since the second would have sold 5 ells for 12 crowns and a half, his 8 ells must have produced him 20 crowns.

The sum is, as before,

35 crowns.

CHAPTER VII.

Of the Nature of Equations of the Second Degree.

573. WHAT we have already said sufficiently shows, that equations of the second degree admit of two solutions; and this property ought to be examined in every point of view, because the nature of equations of a higher degree will be very much illustrated by such an examination. We shall therefore retrace, with more attention, the reasons which render an equation of the second degree capable of a double solution; since they undoubtedly will exhibit an essential property of those equations.

574. We have already seen, it is true, that this double solution arises from the circumstance that the square root of any number may be taken either positively, or negatively; however, as this principle will not easily apply to equations of higher degrees, it may be proper to illustrate it by a distinct analysis. Taking, for an example, the quadratic equation, x x = 12 x 35, we shall give a new reason for this equation being resolvible in two ways, by admitting for æ the values 5 and 7, both of which satisfy the terms of the equation.

575. For this purpose it is most convenient to begin with transposing the terms of the equation, so that one of the sides may become 0; this equation consequently takes the form

x x 12x + 35 = 0;

and it is now required to find a number such, that if we substitute it for x, the quantity xx-12x + 35 may be really equal to nothing; after this, we shall have to show how this may be done in two ways.

576. Now, the whole of this consists in showing clearly, that a quantity of the form x x 12 x + 35 may be considered as the product of two factors; thus, in fact, the quantity of which we speak

is composed of the two factors (x 5) X (x-7). For, since this quantity must become 0, we must also have the product

(x — 5) × (x — 7) = 0 ;

but a product, of whatever number of factors it is composed, becomes =0, only when one of those factors is reduced to 0; this is a fundamental principle to which we must pay particular attention, especially when equations of several degrees are treated of.

577. It is therefore easily understood, that the product

(x — 5) × (x — 7)

may become O in two ways: one, when the first factor x-5= 0; the other, when the second factor x-7= 0. In the first case x = 5, in the other, x = 7. The reason is, therefore, very evident, why such an equation xx 12x35 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. This fundamental principle consists in this, that the quantity xx-12x + 35 may be represented by the product of two factors.

578. The same circumstances are found in all equations of the second degree. For, after having brought all the terms to one side, we always find an equation of the following form xx-a x + b = 0, and this formula may be always considered as the product of two factors, which we shall represent by (x − p) × (x — q), without concerning ourselves what numbers the letters p and q represent. Now, as this product must be 0, from the nature of our equation. it is evident that this may happen in two ways; 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.

x x

=

579. Let us now consider the nature of these two factors, in order that the multiplication of the one by the other may exactly produce xxax+b. By actually multiplying them, we get (p + q) x + p q ; now this quantity must be the same as x x − a x + b, wherefore we have evidently p + q = a, and So that we have deduced this very remarkable property, pq = b. that in every equation of the form xx-ax+b= 0, the two values of x are such, that their sum is equal to a, and their product equal to b; whence it follows that, if we know one of the values, the other also is easily found.

580. We have 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 a become negative. The first takes place when the two factors of the equation give a product of this form (x − p) × (x + q); for then the two values of x are x = p, and x =— q; the equation itself becomes xx+(q − p) x − p q = 0; the second term has the sign, when q is greater than p, and the sign, when q is less than p; lastly, the third term is always negative.

(9

=

The second case, in which both values of x are negative, occurs, when the two factors are (x+p) × (x + q); for we shall then have x and x = q; the equation itself becomes

-P

x x + (p + q) x + p q = 0,

=

in which both the second and third terms are affected by the sign +. 581. The signs of the second and the third term consequently show us the nature of the roots of any equation of the second degree. Let the equation be xx ... ax....b 0, if the second and third terms have the sign+, the two values of x are both negative; if the second term has the sign, and the third term has +, both values are positive; lastly, if the third term also has 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.

582. After what has been said, it will be very 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; thence these x 7 = 0 and x+3= 0, which gives us, in this manner, the factors of the equation required, which consequently becomes xx-4x-21= 0. Applying here, also, the above rule, we find the two given values of x; for if x x = 4 x + 21, we have x 2252 ± 5, that is to say, x = 7, or x = - 3.

583. The values of x may also happen to be equal. Let there be sought, for example, an equation, in which both values may be 5. The two factors will be (x 5) (x5), and the equaX tions sought will be xx 10x250. 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 shows; for we have xx 10 x · 25; wherefore x = 5±0 = 5±0, that is to say, x is in two ways = 5.

584. A very remarkable case, in which both values of a become imaginary, or impossible, sometimes occurs; 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 10 x - x x = 30; wherefore x x = 10 x 30, and x = 5 ± √— 5, which being an imaginary number, shows that the question is impossible.

585. It is very important, therefore, to discover some sign, by means of which he may immediately know, whether an equation of the second degree is possible or not.

Let us resume the general equation a xxx + b = 0. We shall have

This shows, that if b is greater than a a, or 4b greater than a a, the two values of x are always imaginary, since it would be required to extract the square root of a negative quantity; on the contrary, if bis less than a a, or even less than 0, that is to say, is 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 = a, and their product b. In the equation xx-6x+10=0, for example, the sum of the two values of x must be 6, and the product of these two values must be 10; now we find, I. x = 3 + √−1, and II. x = 3√1, quantities whose sum = 6, and the product

= 10.

=

586. The expression, which we have just found, may be represented in a manner more general, and so as to be applied to equations of this form, fxxgx + h = 0; for this equation gives

whence we conclude that the two values are imaginary, and consequently the equation impossible, when 4 fh is greater than gg; that is to say, when, in the equation fxx-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 4 fhx x, and the square of the middle term is g g x x; now, if 4 fhx x is greater than gg xx, 4 fh is also greater than gg, and in that case, the equation is evidently impossible. 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 5; for 100 is as far from being

the value of that root, as 1, or any other number.

587. We have further to observe, that any quantity of the second degree, xxax ±b, must always be resolvible into two factors, such as (xp) × (x± q). For, if we took three factors, such as these, we should come to a quantity of the third degree, and taking only one such factor, we should not exceed the first degree.

It is therefore certain that every equation of the second degree necessarily contains two values of x, and that it can neither have more nor less.

588. We have already seen, that when the two factors are found, the two values of x are also known, since each factor gives one of those values, when it is supposed to be 0. The converse also is true, viz. that when we have found one value of x, we know also one of the factors of the equation; for if x = p represents one of the values of x, in any equation of the second degree, xp is one of the factors of that equation; that is to say, all the terms having been brought to one side, the equation is divisible by x-p; and further, the quotient expresses the other factor.

589. In order to illustrate what we have now said, let there be given the equation xx+4x-21=0, in which we know that x= 3 is one of the values of x, because 3 x 3 + 4x3 -21=0; this shows, that x-3 is one of the factors of the equation, or that x x + 4 x 21 is divisible by x- 3, which the actual division proves.

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