Elements of Algebra

246 ELEMENTS SECT. IV. The second case, in which both values of $x$ are negative, occurs, when the two factors are $(x+p) \times (x+q)$; for we shall then have $x = -p$, and $x = -q$; the equation itself therefore becomes $x^2 + (p+q)x + pq = 0$, in which both the second and third terms are affected by the sign $+$. 697. The signs of the second and the third terms consequently shew us the nature of the roots of any equation of the second degree. For let the equation be $x^2 \dots ax \dots b = 0$. If the second and third terms have the sign $+$, the two values of $x$ are both negative; if the second term have the sign $-$, and the third term $+$, both values are positive: lastly, if the third term also have the sign $-$, one of the values in question is positive. But, in all cases whatever, the second term contains the *sum* of the two values, and the third term contains their *product*. 698. After what has been said, it will be easy to form equations of the second degree containing any two given values. Let there be required, for example, an equation such, that one of the values of $x$ may be $7$, and the other $-3$. We first form the simple equations $x=7$, and $x=-3$; whence, $x-7=0$, and $x+3=0$; these give us the factors of the equation required, which consequently becomes $x^2 - 4x - 21 = 0$. Applying here, also, the above rule, we find the two given values of $x$; for if $x^2 = 4x+21$, we have, by compleating the square, &c. $x = 2 \pm \sqrt{25} = 2 \pm 5$; that is to say, $x=7$, or $x=-3$. 699. The values of $x$ may also happen to be equal. Suppose, for example, that an equation is required, in which both values may be $5$. Here the two factors will be $(x-5) \times (x-5)$, and the equation sought will be $x^2 - 10x + 25 = 0$. In this equation, $x$ appears to have only one value; but it is because $x$ is twice found $=5$, as the common method of resolution shews; for we have $x^2 = 10x - 25$; wherefore $x = 5 \pm \sqrt{0} = 5 \pm 0$, that is to say, $x$ is in two ways $=5$. 700. A very remarkable case sometimes occurs, in which both values of $x$ become imaginary, or impossible; and it is then wholly impossible to assign any value for $x$, that would satisfy the terms of the equation. Let it be proposed, for example, to divide the number $10$ into two parts, such that their product may be $30$. If we call one of those parts $x$, the other will be $10-x$, and their product will be $10x - x^2$.