Elements of Algebra

CHAP. VI. OF ALGEBRA. 223 640. But it is evident that $x^2 + px$ cannot be a square; since we have already seen, (Art. 307.) that if a root consists of two terms, for example, $x + n$, its square always contains three terms, namely, twice the product of the two parts, beside the square of each part; that is to say, the square of $x + n$ is $x^2 + 2nx + n^2$. Now, we have already on one side $x^2 + px$; we may, therefore, consider $x^2$ as the square of the first part of the root, and in this case $px$ must represent twice the product of $x$, the first part of the root, by the second part: consequently, this second part must be $\frac{1}{2}p$, and in fact the square of $x + \frac{1}{2}p$, is found to be $x^2 + px + \frac{1}{4}p^2$. 641. Now, $x^2 + px + \frac{1}{4}p^2$ being a real square, which has for its root $x + \frac{1}{2}p$, if we resume our equation $x^2 + px = q$, we have only to add $\frac{1}{4}p^2$ to both sides, which gives us $x^2 + px + \frac{1}{4}p^2 = q + \frac{1}{4}p^2$, the first side being actually a square, and the other containing only known quantities. If, therefore, we take the square root of both sides, we find $x + \frac{1}{2}p = \sqrt{(\frac{1}{4}p^2 + q)}$; subtracting $\frac{1}{2}p$, we obtain $x = -\frac{1}{2}p + \sqrt{(\frac{1}{4}p^2 + q)}$; and, as every square root may be taken either affirmatively, or negatively, we shall have for $x$ two values expressed thus; $x = -\frac{1}{2}p \pm \sqrt{(\frac{1}{4}p^2 + q)}$. 642. This formula contains the rule by which all quadratic equations may be resolved; and it will be proper to commit it to memory, that it may not be necessary, every time, to repeat the whole operation which we have gone through. We may always arrange the equation in such a manner, that the pure square $x^2$ may be found on one side, and the above equation have the form $x^2 = -px + q$, where we see immediately that $x = -\frac{1}{2}p \pm \sqrt{(\frac{1}{4}p^2 + q)}$. 643. The general rule, therefore, which we deduce from that, in order to resolve the equation $x^2 = -px + q$, is founded on this consideration; That the unknown quantity $x$ is equal to half the co-efficient, or multiplier of $x$ on the other side of the equation, plus or minus the square root of the square of this number, and the known quantity which forms the third term of the equation. Thus, if we had the equation $x^2 = 6x + 7$, we should immediately say, that $x = 3 \pm \sqrt{(9+7)} = 3 \pm 4$, whence we have these two values of $x$, namely, $x=7$, and $x = -1$. In the same manner, the equation $x^2 = 10x - 9$, would give $x = 5 \pm \sqrt{(25-9)} = 5 \pm 4$, that is to say, the two values of $x$ are 9 and 1. 644. This rule will be still better understood, by distin-