the equality to zero of that product, by making each of its factors equal to zero.

We have, therefore, x-n, x=+√n, or x= ±√n.

402. Now, since the square root is taken on both sides of the equation, x2=n, in order to arrive at x=±√n; it is very natural to suppose that, x being the square root of x2, we should also affect x with the double sign; and, therefore, in resolving the equation x2 =n, we would write ±x=±√n; but by arranging these signs in every possible manner, namely:

+x=+√n, +x=−√n,

-X=- ·√n, −x=+√n,

we would still have no more than the two first equations, that is, +x=±√n; for if we change the signs of the equations xn and -x+n, they become +x+n and +x=−√n, or x=±√n.

403. If, in the formula x2-n, n be negative, or which is the same thing, if we have x2-n, where n is positive; then, x=-n=±√n× √−1, and in fact (vn) x ( √ − 1)3 =nX-1=-n; therefore, the two roots of a pure equation are either both real, or both imaginary.

404. All adfected quadratic equations, after being properly reduced according to the rules pointed out in the reduction of simple equations, may be exhibited under the following general forms; namely x2+ nx=o, and x2+nx=n'; where n and n' may be any numbers whatever, positive or negative, integral or fractional,

405. The solution of adfected quadratic equations of the form x2+nxo, is attended with no difficulty; for the equation x2+nx=o, being divided by x becomes x+n=0, from which we find though we find only one value of x, according to this mode of solution, still there may be two values of x, which will satisfy the proposed equation.

In the equation, x2=3x, for example, in which it is required to assign such a value of x, that x may become equal to 3x, this is done by supposing

3,

a value which is found by dividing the equation by x; but besides this value, there is also another which is equally satisfactory; namely, x=0; for then x2=o, and 3x=0.

406. An adfected quadratic equation is said to be complete, when it is of the form x2+nx=n'; that is, when three terms are found in it; namely, that which contains the square of the unknown quantity, as x; that in which the unknown quantity is found only in the first power, as no; and lastly, the term which is composed only of known quantities; and, as there is no difficulty attending the reduction of adfected quadratic equations to the above form by the known rules: the whole is at present reduced to determining the true value of x from the equation x2+nx

=n'.

We shall begin with remarking, that if xnx were a real square, the resolution would be attended with no difficulty, because that it would be only required to extract the square root on both sides, in order to reduce it to a simple equation.

407. But it is evident that x2+nx cannot be a square; since we have already seen (Art. 288), that if a root consists of two terms, for example, x+a, its square always contains three terms, namely, twice the product of the two parts, besides the square of each part, that is to say, the square of x+a is x2+ 2ax+a2.

408. Now, we have already on one side x2+nx ; we may, therefore, consider x2 as the square of the first part of the root, and in this case na must represent twice the product of x, the first part of the root, by the second part: consequently, this second part must be in, and in fact the square of x+n is found to be x2+na+n3.

409. Now xnx+in being a real square, which has for its root x+n, if we resume our equation x2 +nan', we have only to add n to both sides, which gives us x2+nx+n2=n'+n, 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 +‡n=√(‡n2+n'); and as every square root may be taken either affirmatively or negatively, we shall have for a two values: expressed thus ;

a={n±√({n2+n').

410. This formula contains the rule by which all quadratic equations may be resolved, and it will be proper, as EULER justly observes, to commit it to memory, that it may not be necessary to repeat, every time, the whole operation which we have gone through. We may always arrange the equation in such a manner, that the pure square T2 may be found on one side, and the above equation have the form x2=-nx+n', where we see immediately that x=}n±√({n2+n').

411. The general rule, therefore, which may be deduced from that, in order to resolve x2=-nx+n', is founded on this consideration. That the unknown is equal to half the coefficient or multiplier of x on the other side of the equation, plus or minus the 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 x2=6x+7, we should immediately say, that x=3+(9+7)=3±4; whence we have these two values of x; namely, x=7, and x=-1.

412. The method of resolving adfected quadratic equations will be still better understood by the four following forms; in which n and n' may be any positive numbers whatever, integral or fractional.

