Hence the transformed equation must be

y3 + 12y2 + 36y 12 = 0.

Transform the equation -10x2+3x-69460, into another whose roots shall be less by 20. We make x = 20+y.

The three remainders are the numbers just above the double lines, which give the following transformed equation:

y3 + 50y2 + 803y-2886 = 0.

Transform this equation into another whose roots shall be less

by 3.

Hence the transformed equation is

z3+59z2 + 1130% = 0.

This equation may be verified by making z = 0; which gives y = 3, and x = 20 + 3 = 23.

444. If the signs of the alternate terms of any complete equation involving but one unknown quantity be changed, the signs of all the roots will be changed.

In the general equation

xm + Axm¬1 + Bxm¬2 +

... ·

+ Tx+U=0.

(1),

let the signs follow each other in any order whatever. Changing the signs of the alternate terms of this equation, beginning with the second, we have

but if the change begin with the first term, we have

Now, if a be a root of (1), its first member reduces to zero when a is substituted for x; that is, the sum of the positive terms becomes equal to the sum of the negative terms. But if - a be substituted for x in (2) and (3), the numerical values of the terms of these equations will be equal to the values of the corresponding terms of (1), while the signs of the terms in (2), if m is an even number, will be the same, and those of (3), opposite to the signs of the terms of like degree in (1). If m is an odd number, the reverse will be true in respect to signs. In either case, however, if a is a root of (1), a is a root of both (2) and (3).

An obvious consequence of this proposition is, that the roots of an equation are not affected by changing the signs of all its

terms.

EXAMPLES.

1. The roots of the equation 3 — 7x2 + 13x — 3 = 0, are 3, 2√3, and 2-3; what will be the roots of the equation 23+7x2 + 13x+3=0? Ans. 3, 2√3, −2+ √/3.

2. The roots of the equation x — 3x3 + 3x2 + 17x180,

are 1, 2, 2-5, and 2-5; what are the roots of the equation 4 + 3x3 + 3x2-17x180?

− +2,

Ans. 1, 2, 2-√5, −2+ √5.

-

445. If all the coefficients of an equation are rational, surd roots can enter the equation only by pairs, and if all the coefficients are real, imaginary roots can enter the equation only by pairs.

Let the coefficients A, B, .... U, of the equation,

-1

-2

x2+Ax3¬1+Bx3¬2+ Tx+U = 0

....

(1),

be all real and rational, and suppose that a±√b is one of the roots of this equation.

Substituting this value for x, we have

Expanding the several terms of this equation by the binomial

Aam−1±▲ (m—1) am~2. √/ ±b+A (m−1)m7 3 am¬3.(√±6)°±.......

2

Bam−2± B (m—2) am−3. √√±b+B (m—2) m¬3 a

-3

am-3.

2

m-1

observing, in reference to the final terms, ±(√), ±(√±b)", etc., that the sign is to be used before those only which have odd numbers for their exponents; when the exponent is even, the plus sign is to be understood.

If the root of (1) be a + √, the aggregate of these developments will be composed of two parts, the one rational and the other surd. The rational part will be the algebraic sum of those terms which have the even powers of √b for factors, the zero power being included. Represent this part by M.

=

The irrational part will be the algebraic sum of the terms having the odd powers of √ for factors. But since (√õ)3 = b √b, (√b)5 = b2 √, etc., the sum of the terms of this part can be represented by a single term of the form N√b, N being the algebraic sum of the coefficients of vb. Hence (2), under the supposition that a + is a root of (1), becomes

which can be true only when we have separately M=0, N=0 (272, 4).

(4),

Hence if

In reducing (2) to (3), the upper signs in the expansions of the terms of (2) were used. If the lower signs in the equation and the expansions of its terms be used,-which is equivalent to supposing ab to be a root of (1),-the reduced equation will be M-N√b=0 in which M and N are evidently the same as in (3). (1) has a root, a + √ō, it has also the root a Now let us suppose that a + b is a root of (1); then since the even powers of √-b are real and the odd powers imaginary, the developed first member of (2) will be composed of two parts, the one real and the other imaginary. Represent the real part by M'.

The imaginary part is the algebraic sum of the terms having the odd powers of √b for factors. But since (√ b) 3 = √ b2 ( − b) = b √ — b, (√ — b)3 = √ b1 ( — b) = b2 √ — b, etc., the sum of the terms of this part can be represented by a single term of the form N'√-b. Hence, under the supposition that a + √b is a root of (1), (2) becomes

M'+N'√ b=0

=

(5),

which requires that we have separately M' 0, N'=0 (267). By using the lower signs of the terms and their expansions in (2),-which supposes a - √b to be a root of (1),—we find

and by a simple inspection of the expanded terms of (2), we see that M' and N' in (5) and (6) are the same.

Whence we conclude that if (1) has a root, a + √b, it has also the root, a — √ — b.

RULE OF DES CARTES.

446. The number of real positive roots of the equation X = 0 cannot exceed the number of variations in the signs of its terms; and, if the equation X = 0 is complete, the number of real negative roots cannot exceed the number of permanences in the signs of its terms.

NOTE.-In any series of quantities a pair of consecutive like signs is called a permanence of signs, and a pair of consecutive unlike signs is called a variation of signs. Thus, in the expression æ3 — 3x1 — 4x6 + 7x5 + 3x1 + 2x3 — x2 — x + 1, there are four permanences and four variations.

-

by a, b, c...., and suppose (1) to be divided by the product of all the factors x

real positive roots.

· a, x − b, x + c...., corresponding to the Represent the resulting equation by

We shall now show that if (2) be multiplied by the factor xa corresponding to a real positive root, the number of variations in the resulting equation will be at least one greater than in (2).

1st. Suppose (2) to be complete, and let the signs of its terms be ++ The signs of the multiplier are +

+.

A double sign is placed where the sign of any term in the product is ambiguous.

Now, taking the ambiguous signs as we please, the number of variations in the product is greater than in the multiplicand; and this is still true if we suppose some or all of the terms having ambiguous signs to vanish.

2d. If (2) is incomplete, reduce it to a complete form by inserting the missing terms with zero for the coefficient of each; the resulting equation will contain at least as many variations as (2). Multiplying the completed equation by x-a, the num