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Since the equation can have no more than one positive root, we will only make trial of the remaining negative divisors, thus : 1+4-1-16 – 12 1-6

1+4–1–16 – 12 1-4 + 2

+ 3 -14

- 13 Hence – 6 is not a root.

Hence -4 is not a root. 1+4-1-16–12 - 3

1+4-1-16 -12 1-2 -1-1+4+ 4

-1-2+5+ 6 +3+3–12

+2+4-10 Hence - 3 is a root.

Hence – 2 is a root.

EXAMPLES L. i.

Solve the following equations, which have one or more integral roots : 1. 203 — 9x2 +263-24=0. 2. 203 — 2—2x2+2=0. 3. 2x3 +5x2–113+4=0. 4. 4x3 — 16x2 +312—14=0. 5. 203 — 2x2 — 293 +30=0. 6. 23 - 8x2 + 5x+14=0. 7. 24 — 233 — 7x2 +8x+12=0. 8. 34-433 – 14x2+36x+45=0. 9. 34 — 322 — 42% -40=0. 10. 24 - 10x2 – 20x — 16=0. 11. 3+ +8x3 + 9x2 – 8x-10=0. 12. 2+ + 2x3 — 702_8x+12=0. 13. 34 - 3x2 - 6x-2=0.

14. 24 - 2x3 – 12x2 + 10x +3=0. 15. 6x4 – x3 – 17x2 +16x—4=0. 16. X4 — 2x3 -- 13x2 + 14x+24=0. 17. 24 +4x3 — 22x2 - 4x+21=0. 18. 2+ + 2x3 — 702 — 8x+12=0. 19. 24 – 6x2 — 16x+21=0. 20. 3+ +4x3 — 2 — 16x — 12=0. 21. 2x4 — 23 — 29x2 +34x+24=0. 22. 205 — 3x4 -533 + 15x2 +4x–12=0.

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CARVAN'S METHOD FOR THE SOLUTION OF CUBIC EQUATIONS. 588. The general type of a cubic equation is

23 +Px2+ Qx+R=0, but as explained in Art. 556 this equation can be reduced to the simpler form

23 +qx+r=0, which we shall take as the standard form of a cubic equation.

589. To solve the equation 23+qx+r=0.
Let x=y+z; then

x=y8+28+3y=(y+z)=y8+28+3yzx,
and the given equation becomes

38 +23+ (3y2 +9)x+r=0. At present y, z are any two quantities subject to the con: dition that their sum is equal to one of the roots of the given equation; if we further suppose that they satisfy the equation 3yz+q=0, they are completely determinate. We thus obtain 78 +23= -1

....(1), 98

..(2)

27 y: hence

98 y:

+ry 27y3

27 Solving this equation,

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Substituting in (1), 23

p2 . 98
+

.(4).
2 4 27
We obtain the value of x from the relation x=y+z; thus

}
ir2 q?

93 ?
+ V +
+

+ .....(5) . 4 27

2

27 S 4

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r

V

The above solution is generally known as Cardan's Solution, as it was first published by him in the Ars Magna, in 1545. Cardan obtained the solution from Tartaglia; but the solution of the cubic seems to have been due originally to Scipio Ferreo, about 1505. In this solution we assume x=y+z, and from (2) find

9, hence to solve a cubic equation of the form x8+2x+r=0

Зу
we substitute y-

q

for x. Зу

Example. Solve the equation 23 – 15x = 126. -15

5 or y +

for x, then Зу

y

Put y

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or

125 43+

=126,

43 whence

y6 — 126y3= -125.

.. y3=125,

.: y=5. 5 But x=y+2=6.

y Dividing the given equation 23 — 15x — 126=0 by x-6, we obtain the depressed equation

2c2 +6x+21=0,

the roots of which are – 3+2V--3, and —3—2V-3.

Thus the roots of 23 — 15x=126 are 6, –3+2V-3, and 43 -2V-3.

EXAMPLES L. k. Solve the following equations: 1. 23–18x=35.

2. 23+72x – 1720=0. 3. 23 +63x - 316=0.

4. 23+213 +342=0. 5. 28x} — 932+1=0.

6. 23 — 15x2 – 33x+847=0. 7, 2x3 + 3x2 + 3x+1=0. 8. 203-6x2 + 3x – 18=0. 9. 8x3 – 36x+27=0.

10. 203 - 15x —4=0.

INCOMMENSURABLE Roots. 590. The incommensurable roots of an equation cannot be found exactly. If, however, a sufficient number of the initial figures of the root have been found to distinguish it from the other roots we may carry the approximation to the exact value to any required degree of accuracy by a method first published in 1819 by W. G. Horner.

HORNER'S METHOD OF APPROXIMATION.
591. Let it be required to solve the equation
23 – 3x2 – 2x+5=1....

....(1) By Sturm's Theorem there are 3 real roots and one of them lies between and 2; we will find its value to four places of decimals, which will sufficiently illustrate the method.

Diminishing the roots of the equation by 1 [Arts. 554, 555), we have 1 -3 -2

11 1

-4 -2

1 1

+5

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+1

The transformed equation is
y3 5y +1=0..

(2) Equation (1) has a root between 1 and 2. The roots of equation (2) are each less by 1 than those of equation (1); hence equation (2) has a root between 0 and 1. This root being less than unity the higher powers of y are each less than y. Neglecting them, we obtain an appropriate value of y

from

- 5y+1=0, or y=2. Diminishing the roots of (2), the first transformed equation, by ·2, we have 1 +0 -5

1:2 .04

.992 -4.96

·008 •08

-4.88
2

.6
The transformed equation is
23 + .622 -4.88z+.008 ..

(3). Equation (2) has a root between •2 and 3; the roots of equation (3) are less by 2 than those of equation (2); hence equation (3) has a root between 0 and ·1. Neglecting the higher powers pf z, we have

- 4.88z+.008=0, or z=:001.

Glülisliott

+.6

+.008

Diminishing the roots of (3), the second transformed equation, by .001, we have 1 -4.88

1.001 ·001

•000601 - .004879399 .601 -4.879399 .003120601 ·001

.000602 602 -4.878797 ·001

.603 The transformed equation is

23+ •603v2 —4.8787970+ •003120601..........(4). Equation (3) has a root between .001 and •002; the roots of equation (4) are less than those of equation (3) by.001; therefore equation (4) has a root between 0 and ·001. Solving

-4.878797v+.003120601=0, we have v=·0006. Diminishing the roots of (4), the third transformed equation, by :0006, we find this to be the correct figure for the fourth decimal place; hence the value of x to four places of decimals is 1.2016.

Denoting the coefficients of successive transformed equations by (A), (B), (C), etc., the work is more compactly arranged thus: 1 -3 -2

11.2016
1
-2

-4
-2
-4

(A) 1
-1

.992
- 1
(A) - 5

(B) ·008
·04

.004879399 (A)

-4.96 (C) .003120601

.08 (B) – 4.88 .2

.000601 -4.879399

.000602
(B) :6 (C) -4.878797

.001
.601
·001
.602

·001
(C) •603

+5

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