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EXAMPLES.

1. Does the equation x_2x'—7*+20x—12 = 0, contain equal roots, and if so, what are they? The first derived polynomial of the first member is

4x'_-62-14.0+20. The greatest common divisor between this and the first member of this equation is x— _2; therefore x = 2 is twice a root of the equa

tion, and

2 and 2 ;

x*—24_7**+20x—12 may be divided twice by 2-2, or once by (x-2)' = x'—4x+4. Performing the division, we find the quotient to be x*+2x–3, and the original equation may now be written

(x*--4x+4)(x+2x-3) = 0. This equation will be satisfied by the values of x found by placing each of these factors equal to zero. From the first we get a = - 2,

= 2, and from the second x = 1 x= -3; hence the four roots of the given equation are 1,2,2,-3. 2. Find the equal roots of the equation 2+2 -11x-82" +20+16=0. Ans.

-laud -1. 3. What are the equal roots of the equation

x-2x+3x*—7.0*+82—3 = 0?

Ans. It has three roots, each equal to 1. 4. What are the roots of the equation x*—2x9-112+12x+36 = 0?

Ans. Its roots are 3, 3, -2,-2. 5. What are the equal roots of the equation

X= x'—5x92x+38x*—313-61c* +96436 = 0?
We find
X = 7x®-30x10x* +152x93x'-122x+96,

D= x -3x'—+11x-6,
D, = 4.-9.x-2x+11,

D' =.X-1. Hence x = 1 is twice a root of the equation D = 0, and three times a root of the given equation.

Dividing D=x*—3x® —*+11x—6 by D" = æ–2x+1, we find for the quotient x*—*—6 = (x-3)(x+2). Therefore,

D= (x-3)(x+2)(x-1)', and

X=(x-3)*(0+2)*(x-1)". Hence the roots of the given equation are

Ans. 3, 3, -2,-2, +1, +1, +1. 436. Having an equation involving but one unknown quantity, to transform it into another, the roots of which shall differ from those of the proposed equation by a constant quantity.

Assume 30++Ax*-+B.:"-"+0.2m-*+....+Tx+U=0, and denote the new unknown quantity by y, and by x' the arbitrary but fixed difference which is to exist between the corresponding values of x and y; we shall then have x = yta'.

Substituting this value of x in the given equation, it becomes

(y+x')"+A(y+x')*-*+B(y+x')*+(y+x')"-+.... +Tyt.')+U= 0.

Developing the terms separately, by the binomial formula, and arranging the

aggregate of the results with reference to the ascending powers of y, we have

y*+.. x' y+ mx'm

y+ +Am-1).

-2 +B(m2)2m +Am-1)

2 + C(m-3).

-3 +B(m—2)

m-1

m

m-1

æ'm—

1

m

x'm_3

Im-3

Im

+ Ax'm— + Bx'm— + C'x'm : + Tx' +U

m

m

= 0.

(1)

ym-' +mx' lym-' tym
+m
2

+A
+Am-1).

+B An examination of this developed first member leads to these conclusions :

1.—The absolute term of the transformed equation, or the coefficient of yo, is what the first member of the given equation becomes when ac' is substituted for x,

2.—The coefficient of y, the first power of the unknown quantity, is what the first derived polynomial of the first member of the given equation becomes, when in it x' takes the place of x.

3.- The coefficient of ye is what the second derived polynomial of the first member of the given equation becomes when it is divided by 2, and x' takes the place of x.

4.-And in general, the coefficient of yn is what the nth derived polynomial of the first member of the given equation becomes when it is divided by the product of the natural numbers from 1 to n inclusive, and is replaced by x'.

Representing the first member of the given equation, and its successive derived polynomials, after x' has been substituted for x, by X', X'ı, X'2, X'z, etc. respectively, the transformed equation may be written

X2

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m-1

+1:2..(m-1)

-1

m-2

m-1

X'+X'1y+1:2

t.
+ +
1.2..(m-

-2)
X'.

-you-i tym = 0.

rity Or, by inverting the order of terms, Xo

X'. ym+

! + 1.2..(m-1)

1.2..(m—2)Y? +..Xy+X'= 0. (19) 437. By comparing eqs. (1) and (1') of the preceding article it is shown that,

X'.
= ma'+A,

(2)
1.2..(m1)

X and

X'? + A(m-1)+B; (3)
1.2..(m-
—2)

2 the degree of the coefficients of the equation, with respect to a', increasing by at least one from term to term as we pass from left to right, the absolute term being of the mth degree.

Now, since ' is an arbitrary quantity, such a value may be assumed for it as will cause it to satisfy any reasonable condition. We may therefore form an equation, by placing any one of these coefficients equal to zero, regarding as the unknown quantity, and any root of this equation will cause the corresponding term of the transformed equation (1'), (436), to disappear.

m

m-2

= m

m

m-2

m

m

Suppose

A mx'+A=0; whence x' = If this value of x' be substituted in the equation just referred to, it takes the form

X'.
ym +
1.2..(m-2)

to...X'y+X'= 0. Hence, to transform an equation into another which shall be incomplete in respect to the second term: Substitute for the unknown quantity another, minus the coefficient of the second term divided by the erponent of the degree of the equation.

438. The third term will disappear from the transformed equation when x' is made equal to either of the roots of the equation,

-1

ad" +A(m-1)+B=0.

2 But there may exist such a relation between m, A, and B, that the

A value, x'

will satisfy this equation; in which case the vanishing of the second term of the transformed equation will involve that of the third. To find what this relation is, substitute this value of x' in the above equation, and it becomes

-1 A

(m-1) +B=0

2 This, reduced as follows,

1 AP

A
(m -1)

+B=0,
2
(m-1)A?—2(m-1)AP+2mB = 0,
(m-1)A' = 2mB,

2mB gives finally

-1' When the values m, A, and B, will satisfy this equation, the third term of the transformed equation will disappear with the second. In general, to find the value of x' which will free the transformed equation of the third term, an equation of the second degree must be solved; and to free it of the fourth term, the equation to be solved would be of the third degree; and finally, to make the absolute term disappear, would require the solution of the original equation,

m m.

.

ma

m

m

m

m

A' =

m

EXAMPLES.

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1. Transform the equation r’+2px-9 = 0, into another which which shall not contain the second term.

2p This is done by making x = y = ур (437);

2 whence, by (436),

X'=(-2p(p)-9=p';
X', = 2(P)-2p=0;
2

1.
2 2

X2

Therefore, the required equation is

y-9-p= 0, from which we find y= +Vatp; and since x = yếp, the values of x are given by the formula,

x=-p+Va+p'.

?. the same as that found by the rule for quadratics.

2. Transform the equation x+px'+qx+r= 0, into one not having the second term.

P
3
P

р

2p PI X

+y =
3
9

+r, 3

3

27

Make x = y

; then

x, =(-) +2()+9 = +9

-3

X'

2P

2

+

2

= 0,

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X 3

=1.

2:3 2.3 2:3 Hence, the equation sought is

2p

P9

+ 3

27 3

2po Pa or, by making

+ = 0,
27 3
go-my+=0.

-9=m, and

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