Elements of Algebra

CHAP. IX. OF ALGEBRA. 247 $x^2 = 30$; wherefore $x^2 = 10x - 30$, and $x = 5 \pm \sqrt{-5}$, which, being an imaginary number, shews that the question is impossible. 701. It is very important, therefore, to discover some sign, by means of which we may immediately know whether an equation of the second degree be possible or not. Let us resume the general equation $x^2 - ax + b = 0$. We shall have $x^2 = ax - b$, and $x = \frac{1}{2}a \pm \sqrt{(\frac{1}{4}a^2 - b)}$. This shews, that if $b$ be greater than $\frac{1}{4}a^2$, or $4b$ greater than $a^2$, the two values of æ are always imaginary, since it would be required to extract the square root of a negative quantity; on the contrary, if $b$ be less than $\frac{1}{4}a^2$, or even less than $0$, that is to say, if it be a negative number, both values will be possible or real. But, whether they be real or imaginary, it is no less true, that they are still expressible, and always have this property, that their sum is equal to $a$, and their product equal to $b$. Thus, in the equation $x^2 - 6x + 10 = 0$, the sum of the two values of $x$ must be 6, and the product of these two values must be 10; now, we find, 1. $x = 3 + \sqrt{-1}$, and 2. $x = 3 - \sqrt{-1}$, quantities whose sum is 6, and the product 10. 702. The expression which we have just found may likewise be represented in a manner more general, and so as to be applied to equations of this form, $fx^2 \pm gx + h = 0$; for this equation gives $x^2 = \mp \frac{gx}{f} - \frac{h}{f}$, and $x = \mp \frac{g}{2f} \pm \sqrt{(\frac{g^2}{4f^2} - \frac{h}{f})}$, or $x = \frac{\mp g \pm \sqrt{(g^2 - 4fh)}}{2f}$; whence we conclude, that the two values are imaginary, and, consequently, the equation impossible, when $4fh$ is greater than $g^2$; that is to say, when, in the equation $fx^2 - gx + h = 0$, four times the product of the first and the last term exceeds the square of the second term: for the product of the first and the last term, taken four times, is $4fhx^2$, and the square of the middle term is $g^2x^2$; now, if $4fhx^2$ be greater than $g^2x^2$, $4fh$ is also greater than $g^2$, and, in that case, the equation is evidently impossible; but in all other cases, the equation is possible, and two real values of $x$ may be assigned. It is true, they are often irrational; but we have already seen, that, in such cases, we may always find them by approximation: whereas no approximations can take place with regard to imaginary expressions, such as $\sqrt{-5}$; for 100 is as far from being the value of that root, as 1, or any other number. 703. We have farther to observe, that any quantity of