And, since a-b√-1 substituted for a gives a result equal to zero, it must be a root of the equation.

Ex. 1. Find the roots of the equation

Hence every equation of the third degree whose coefficients are all real, must have one real root. The same is true of every equation of an odd degree.

PROPOSITION VII.

(314.) Every equation must have as many variations of sign as it has positive roots, and as many permanences of sign as there are negative roots.

To prove this Proposition, it is only necessary to show that the multiplication of an equation by a new factor, x-a, corresponding to a positive root, will introduce at least one variation, and that the multiplication by a factor x+a will introduce at least one permanence.

For an example, take the equation

x+3x-10x-24=0,

a which the signs are ++−−, giving one variation. Multiply this equation by x-2=0, as follows:

7

x3-3x2-10x -24

x-2

x*+3x3-10x2 - 24x

-2x3- 6x2+20x+48

x*x3-16x2- 4x+48=0.

In this last product the signs are ++−−+, giving two va

itions; that is, the introduction of a positive root has introtuced one new variation in the signs of the terms.

To generalize this reasoning, we perceive that the signs in the upper line of the partial products are the same as in the given equation; but those in the lower line are all contrary to those of the given equation, and advanced one term toward the right.

Now, if each coefficient of the upper line is greater than the corresponding one in the lower, the signs of the upper line will be the same as in the total product, with the exception of the last term. But the last term introduces a new variation, since its sign is contrary to that which immediately precedes it; that is, the product contains one more variation than the original equation.

When a term in the lower line is larger than the corresponding one in the upper line, and has the contrary sign, there is a change from a permanence to a variation; for the lower sign is always contrary to the preceding upper sign. Hence, whenever we are obliged to descend from the upper to the lower line in order to determine the sign of the product, there is a variation which is not found in the proposed equation; and as all the remaining signs of the lower line are contrary to those of the proposed equation, there must be the same changes of sign in this line as in the given equation. If we are obliged to reascend to the upper line, the result may be either a variation or a permanence. But even if it were a permanence, since the last sign of the product is in the lower line, it is necessary to go once more from the upper line to the lower, than from the lower to the upper. Hence each factor, corresponding to a positive root, must introduce at least one new variation; so that there must be as many variations as there are positive

roots.

In the same manner, we may prove that the multiplication by a factor x+a, corresponding to a negative root, must introduce at least one new permanence; so that there must be as many permanences as there are negative roots.

Ex. 1. The roots of the equation

x-3x-5x+15x+4x-12=0

are 1, 2, 3, -1, and −2. There are also three variations of sign, and two permanences, as there should be, according to the Proposition.

Ex. 2. The equation

x-3x2-15x2+49x-12=0

has four real roots. How many of these are negative?

Ex. 3. The equation

x+3x-41x1-87x2+400x2+444x-720=0

has six real roots. How many of these are positive?

If all the roots of an equation are real, the number of positive roots must be the same as the number of variations, and the number of negative roots must be the same as the number of permanences. If any term of an equation is wanting, we must supply its place with ±0 before applying the preceding Rule.

PROPOSITION VIII.

(315.) If two numbers, when substituted for the unknown quantity in an equation, give results with contrary signs, there is at least one root comprised between those numbers.

Take, for example, the equation

x3-2x2+3x-44-0.

If we substitute 3 for x in this equation, we obtain -26; and if we substitute 5 for x, we obtain +46. There must, therefore, be a real root between 3 and 5; for, when we suppose x=3, we have

x2+3x<2x2+44.

But when we suppose x=5, we have

x2+3x>2x2+44.

Now both the quantities

x+3x and 2x2+44

increase while x increases. And since the first of these quantities, which was originally less than the second, has become the greater, it must increase more rapidly than the second. There must, therefore, be a point at which the two magnitudes are equal, and that value of x which renders these two magni tudes equal must be a root of the proposed equation.

In general, if two numbers, p and q, substituted for x in an equation, give results with contrary signs, we may suppose the less of the two numbers to increase by imperceptible degrees

until it becomes equal to the greater number. The results of these successive substitutions must also change by imperceptible degrees, and must pass through all the intermediate values between the two extremes. But the two extreme values are affected with opposite signs; there must, therefore, be some number between p and q which reduces the given equation to zero, and this number will be a root of the equation.

In the same manner, it may be proved that if any quantity p, and every quantity greater than p, substituted in an equation, renders the result positive, then p is greater than the greatest

root.

Hence, also, if the signs of the alternate terms are changed, and if q, and every quantity greater than q, renders the result positive, then -q is less than the least root.

If the two numbers, which give results with contrary signs, differ from each other only by unity, it is plain that we have found the integral part of a root.

Ex. 1. Find the integral part of one of the roots of the equa

tion

2x-11x+8x-16=0.

When x=2, the equation reduces to -12; and when x=3, it reduces to +71. Hence there must be a root between 2 and that is, 2 is the first figure of one of the roots.

3;

Ex. 2. Find the first figure of one of the roots of the equa

tion

x+x+x-100=0.

Ans. 4.

Ex. 3. Find the first figure of each of the roots of the equa

tion

x3-4x-6x+8=0.

PROPOSITION IX.

(316.) Every equation may be transformed into another, whose roots are greater or less than those of the former by any given quantity.

Let it be required to transform the general equation of the mth degree into another whose roots are greater by r thau those of the given equation.

and substitute y-r for x in the proposed equation; we shall

y is greater than x by r.

If we take y=x−r, or x=y+r, we shall obtain in the same way an equation whose roots are less than those of the given equation by r.

Ex. 1. Find the equation whose roots are greater by 1 than those of the equation

x2+3x2-4x+1=0.

We must here substitute y-1 in place of x.

Ans. y3-7y+7=0.

Ex. 2. Find the equation whose roots are less by 1 than those of the equation

x3-2x2+3x-4=0.

Ans. y+y+2y-2=0.

Ex. 3. Find the equation whose roots are greater by 3 than those of the equation

x2+9x+12x2-14x=0.

Ans. y-3y-15y+49y-12==0. Ex. 4. Find the equation whose roots are less by 2 than those of the equation

5x*— 12x3-3x2+4x-5=0.

Ans. 5y+28y3+51y2+32y-1=0.

Ex. 5. Find the equation whose roots are greater by 2 than those of the equation

x+10x+42x2+86x2+70x+12=0.

Ans. y+2y-6y'-10y+8=0.

PROPOSITION X.

(317.) Any complete equation may be transformed into another whose second term is wanting.