by making $4p = 3$, or $p = \frac{3}{4}$; and the fourth, by making $4q + p^2 = 3$, or $q = \frac{3-p^2}{4} = \frac{39}{64}$; so that $1 = 2pq + q^2y$; whence we obtain $y = \frac{1-2pq}{q^2}$, or $y = \frac{352}{1521}$; and, consequently, $x = \frac{1873}{1521}$. 120. In general, if we have the formula $a + bx + cx^2 + dx^3$, and know also that it becomes a square when $x=f$, or that $a + bf + cf^2 + df^3 = g^2$, we may make $x = f+y$, and shall hence obtain the following new formula: $a$ $+ bf + by$ $+ cf^2 + 2cfy + cy^2$ $+ df^3 + 3df^2y + 3dfy^2 + dy^3$ $g^2 + (b+2cf+3df^2)y + (c+3df)y^2 + dy^3$. In this formula, the first term is a square; so that the two methods above given may be applied with success, as they will furnish new values of $y$, and consequently of $x$ also, since $x = f + y$. 121. But often, also, it is of no avail even to have found a value of $x$. This is the case with the formula $1 + x^3$, which becomes a square when $x=2$. For if, in consequence of this, we make $x=2+y$, we shall get the formula $9 + 12y + 6y^2 + y^3$, which ought also to become a square. Now, by the first rule, let the root be $3+py$, and we shall have $9+12y+6y^2+y^3 = 9+6py+p^2y^2$, in which we must have $6p=12$, and $p=2$; therefore $6+y=p^2=4$, and $y=-2$, which, since we made $x=2+y$, this gives $x=0$; that is to say, a value from which we can derive nothing more.