Elements of Algebra

CHAP. IX. OF ALGEBRA. 245 $(x-5) \times (x-7)$ may become 0 in two ways: first, when the first factor $x-5=0$; and also, when the second factor $x-7=0$. In the first case, $x=5$, in the second $x=7$. The reason is therefore very evident, why such an equation $x^2 - 12x + 35 = 0$, admits of two solutions; that is to say, why we can assign two values of $x$, both of which equally satisfy the terms of the equation; for it depends upon this fundamental principle, that the quantity $x^2 - 12x + 35$ may be represented by the product of two factors. 694. The same circumstances are found in all equations of the second degree: for, after having brought the terms to one side, we find an equation of the following form $x^2 - ax + b = 0$, and this formula may be always considered as the product of two factors, which we shall represent by $(x-p) \times (x-q)$, without concerning ourselves what numbers the letters $p$ and $q$ represent, or whether they be negative or positive. Now, as this product must be $=0$, from the nature of our equation, it is evident that this may happen in two cases; in the first place, when $x=p$; and in the second place, when $x=q$; and these are the two values of $x$ which satisfy the terms of the equation. 695. Let us here consider the nature of these two factors, in order that the multiplication of the one by the other may exactly produce $x^2 - ax + b$. By actually multiplying them, we obtain $x^2 - (p+q)x + pq$; which quantity must be the same as $x^2 - ax + b$, therefore we have evidently $p+q=a$, and $pq=b$. Hence is deduced this very remarkable property; that in every equation of the form $x^2 - ax + b = 0$, the two values of $x$ are such, that their sum is equal to $a$, and their product equal to $b$: it therefore necessarily follows, that, if we know one of the values, the other also is easily found. 696. We have at present considered the case, in which the two values of $x$ are positive, and which requires the second term of the equation to have the sign $-$, and the third term to have the sign $+$. Let us also consider the cases, in which either one or both values of $x$ become negative. The first takes place, when the two factors of the equation give a product of this form, $(x-p) \times (x+q)$; for then the two values of $x$ are $x=p$, and $x=-q$; and the equation itself becomes $x^2 + (q-p)x - pq = 0$; the second term having the sign $+$, when $q$ is greater than $p$, and the sign $-$, when $q$ is less than $p$; lastly, the third term is always negative.