therefore every equation of an even order, with real coëfficients, is composed of real factors of the second degree.

Ex. 1. One root of the equation x3-26x+60-0 is -6; required the other roots. Ans. 3±√1. Ex. 2. One root of the equation x3-15x+4=0 is -4; required the other roots. Ans. 2√3. Ex. 3. Two of the roots of the equation x-4x3-7x2+26x -14=0 are 3+√2, and 3−√2; required the remaining roots. Ans. -1/3.

Ex. 4. One root of x-7x+13x-3=0, is 2-√3; find the other roots. Ans. 2+3 and 3.

Ex. 5. One root of x3-3x2-42x-40—0 is −1(3+√—31); find the other roots. Ans. (3--31), 4, and -1.

Ex. 6. Two roots of x5-10x4+29x3-10x2-62x60=0 are 3 and 2; find the other roots.

Ans. —√2, 2, and 5.

ART. 402. PROP. VIII.- DESCARTES' RULE OF THE SIGNS.No equation can have a greater number of POSITIVE roots than there are VARIATIONS of sign; nor a greater number of NEGATIVE roots than there are PERMANENCES of sign.

In the equation x-a=0, where the value of x is +a, there is one variation, and one positive root.

In the equation x+a=0, where the value of x is —a, there is one permanence, and one negative root.

In the equation x2-(a+b)x+ab=0, where the values of x are +a and b, there are two variations and two positive roots.

In the equation x2+(a+b)x+ab=0, where the values of x are —a, and —b, there are two permanences, and two negative roots.

In the equation x2-x-12=0, where x=+4, and -3, there is one variation, and one positive root, and one permanence, and one negative root.

If we form an equation of the third degree, (Art. 397), whose roots are +2, +3, +4, we shall have x3-9x2+26x-24=0, where there are three variations, and three positive roots.

But if we form an equation whose roots are -2, −3, +4, we shall have x3+x2-14x-24=0, where there is one variation, and one positive root, and two permanences, and two negative roots.

To prove the proposition generally, let the signs of the terms in their order, in any complete equation, be

++-+-+ + +, and let a new factor x-a=0, corresponding to a new positive root be introduced, the signs in the partial and final products will be

Now in this product, it is obvious, that each permanence is changed into an ambiguity; hence, the permanences, take the ambiguous sign as you will, are not increased in the final product of the introduction of the positive root +a; but the number of signs is increased by one, and therefore the number of variations must be increased by one. Hence, the introduction of any positive root introduces, at least, one additional variation of sign.

Now the equation x-a=0, contains one positive root, and has one variation of sign. Therefore, since every additional positive root introduces, at least, one additional variation of sign, the number of positive roots can never exceed the number of variations of sign.

Again, if we change the signs of the alternate terms, the roots will be changed from positive to negative, and conversely (Art. 400). Hence, the permanences in the proposed equation will be replaced by variations in the changed equation, and the variations in the former by permanences in the latter; and since the changed equation cannot have a greater number of positive roots than there are variations of sign, the proposed equation cannot have a greater number of negative roots than there are permanences of sign.

Cor. 1. Since the whole number of variations and permanences is evidently equal to the degree of the equation, (the equation if not complete being rendered so by the introduction of ciphers). Therefore, if the roots of an equation be all real, the number of positive roots must be equal to the number of variations, and the number of negative roots to the number of permanences. (See examples, pages 343, 345.)

2. By means of this theorem we can often determine whether there are imaginary roots in an equation.

Now, if we take the upper sign there are no variations, hence there is no positive root; and if we take the lower sign there are no permanences, hence there is no negative root. But since the equation has two roots (Art. 396), they must, therefore, both be imaginary.

In like manner the cubic equation

Now if we take the upper sign there are no variations, and consequently no positive root. But if we take the lower sign, there is one permanence, hence there can be but one negative root. Therefore, the other two roots must be imaginary.