I. In the case x2+nx=n', where x = −}n+√({n2 +n'), or in—√(‡n2+n'), the first value of a must be positive, because (n2 +n') is >√/‡n2, or its equal n; and its second value will evidently be negative, because each of the terms of which it is composed is negative.

II. In the case x2-nx=n', from which we find x= {n+√(in2+n'), or {n—√(‡n2+n'), the first value of x, is manifestly positive, being the sum of two positive terms; and the second value will be negative, because ✔(‡n2 +n') is >√(‡n2), or its equal an

III. In the case xa. -nx-n', we have xn+ (nn), or in- V (in2-n'); both the values of x will be positive, when n2 is >n'; for its first value is then evidently positive, being composed of two positive terms; and its second value will also be positive, because √(‡n2 —n') is less than (‡n3),

or its equal n. But if n2, in this case, be less than n' both the values of x will be imaginary; because the quantity n2 -n', under the radical sign, is then negative; and consequently (}n3 —n') will be imanary, or of no assignable value.

IV. Also, in the fourth case, x2+nx=-n', where x=−{n+√ (‡n3—n'), or —±n−√(‡n2 —n'), the two values of x will be both negative, or both imaginary, according as n2 is greater or less than n'

413. Hence we may conclude, from the constant occurrence of the double sign before the radical part of the preceding expressions, that every quadratic equation must have two roots; which are both real, or both imaginary; and though the latter of these cannot be considered as real quantities, but merely as pure algebraic symbols, of no determinate value, yet when they are submitted to the operations indicated by the equation, the two members of that équation will be always identical, or which is the same, it shall be always reduced to the form 0=0.

414. It may here also be further observed that, in some equations involving radical quantities of the form (ax+b) both values of x, found by the ordinary process, will not answer the proposed equation, except that we take the radical quantity with the double sign. Let, for example, the values of x be found in the equation x+(5x+10)=8.

Here, by transposition,

(5x+10)=8—x; therefore by squaring, 5x+10=64-16x+x3,

or x2-21x=-54; and .. x=18, or 3. Now, since these two values of x are found from the resolution of the equation x2-21x=-54; it necessarily follows, (Art. 413), that each of them, when substituted for a, must satisfy that equation ;

which may be verified thus; in the first place, by substituting 18 for x, in the equation x2-21x=-54, we have (18)2-21 X18=-54, or 324-378=-54; that is, 54-54, or 0=0.

Again, substituting 3 for x, we have (3) - 21 × 3— -54, or 9-63=-54; -54-54, or 54-54-0; ..0=0.

415. And as the equation x2-21x=-54, may be deduced from the equation +(5x+10)= 8-x, or (5x+10)=8−x; it is evident that the radical quantity (5x+10), must be taken, with the double sign ±, in the primitive equation, in order that it would be satisfied by the values, 18 and 3, of x, above found; that is, 18 answers to the sign and 3 to the sign +. But if one of these signs be excluded by the nature of the question; then only one of the values will satisfy the original equation; for instance, if in the equation x2+(5x+10)=8, the sign be excluded from the radical quantity, then the square root of 5x+10 must be considered as a positive quantity; and because it is equal to 9-x; the value of x, since both are positive, which will answer the proposed equation, must be less than 8 ; therefore, 3 is the value of x, which will satisfy the equation x+(5x+10) 8, which can be readily verified thus; substituting 3 for x, we have 8 +√(15+10)=8, or 3+5=8. And for a similar reason, 18 is the value of x, which will answer the equation (5x+10)=8; for 18-(90+10)= 18-10=8; .. 8=8, or 0=0.

x

416. It is proper to take notice here of the following method of resolving quadratic equations, the principle of which is given in the Bija Ganita, before mentioned: thus, if a quadratic equation be of the form 4a2x2+4abx+4ac, it is evident that, by adding b2 to both sides, the left-hand member will be a complete square, since it is the square of 2ax+b; and, therefore, by extracting the square root of both sides, there will arise a simple equation, from which the va laes of may be determined.