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8. 3x3 – 26x2 +522—24=0, the roots being in geometrical progression.

9. 2003 — 22 — 223—24=0, two of the roots being in the ratio of 3: 4.

10. 24x3+46x2 +9—9=0, one root being double another of the roots.

11. 8x4 — 223 — 2722 +63 +9=0, two of the roots being equal but opposite in sign.

12: 54x3 – 39x2 – 26x+16=0, the roots being in geometrical progression.

13. 32203 — 48x2 + 22x—3=0, the roots being in arithmetical progression.

14. 6x4 — 2933 +40x2 — 73 — 12=0, the product of two of the roots being 2.

15. 24 — 223 — 21x2 +22x+40=0, the roots being in arithmetical progression.

16. 27x1 – 19523 +494x2 – 520x+192=0, the roots being in geometrical progression.

17. 1823 +81x2 +121x+60=0, one root being half the sum of the other two.

18. Find the sum of the squares and of the cubes of the roots of 24 +qx2 +rx+8=0.

FRACTIONAL Roots.

546. An equation whose coefficients are integers, that of the first term being unity, cannot have a rational fraction as a root. If possible suppose the equation

2" +22%"-1+Pox"-? +...... +Px-2x2 + Pn-13+Pn=0, has for a root a rational fraction in its lowest terms, represented by a Substituting this value for x and multiplying through by In-1, we have

+Pan-1+pan-2b+...... +Pn-1 ab*-?+Pmon-1=0.

b Transposing,

=pan-1+p,an-2b+...... +Pm-1 abn-2+pnin-1. 6

an

an

This resulü is impossible, since it makes a fraction in its lowest terms equal to an integer. Hence a rational fraction cannot be a root of the given equation.

IMAGINARY Roots.

547. In an equation with real coefficients imaginary roots occur in pairs.

Suppose that f(x)=0 is an equation with real coefficients, and

suppose that it has an imaginary root a +ib; we shall show that a- ib is also a root. The factor of f(x) corresponding to these two roots is

(x - a - ib)(x a +ib), or (x - a)+62. Let f(x) be divided by (x-a)+62; denote the quotient by Q, and the remainder, if any, by Rx+R'; then

f(x)=Q{(x-a) +62} + Rx+R'. In this identity put x=a+ib, then f(x)=0 by hypothesis; also (x-a)2+b=0; hence R(a+b) + R"=0. Equating to zero the real and imaginary parts,

Ra+R' =0, Rb=0; and b by hypothesis is not zero,

... R=0 and R' = 0. Hence f(x) is exactly divisible by (x-a)2 +62, that is, by

(x-a-ib)(x-a+ib); hence x=a- ib is also a root.

548. In the preceding article we have seen that if the equation f(x)=0 has a pair of imaginary roots a+ib, then (x – a)2 +62 is a factor of the expression f(x).

Suppose that a+ib, c+id, e+ig,... are the imaginary roots of the equation f(x)=0, and that $(x) is the product of the quadratic factors corresponding to these imaginary roots; then

$(x)={(x—a)+b7}{(x–c)2+d2}{(x —e)2+g}... Now each of these factors is positive for every real value X; hence $ (x) is always positive for real values of x.

549. As in Art. 547 we may show that in an equation with rational coefficients, surd roots enter in pairs; that is, if a+ vb is a root then a - -vb is also a root.

Example 1. Solve the equation 6x2 – 1323 – 35x2 – 2 + 3 = 0, having given that one root is 2-V3.

Since 2-V3 is a root, we know that 2+ V3 is also a root, and corresponding to this pair of roots we have the quadratic factor 2c2-4x+1.

Also 6x4 – 13x3 — 35x2 — +3=(x2—4x+1)(6x2 +11x+3); hence the other roots are obtained from

6x2 +11x+3=0, or (3x+1)(2x+3)=0;

1 3 thus the roots are

2+13, 2-V3.

2 Example 2. Form the equation of the fourth degree with rational coefficients, one of whose roots is V2+V-3.

Here we must have V2+V–3, 12-V-3 as one pair of roots, and — V2+V-3, -V2-V-3 as another pair.

