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VARIATION.

PROPOSITION XIII.

y±x = ax±x = (a ±1) x = (a ± 1) • —
— • y ; i. e.,

If y vary as x, x±y varies as x or y.

PROPOSITION XIV.

a

y = ax,

y” = aTMx” ; i. e.,

If y vary as x, yTM varies as xTM.

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y abx; i. e.,

If y vary as z, and z vary as x; then y varies as x.

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It will be observed that the above forms embrace, very briefly and simply, not only essentially the whole doctrine of proportion, but a vastly wider field, by reason of the unlimited number of values of which y and x are susceptible.

SECTION THIRD.

Analysis of Equations.

The following is a principle of the first importance in analytical investigations :

PROPOSITION I.

Certain equations are so constituted that they necessari- (56) ly resolve into, and are consequently equivalent to, several independent equations.

We do not propose, in the present article, to enter into a full developement of the boundless resources which this principle affords, but simply to illustrate it by examples of such particular cases as will be serviceable to us as we proceed.

Required two numbers such that if they be diminished severally by a, b, and the remainders squared, the sum of these

be equal to zero.

Denoting the numbers by x, y, we have

*

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squares shall

and it follows from (63) that (x — a)2, (y — b)2, must both be +, whether xa, y≥b; but it is obvious that, since neither of the terms (x − a)2, (y — b)2 can be minus, neither can be greater than 0; for if either term, (x-a)2 for instance, have an additive value, the other (y—b)2 must possess the same value and be subtractive, in order to satisfy the equation, or that their sum may = 0; whence it is necessary that

or

(x — a)2 = 0, and (y — b)2 = 0 ;
xa=0, and y—b=0,

x=

c = a, and

y = b.

As a second example, what numbers are those from which if a and b be severally subtracted, the product of the remainders will be = 0? Representing the numbers by x, y, there results

... dividing by y—b, and dividing by x-a,

*

(xa) (y-b) = 0;
x — a= 0, or x = a,
y-b=0, or y = b.

ra, a greater or less than a.

QUADRATIC EQUATIONS.

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Every equation of the second degree, or, embracing no other unknown quantities than 2 and x, may, by transposing, uniting terms, and dividing by the coëfficient of x2, be readily reduced to the form

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understanding that a, b, may be either + or - .

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In order to find x,

we observe that the first member of the equation will become a binomial square (8) by the addition of a2;

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To solve a QUADRATIC EQUATION :

1o. Reduce it to the form of (57);

2°. Add the square of half the coefficient of x to both sides; 3°. Take the square root of the members, prefixing the double sign [+] to the second;

4°. Transpose.

It will be observed that (57) resolves into two independent equations (58),

x=−a+√(b+a2), and x =—a— √ (b+a3),

thus illustrating (56).

Adding these values of x, we have

[− a+ √(b+a2)]+[−a−√ (b+ a2)] = — 2a,

and their product is (?)

[− a + √(b+a2)] [− a−√√(b+a2)] = [− a]2 − [√(b+a2)]2 = a2 − (b + a2) = — b.

Cor. In a quadratic of the form (57) the sum of the values (59) of x, is equal to the coëfficient of x1 taken with the contrary sign, and their product to the second member also taken with the contrary sign.

Queries. What will (57) and (58) become, when a is changed into — a ? b into-b? When b is minus, what must be its value compared with a2 in order that the value of x in (58) be impossible? [See (6.)] Will change of value or sign in a ever render x imaginary? Why not?

Scholium I. We naturally inquire if the Cubic Equation can be

resolved into three independent equations. Every equation of the third degree reduces to the form

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As we must suppose x to have some value-one at least-let r be that value; then must r satisfy the equation,

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dividing by xr, see examples under (16);

x2+(a+r) x + r2 + ar + b = 0,

(c)

whence [(57), (58)] (a) is resolvable into three equations, giving three values for x.

Comparing equations (a), (b), (c), and observing that (b) is the same as (a), we learn that, if r be a root of the cubic equation (a), that the equation is divisible by x − r, giving a quadratic (c).

The student is requested to prove that the equation x2 + px3 + qx2+rx+s=0,

is resolvable into the factors

x-a, x-b, xc, x-d,

or that (a) (x — b) (x − c) (x - d) = x2 + px3+qx2+rx+8=0, a, b, c, d, being values of x, or x = = a, x = b, x = =C, x =

d.

Scholium II. Many Biquadratic Equations may be reduced as Quadratics; e. g.

10. When reducible to the form

(x2+Ax+B) (x2 + A'x + B') = 0,

where the conditions are

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x2+(A+A')x3 + (AA' + B+B')x2 + (AB' + A'B)x + BB' = 0, or s BB', q = AA' + B+ B', p= A+ A', r = AB' + A'B.

=

2o. When reducible to the form

(x2+Ax)2+B(x2+Ax)1=C,

where the conditions are

x2+2Ax3 + (A2 + B)x2 + ABx = C,

or p=2A, r=AB, q=A2+B.

3°. When reducible to the form

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Example. Given x+4x3+6x2+4x= 15, to find x.

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42+B=q, or 22+2=6; S

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How many values has x? What values are imaginary? Verify by substituting these values in the first equation.

Scholium III. When a problem embraces several unknown quantities, it may be solved by representing these quantities in different ways; but the elegance and facility of the solution will frequently depend upon the notation which we employ. The following rule, taken from the Cambridge Mathematics, (application of Algebra to Geometry,) will be serviceable.

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If among the quantities which would, when taken each (62) for the unknown quantity, serve to determine all the other quantities, there are two which would in the same way answer this purpose, and it could be foreseen that each would lead to the same equation (the signs + and - excepted); then we ought to employ neither of these, but take for the unknown quantity one which depends equally upon both; that is, their half sum, or their half difference, or a mean proportional between them, or, &c."

Thus, suppose it were required to find two numbers such that their sum should be 4 and the sum of their 4th powers 82. Instead of taking x and y for the numbers, we may make them both depend equally upon by putting

2+x= greater No.,
x = less No. ;

2

since the sum would be 4

and

or

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2x1 +2.24x2+2•16= (2+ x) * + (2 − x) 1 = 82, x1+24x2 = 41-16=25;

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