ART. 403. PROP. IX.- If two numbers, when substituted for the unknown quantity in an equation, give results affected with different signs, one root at least of this equation lies between these numbers.

Let the equation, for example, be

x3-x2+x-8=0.

If we substitute 2 for x in this equation, the result is -2; and if we substitute 3 for x, the result is +13. These results have different signs, and it is required to show that there must be one real root, at least, between 2 and 3.

The equation may evidently be written thus,

(x3+x)—(x2+8)=0.

Now in substituting 2 for x, x3+x=10, and x2+8=12, ·. x3+x<x2+8;

also, in substituting 3 for x, x3+x=30, and x2+8=17,

Now both these quantities increase while x increases, but the first increases more rapidly than the second, since when x=2, it is less than the second, but when x=3 it is greater. Consequently, for some value of x between 2 and 3, we must have x3+x=x2+8, and this value of x is, therefore, a real root of the equation.

In general, suppose we have an equation X=0, where X represents a polynomial involving x, and that two numbers, p and q, when substituted for x, give results with contrary signs. Let P be the sum of the positive, and N the sum of the negative terms;

also suppose that when x=p, P-N is negative, or P<N, and that when x=q, P—N is positive, or P>N.

Suppose to change by imperceptible degrees from p to q, then P and N must also change by imperceptible degrees, and both increase, but P must increase faster than N, otherwise from having been less it could never become greater; there must, therefore, be some value of x between p and q, which renders P=N, or satisfies the equation X=0, and this value of x is, therefore, a real root of the equation.

Cor. If the difference of the two numbers, p and 9, which give results with contrary signs, is equal to unity, it is evident that we have found the integral part of one of the roots.

Ex. 1. Find the integral part of one value of x in the equation x2-4x3-3x2+x-5=0.

If x=3, the value of the equation is -2, but if x=4, the value is 47. Hence, a root lies between 3 and 4; that is, 3 is the first figure of one of the roots.

2. Required the first figure of one of the roots of the equation x3-5x2-x+1=0. Ans. 5.

TRANSFORMATION OF EQUATIONS.

ART. 404. The transformation of an equation is the changing it into another of the same degree, whose roots shall have a specified relation to the roots of the given equation.

Thus, in the general equation of the nth degree

if -y be substituted for x, the equation will be transformed into another whose roots are the same as those in (1), but with contrary signs, for y=-x.

If1 be substituted for x, the roots of the new equation in y

y

will be the reciprocals of those of equation (1), for y=1

ART. 405. PROP. I.- To transform an equation into one whose roots are the roots of the given equation multiplied or divided by any given quantity.

Let a, b, c, &c., be the roots of the equation

+Tx+V=0;

assume y=kx, or x=; then substitute this value for x, and th

k

Since y=kx, the roots of this equation are ka, kb, kc, &c.

It is evident this equation may be derived from the proposed equation, by multiplying the successive terms by 1, k, k2, h3, &c., and changing x into y.

In the case of division, assume y= or x=ky, and substitute.

',

Cor. By this transformation an equation may be cleared of fractions, or if the first term be affected with a coëfficient, that coëfficient may be removed.

Ex. Let it be required to transform the equation

x2+1 px2+qx+r=0,

into one which is clear of fractions, and which has unity for the coëfficient of the highest term.

By multiplying by 6, we have

6x+3px2+2qx+6r=0.

Let y=6x, or x=ty, and the equation becomes

y3+3py2+12qy+216r=0,

an equation of the required form.

Ex. 1. Find the equation whose roots are those of the equation x2+7x2-4x+3=0 multiplied by 3.

Ans. y+63y2-108y+243=0.

2. Find the equation whose roots are each 5 times those of the equation 1+2x3-7x-1=0.

Ans. y1+10y3-875y-625—0.

3. What is the equation whose roots are each of those of 3-3x2+4x+10=01 Ans. 4y3-6y+4y+5=0.