2. Required the commensurable roots of the equation x2-6x2+11x-6=0.

Ans. +1+2+3. 3. Required the commensurable roots of the equation

x2-6x2-17x+21=0.

Ans. 3 and 1.

Here the student might hesitate, as one regular term of the equation is wanting, or rather the coefficient of r3 is 0; hence, the equation is x4±0x36x2-17x+21=0.

Go through the form of adding 0.

4. Required the commensurable roots of the equation x2-6x3+5x2+2x-10=0. Ans. -1, +5.

As the commensurable roots are only two, there must be two incommensurable roots, and they can be found by dividing the given equation by x+1, and that quotient by x-5, and resolving the last quotient as a quadratic.

EQUAL ROOTS.

(Art. 168.) In any equation, as

(1)

x+Ax+Bx3+Cx2+Dx+E=0,* the roots of a may be represented by a, b, c, d, e, and either one put in the place or a will verify the equation..

Now, let y represent the difference between any two roots, as a-b; then y=a-b, and by transposition b+y=a. But as a will verify the equation, it being a root, its equal, (b+y) substituted for x, will verify it also. That is,

(b+y)+A(b+y)+B(b+y)2+C(b+y)2+D(b+y)+E=0

• We might have been more general, and have taken z+Axm-1 &c., for the equation; but in our opinion we shall be better comprehended by taking an equation definite in degree; the reasoning is readily understood as general.

By expanding the powers, and arranging the terms ac

cording to the powers of y, we have

b5+5b4y+10b3y2+10b2y3+5by-tys

Ab+4 Aby+6Ab2y2 +4 Aby3+Ay
Bb3+3Bb2y+3Bby2+By
Cb2+2Cby+Cy

4

=0

Now, as b is a root of the equation, the first column of this transformation is identical with the proposed equation, on substituting the root b for x. Hence, the first column is equal to zero; therefore, let it be suppressed, and the remainder divided by y.

On the supposition that the two roots a and b are equal,

y becomes nothing, and this last equation becomes

564+4Ab3+3Bb2+2Cb+D=0

As b is a root of the original equation, x may be written

in place of b; then this last equation is

5x4+4Ax3+3Bx2+2Cx+D=0

(2)

This equation can be derived from the primitive equation by the following

RULE. Multiply each coefficient by the exponent of x, and diminish the exponent by unity.

Equation (2) being derived from equation (1), by the above rule, may be called a derived polynomial, and by the same rule a second derived polynomial can be derived from equation (2), &c., &c.

(Art. 169.) We again remind the reader that b will verify the primitive equation (1), it being a root, and it must also verify equation (2); hence, b at the same time must verify the two equations (1) and (2).

But if b will verify equation (1), that equation must divide by (x-b), (Art. 164,) and if it will verify equation (2), that equation also, must divide by (x-b), and (x-b) must be a common measure between the two equations (1) and (2). That is, in case the primitive equation has two equal roots equal to b.

(Art. 170.) To determine whether any equation contains equal roots, take its derived polynomial by the rule in (Art. 168.) and seek the greatest common divisor (Art. 27.), [which designate by (D),] between the given equation and its derived polynomial, and if the divisor Dis of the first degree, or of the form of x-h, then the equation has two equal roots each equal to h.

If no common measure can be found, the equation con. tains no equal roots. If Dis of the second degree with reference to x, put D=0, and resolve the equation, and if D is found to be in the form of (x-h)2; then the given equation has three roots equal to h.

If D be found of the form of (x-h)(x-h'), then the given equation has two roots equal to h, and two equal to h'. Let D be of any degree whatever; put D=0, and if possible completely resolve the equation; and every simple root of D is twice a root in the given equation; every double root of D will be three times a root in the given equation, and so on.

EXAMPLES.

1. Does the equation 24-2x37x2+20x-12=0 contain any equal roots, and if so, find them?

Its derived polynomial is 4x3-6x2-14x+20. The common divisor, by (Art. 27.), is found to be x-2, therefore, the equation has two roots, equal to 2.

The equation will then divide twice by x-2, or once by (x-2), or by x2-4x+4. Performing the division, we find the quotient to be x2+2x-3, and the original equation is now separated into the two factors,

(x2-4х+4)(x2+2x-3)=0

The equation can now be verified by putting each of these factors equal to zero. From the first we have already x=2, and 2, and from the second we may find =1 or -3; hence, the entire solution of the equation gives 1, 2, 2, -3 for the four roots.

2. The equation x5+2x4-11x38x2+20x+16=0 has two equal roots; find them.

Ans. 2 and 2.

3. Does the equation x5-2x4+3x37x2+8x-3=0 contain equal roots, and how many?

Ans. It contains three equal roots, each equal to 1.

4. Find the equal roots, if any, of the equation

x2+x2-16x+20=0.

5. Find the equal roots of the equation x2+2x3-3x2-4+4x=0.

Ans. 2 and 2.

Ans. Two roots equal to 1 and two roots equal to 2.

6. Find the equal roots of the equation

x3-5x2+10x8=0..

Ans. It contains no equal roots.

(Art. 171.) Equations which have no commensurable roots, or those factors of equations which remain after all the commensurable and equal roots are taken away by division, can only be resolved by some method of approximation, if they exceed the third or fourth degree. It is possible to give a direct solution in cases of cubics and in many cases of the fourth degree; but, in practice, approximate methods are less tedious and more convenient.

Methods of approximation can only be illustrated by examples.

Required a root of the equation x2+2x2-8x-24-0. We find by a few trials that one root or a must be a little over 3. Represent 3 by a, and put y to represent the remaining part of the root. Then x=a+y. Then the equation becomes

(a+y)+2(a+y)2-8(a+y)-24=0.

Expanding and arranging the powers in reference to y, as

Making a=3, as previously determined, and we have

y3+1ly2+3ly-3=0

As y is a very small fraction, (which we perceive by inspection,) the higher powers of y are comparatively small, and by disregarding them we have

y==0.096

Thus, we have the approximate value of x=3.096. Now a may be taken equal to 3.096, in place of 3 in equation (A), and as a becomes nearer to the true value of x, y will become a smaller and smaller quantity, and the errors arising from rejecting the higher powers of y will become less and less, and at length will become very inconsiderable.

After taking a=3.09, and substituting that value for a in equation (A), and finding a corresponding value of y, we shall have a further approximate value of x, which will be true to six or seven places of decimals, and by repeating the process we may obtain the value of x to any required degree of exactness.