1+ a

Hence if the proposed cubic equation be x3+ a x2 + bx + c = 0, the type of solution will be as in the margin, where the values of a and ẞ are obtained from the coefficients of the transformed equation whose roots are less by r than those of the proposed cubic, and r, a and ß being thus found, the three roots of the equation are readily determined. When the value of ẞ is positive, the roots are all real; and when it is negative, two of the roots are impossible. In the following example the additions are performed mentally, and the results alone are written

down.

+ b + c (r

ra'

rb'

a

b'

0

ra"

a"

r

36

a2

2) a"

1

- 2600 = 0 to find all the roots.

+5

-2600(11·00679933972

The three roots of the given cubic equation are thus found to be

SOLUTION OF EQUATIONS BY DOUBLE POSITION.

130. The roots of equations of all degrees may be determined to any degree of accuracy by the method of Double Position. It is equally applicable to all forms of equations.

Let xa x2 + bx = c be an equation of the third degree, and let s and s' be any two near values of x, found by trials; then if s and s be substituted for x in the given equation, we shall have

x2 + a x2 + bx = c

s3 + a s2 + bs = c'

s13 + as22+bs' = c"

Subtract (3) from (2) and (2) from (1); then we get (s3 − s13) + a (s2 — s12) + b ( s

· s') = c' — c'',

(28) a (x2 - 82) + b (x − s) = c - c',
+

or, (s

(1),

(2),

(3).

− s') { s2 + s s' + s'2 + a (s + s') + b } = c' — c'. . . (4), (x 8) {x2 + xs + s2 + a (x + s) + b } = c − c' ... (5). Now, since the values of x, s, and s' are nearly equal to each other, the bracketted expressions in (4) and (5) may be considered nearly equal to each other, and dividing (5) by (4) on this supposition, we get

s - s'

(c — c') . . . (6).

This expression affords the following rule:

الله

...

Find by trials two numbers nearly equal to the root required, and substitute them for the unknown quantity in the given equation, noting the results that arise from each substitution. Then the difference of these results (c' — c') is to the difference of the assumed numbers (s · s'), as the difference between the true result given by the equation and either of the former (c-c') to the correction of the number belonging to the result used (x s). Apply this correction to the assumed number, according as that number is too little or too great, and an approximate value of the root will be obtained.

With this value and the nearer of the two former values, or with any other values that appear to be more accurate, repeat the operation, and a second approximation to the true value will be obtained; and so on.

EXAMPLES.

1. Given 3+3x2= 500, to find an approximate value of x.

By trials it is found that the value of x is greater than 7, but less than 7.1. Take these as the assumed numbers; then

whence x=

of x.

7·1 — ·047 = 7·053, which is the first approximate value

Taking 7.05 and 7.06 for the assumed numbers, then

.

whence 7·05 + 00255 = 7.05255, which is a near value of x, and correct as far as the last place of decimals inclusive.

Let the root thus found be denoted by r, then x = r, and if this value be substituted for x in the given equation, we have 2+3x2=500 . . . . (1), p3 +3 r2 = 500

Subtracting (2) from (1) gives

or

(7)+3(x22)=0,"

(2).

(x − r) (x2+x r + r2) + 3 (x − r) (x + r) = 0; :. (x − r) {x2 + (r + 3) x + r (r + 3)} = 0 . . Now this equation is fulfilled by making

x2 + (r+3) x + r (r+3)=0;

(3).

we substitute the value of r, viz., 7·05255, the values of the remaining roots may be found.

Find the roots of the following equations:

2. x2+10x2+5x=2600.

3. x3 + x2 + x = 100.

4. 2x3-3x2

4 x 10.

x= 4.

CUBIC EQUATIONS.-CARDAN'S METHOD OF SOLUTION. 131. Let the general form of a cubic equation +p+q x + r 0 be transformed into the form x+ax+b: = = 0, where the second power of the unknown quantity is absent; then if x = y +2, we get, by substitution,

or

(y + z)3 + a (y + z) + b = 0,

y3 + 3y z (y + z) + z3 + a (y + z) + b = 0.

Let 3 y z = a; then the last equation becomes y3 +23 = − b ; whence y+2 y3 z3 + zo = b2 ;

y3 +

Combining this equation with 2b, by addition and subtraction, we get

This method of solution applies only to those cubic equations which have one real root and two impossible ones.

For let h+k, h − ✅ k, and 2h be the three roots of a cubic equation; then (125) that equation will be 3 − (3 h2 + k) x + 2h (h1 - k) = 0. (h1-k) Now if we substitute (3h+k) for a and

2h (ha — k) for b in the expression

terms of the value of x, we get

b2 as

(+) which occurs in both

2

27

+

81

=

Now this expression is always impossible when k is positive, and when k is positive, the roots h+k and h√k are real; and on the contrary, when k is negative, the radical value above is real, and the roots h + √k and hk are impossible. Cardan's method of solution is hence useless in practice when the roots are all real, as they cannot be computed by means of it, on account of the occurrence of an imaginary or impossible quantity, and it is only when the cubic equation has one real root that it is applicable.

BIQUADRATIC EQUATIONS.-SIMPSON'S METHOD OF SOLUTION. 132. Let the general form of a biquadratic equation x + px3 + q x2 +rx+s=0 be transformed into the form x1 + a x2 + bx + c = 0, where the third power of the unknown quantity is absent; then if we

assume

x2+ax2 + bx + c = (x2 + y)2 - (hx + k)2

= x + (2 y h2) x2 - 2 h kx + y2 − k2, and equate the coefficients of the same powers of x on both sides, then 2y-h2 = a, or h2 = 2y

- 2hk = = b, or 2 h k-b y2 - k2 = c, or

- α

k2 y2 - c.

=

(1)
(2)

(3).

Now it is obvious that the square of (2) is equal to four times the pro

duct of (1) and (3); hence

c)

Remove the second term of this cubic, and find the value of y by the previous method; hence h = ± √ (2 y − a) and k = √(y2 will be known, and they must have opposite signs, for 2 hk is a negative quantity by (2). Then the first side 1 + a x2 + bx + c of the first equation being equal to 0, we have

(x2 + y)2= (hx + k)2, or x2 + y = ± (h x + k) ; hence, since y, h, and k are all known, the solution of the quadratics k=0, and x2+hx+y+k=0

These methods may be applied to any of the examples which have already been given in (129) and (130).

INDETERMINATE ANALYSIS.

133. We have seen that, when the conditions of a question furnish as many independent and consistent equations as there are unknown quantities to be determined, the values of these unknowns may be found; but if the number of unknown quantities exceeds the number of equations, the question will admit of various solutions, and it is hence said to be indeterminate. If integer values only are required, the number of solutions will be greatly restricted; and if negative values be excluded, the number of solutions in some cases may be very limited, while in others the number of positive integer solutions is unlimited.

Indeterminate analysis of the first degree.

Indeterminate equations of the first degree are of the form ax+by=c, or ax+by+c z = d,

where a, b, c, d are given whole numbers, and x, y, z are limited to positive integer values. When the formula a x + by = c admits of integer values of x and y, the coefficients a and b cannot have a common divisor which is not also a divisor of c. For if a = md, and b = me, then

c

ax±by=m (dx ± ey) =c, hence dx ±ey= ;

c

m

but d, e, x, y are all integers; therefore must be an integer, and m is

a divisor of c.

m

If a and b are prime to.each other, the solution of an equation of the cis always possible, and the number of solutions