Elements of Algebra

362 ELEMENTS PART II. first term is itself a square already, and the formula may be expressed thus, $f^2 + bx + cx^2 + dx^3$; for it evidently becomes a square, if $x = 0$. We shall therefore enter upon the subject, by considering this formula; and shall endeavour to see how, by setting out from the known case $x = 0$, we may arrive at some other value of $x$. For this purpose, we shall employ two different methods, which will be separately explained: in order to which, it will be proper to begin with particular cases. 115. Let, therefore, the formula $1 + 2x - x^2 + x^3$ be proposed, which ought to become a square. Here, as the first term is a square, we shall adopt for the root required such a quantity as will make the first two terms vanish. For which purpose, let $1 + x$ be the root, whose square is to be equal to our formula; and this will give $1 + 2x - x^2 + x^3 = 1 + 2x + x^2$, of which equation the first two terms destroy each other; so that we have $-x^2 + x^3 = x^2$, or $x^3 = 2x^2$, which, being divided by $x^2$, gives $x = 2$; so that the formula becomes $1 + 4 - 4 + 8 = 9$. Likewise, in order to make a square of the formula, $4 + 6x - 5x^2 + 3x^3$, we shall first suppose its root to be $2 + nx$, and seek such a value of $n$ as will make the first two terms disappear; hence, $4 + 6x - 5x^2 + 3x^3 = 4 + 4nx + n^2x^2$; therefore we must have $4n = 6$, and $n = \frac{3}{2}$; whence results the equation $-5x^2 + 3x^3 = n^2x^2 = \frac{9}{4}x^2$, or $3x^3 = \frac{29}{4}x^2$, which gives $x = \frac{29}{12}$; and this is the value which will make a square of the proposed formula, whose root will be $2 + \frac{3}{2}x = \frac{45}{8}$. 116. The second method consists in giving the root three terms, as $f + gx + hx^2$, such, that the first three terms in the equation may vanish. Let there be proposed, for example, the formula $1 - 4x + 6x^2 - 5x^3$, the root of which we will suppose to be $1 - 2x + hx^2$, and we shall thus have $1 - 4x + 6x^2 - 5x^3 = 1 - 4x + 4x^2 - 4hx^3 + h^2x^4 + 2hx^2$. The first two terms, as we see, are immediately destroyed on both sides; and, in order to remove the third, we must make $2h + 4 = 6$; consequently, $h = 1$; by these means, and transposing $2hx^2 = 2x^2$, we obtain $-5x^3 = -4x^3 + x^4$, or $-5 = -4 + x$, so that $x = -1$. 117. These two methods, therefore, may be employed, when the first term $a$ is a square. The first is founded on expressing the root by two terms, as $f + px$, in which $f$ is