CHAPTER XII.

GENERAL THEORY OF EQUATIONS.

187. EVERY equation containing but a single un known quantity, whose exponents are whole numbers, may be reduced to the form,

-1

x2 + px2 - 1 + qx2· -2+ &c. + sx2 + tx + u

= 0 (1.)

In this equation, n is a positive whole number, but the coefficients p, q, &c., may be either positive or negative, entire or fractional, real or imaginary. The method of reducing equations to the form (1), is entirely analogous to that given for reducing equations of the second degree to the form,

x2 + 2px = 9;

and since the reduction can always be made, we shall hereafter, in speaking of equations, suppose them reduced to the form of Equation (1), unless the contrary is expressly mentioned.

188. Any value of x, either real or imaginary, which when substituted for a, in Equation (1), will satisfy it, that is, make the two members equal, is a root of the equation

It is evident, that Equation (1) has at least one root, for if the first member is equal to 0, it must be so for some value of x, either real or imaginary, and this value of is, by definition, a root.

PROPERTIES AND TRANSFORMATIONS OF EQUATIONS.

I. In every equation of the form (1), if a is a root, the first member is divisible by x -α.

189. For, by applying the rule for division, and continuing the operation till a remainder is found which does not contain x, denoting the quotient by m and the remainder by n, we shall have,

- 1 + &c. + tx + u = (x-a)m + n (2.)

Now, if a is a root of the proposed equation, it will reduce the first member of (2) to 0, when substituted for ; it will also reduce the first term of the second member to 0; hence, n is also equal to 0, that is, the remainder is 0, and consequently the first member is exactly divisible by x-a.

II. Conversely, if the first member of an equation of the form (1) is exactly divisible by x-a, then is a a root of the equation.

190. For, in this case, the remaindern is equal to 0, which reduces Equation (2) to the form,

·-1 + &c. + tx + u = (x − a)m (3.)

If, in (3), we make x = a, the second member reduces to 0; consequently, the first member also reduces to 0, which satisfies Equation (1). Hence, a is a root of (1),

as was enunciated.

It follows, from the preceding propositions, that we can ascertain whether a polynomial containing 2 is exactly divisible by a binomial of the form, x u, by simply

substituting a for x in the polynomial. If the result is 0, the polynomial is divisible by xa; if it is not 0, the polynomial is not exactly divisible by x-a.

III. Every equation of the form (1), has as many roots as there are units in n, and no more.

least one root, let

a

be a

of the other

191. Since the equation has at that root be denoted by a, then will factor of the first member; the first term factor is -1, and the exponent of x, in each of the other terms, is less than -1; hence, the given equa tion may be written under the form,

-2

n

(x − a) (xn−1 + p'xn−2+ &c. + t'x + u') = 0

= 0 . (4.)

Now, Equation (4) may be satisfied in two ways, viz. by placing the first factor equal to 0, or by placing the second factor equal to 0. In the latter case, we shall have the equation,

xn ·− 1 + p2x2 − 2 + &c. + t'x + u' = 0

Now, Equation (5) has at least one root; let that root be designated by b; then it may be shown as before. that Equation (5) can be written under the form,

b) (x2+p''x-3+ &c. + t'x+u") 0; (6.)

(x - b) (xn

which reduces Equation (4) to the form of,

-3

(ra) (x-b) (xn−2+p'xn-3+ &c. + t'x+u') = 0 (7.)

In the same manner as before, it may be shown that the second factor of the first member of (6) can be placed equal to 0, and factored, giving,

(x − c) (2,7 −3 + p'''xn−4 + &c. + t'''x + u''')

= 0;

which reduces Equation (4) to the form,

-3

(x-α) (x-b) (x−c) (x2¬3+p'''x2++&c. +t'''x+-u''') = 0.

By continuing the process, it may be shown that the first member will ultimately be resolved into just as many binomial factors, of the form, (xa), (x − h), &c., as there are units in n. Hence, Equation (1) may be written under the form,

(x-α) (x-b) (x−c)... (x-k) (x-1)

x

= 0.

(8.)

Now,

Equation (8) may be satisfied, by placing either of the factors, aα, xb, &c., equal to 0, and either factor being placed equal to 0, gives a root. since there are n factors, the equation has n roots, and since the equation cannot be satisfied except by making one of the factors equal to 0, there are only n roots which was to be shown.

If both members of Equation (8), be divided by either of the factors, x- a, x-b, &c., it will be reduced in degree by 1. It is to be observed, that some of the roots, and consequently, some of the binomial factors, may be equal to each other.

EXAMPLES.

1. One root of the equation, x3 — 9×2+ 26x 24 0, is 4; what does the equation become when freed of this root?

an equation that may be solved by known rules.

0, 18

2. One root of the equation, 3 37x 84 = +, what does it become when freed of this root? Ans. x2 + 7x+12= 0.

3. One root of the equation, x3-11x2+16x+84 = 0, is 2; what does the equation become when freed of this root?

is

Ans. x2

13x+42 = 0.

4. One root of the equation, x3+ 7x2 — 4x

7; what does the equation become when freed of this root?

5. One root of the equation, is 3; what are the other roots?

Ans. x2 4

0

3-12x2+47x-60 = 0,

Ans. 4 and 5.

6. One of the roots of x3+ 9x2 + 26x + 24 = 0, is

7. Two roots of the equation, æa. 12x348x2-68x 150, are 3 and 5; what are the other two? Ans. 2+√3, and 2√3.

8. One root of the equation, is 1; what are the other two.

x3-6x2+11x-6 == 0, Ans. 3 and 2.

192. Equation (8) indicates the method of forming an equation whose roots are given.

RULE.

Subtract each root from the unknown quantities; multiply the resulting binomials together, and place the product equal to 0.