Corresponding to the first pair we have the quadratic factor 22 – 2 2x+5, and corresponding to the second pair we have the quadratic factor

2c +223+5. Thus the required equation is

(2c2+2 V220+5)(x2–2 2x+5)=0,

(22+5)2—8x2=0,
34 +232 +25=0.

or

or

EXAMPLES L. C. Solve the equations :

1+1-3 1. 3x4 — 10x3+4.2 — 3—6=0, one root being

2 2. Xt – 36x2+72x-36=0, one root being 3–V3. 3. 24+4x3+5x2+2x–2=0, one root being -1+ V-1. 4. 204 +423 +6x2 +42+5=0, one root being V-1.

TRANSFORMATION OF EQUATIONS. 550. The discussion of an equation is sometimes simplified by transforming it into another equation whose roots bear some assigned relation to those of the one proposed. Such transformations are especially useful in the solution of cubic equations.

551. To transform an equation into another whose roots are those of the original equation with their signs changed.

Let f (x)=0 be the equation.
Put - y

for X; then the equation f(-y)=0 is satisfied by every root of f(x)=0 with its sign changed; thus the required equation is f(-y)=0. If the given equation is

2 +P22n-1+P22n-2+ +Pn-\*+Pn=0, then it is evident that the required equation will be

y" Puyn-1+Payn-2 - ......+(-1)n-Pn-1Y+(-1)"pn=0; therefore the transformed equation is obtained from the original equation by changing the sign of every alternate term beginning with the second.

Note. If any term of the given equation is missing it must be supplied with zero as a coefficient.

Example. Transform the equation 24 — 1722 — 20x—6=0 into another which shall have the same roots numerically with contrary signs. We may write the equation thus :

24 +0x3 - 17x2 — 20x — 6=0. By the rule, we have

24 – 033 — 172c2 +20% — 6=0,

204 – 1702 +20x — 6=0. To transform an equation into another whose roots are equal to those of the original equation multiplied by a given factor.

Let f(x)=0 be the equation, and let q denote the given quantity. Put y=qx, so that when x has any particular

value, y

is 9 times as large; then x= and the required equation is

9

(y +P1

to.. ·+Pn-1

9 Multiplying by q”, we have

y" +P297"-1+Polyn-2+...... +Pr-19"-ly+Paq" =0. Therefore the transformed equation is obtained from the original equation by multiplying the second term by the given factor, the third term by the square of this factor, and so on.

or

552.

--(%)+

+Pn=0.

553. The chief use of this transformation is to clear an
equation of fractional coefficients.
Example. Remove fractional coefficients from the equation

3 1 3
2x3. 22 3+ -=0.

16
y
Put <= and multiply each term by q3; thus
a

3 1 3
2y3 qy2 q2y+ q=0.

2 8 16
By putting q=4 all the terms become integral, and on dividing
by 2, we obtain

y3 3y2-y+6=0. 554. To transform an equation into another whose roots exceed those of the original equation by a given quantity.

Let f(x)=0 be the equation, and let h be the given quantity. Assume y=x+h, so that for any particular value of x, the value of y is greater by h; thus x=y-h, and the required equation is f(y-2)=0.

Similarly if the roots are to be less by h, we assume y=x-h, from which we obtain x=y+h, and the required equation is fly+h)=0.

555. If n is small, this method of transformation is effected with but little trouble. For equations of a higher degree the following method is to be preferred : Let

f(x)=P,x" +p, 29-1+p, 2n-2+ +Pn-1 X + Pni put x=y+h, and suppose that f(x) then becomes

9.7+91 yn-1+42yn-2+ ... +9n-1y+qnu Now y=x-h; hence we have the identity

Po2n +P, 2n-1+P22n-2+ + Pn-1X + Pn

=90(x-h)" +91(x-h)n-1+ ... +9n-1(x-1)+qni therefore 9n is the remainder found by dividing f(x) by x-h; also the quotient arising from the division is

9.(x-h)"-1+(x-2)^-2+ ... +9n-1. Similarly 9n-1 is the remainder found by dividing the last expression by x - h, and the quotient arising from the division is

9.(x - 1)n-2+91(x-2)^-8+ ... + 9n-2;

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