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III. Solution of Biquadratic Equations.

Let the proposed biquadratic be r* + 2px3 = qx2 + rx + s. Now (22+px + n)2 = x2 + 2px3 + (p2+2n)x2 +2pnx+n2: if therefore (p + 2n) x2 + 2pnx + n2 be added to both sides of the proposed biquadratic, the first will become a complete square (x2+ pr+n), and the latter part (p2 + 2n + g) x2 + (2pm + r)x + n2+s, is a complete square if 4(p + 2n +q)·(n2+s)=2pn + r2; that is, multiplying and arranging the terms according to the dimensions of n, if 8n3 +4qn2 + (8s - 4rp)n + 4qs + 4p2s — 2 = 0. From this equation let a value of n be obtained, and substituted in the equation (x2 + px + n)2 = (p2 + 2n + q).x2 + (2pn + r)x + n3 + s; then extracting the square root on both sides

x2+px+n=± {√(p2+2n+q)x‍+√(n2‍+s) { when 2pn+r is positive; orx2+px+n=± {} √(p2+2n+q).x— √(n2+s) { “is negative. when 2pm+r And from these two quadratics, the four roots of the given biquadratic may be determined*.

Note. Whenever, by taking away the second term of a biquadratic, after the manner described in cor. th. 3, the fourth term also vanishes, the roots may immediately be obtained by the solution of a quadratic only.

A biquadratic may also be solved independently of cubics, in the following cases:

1. When the difference between the coefficient of the third term, and the square of half that of the second term, is equal to the coefficient of the fourth term, divided by half that of the second. Then if p be the coefficient of the second term, the equation will be reduced to a quadratic by dividing it by ripx.

2. When the last term is negative, and equal to the square of the coefficient of the fourth term divided by 4 times that of the third term, minus the square of that of the second: then to complete the square, subtract the terms of the proposed biquadratic from (x2± px)2, and add the remainder to both its sides.

3. When the coefficient of the fourth term divided by that of the second term, gives for a quotient the square root of the last term: then to complete the square, add the square of half the coefficient of the second term, to twice the square

This rule, for solving biquadratics, by conceiving each to be the difference of two squares, is frequently ascribed to Dr. Waring; but its original inventor was Mr. Thomas Simpson, formerly Professor of Mathematics in the Royal Military Academy.

root

root of the last term, multiply the sum by 2, from the product take the third term, and add the remainder to both sides of the biquadratics.

4. The fourth term will be made to go out by the usual operation for taking away the second term, when the difference between the cube of half the coefficient of the second term and half the product of the coefficients of the second and third term, is equal to the coefficient of the fourth term.

IV. Euler's Rule for the Solution of Biquadratics.

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Let x-ax2- bx c = 0, be the given bìquadratic equation wanting the second term. Take ƒ = 4a, g = +α2 + 1c, and hb, or h= b; with which values of f, g, h, form the cubic equation, z3 —ƒz2 + gz - h = 0. Find the roots of this cubic equation, and let them be called p, q, r. Then shall the four roots of the proposed biquadratic be these following: viz.

When bis positive.

1. x =

2. x =

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3. x =−√p + √ q = √r. 4. x = √p -√q+√r.

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Note 1. In any biquadratic equation having all its terms, if of the square of the coefficient of the 2d term be greater than the product of the coefficients of the 1st and 3d terms, or of the square of the coefficient of the 4th term be greater than the product of the coefficients of the 3d and 5th terms, or of the square of the coefficient of the 3d term greater than the product of the coefficients of the 2d and 4th terms; then all the roots of that equation will be real and unequal: but if either of the said parts of those squares be less than either of those products, the equation will have imaginary

roots.

2. In a biquadratic + + ax3 + bx2 + cx + d = 0, of which two roots are impossible, and d an affirmative quantity, then the two possible roots will be both negative, or both affirmative, according as a3 - 4ab8c, is an affirmative or a negative quantity, if the signs of the coefficients a, b, c, d, are neither all affirmative, nor alternately and +*.

Various general rules for the solution of equations have been given by Demoivre, Bezout, Lagrange, &c; but the most universal in their application are approximating rules, of which a very simple and useful one is given in our first volume.

EXAMPLES.

EXAMPLES.

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Er. 1. Find the roots of the equation x2 +

by tables of sines and tangents.

449=

1695

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and the equation agrees with the

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10=

1.9444827

= 9.1549020

log tan A = 10.6617943 = log tan 77°42′31′′;

log tan A

= 9.9061115= log tan 38°51'15"};

log 9, as above =

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9.5624096

sum 10 = log x = 1·4685211 = log 2941176. This value of x, viz •2941176, is nearly equal to 5. To find

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whether that is the exact root, take the arithmetical compliment of the last logarithm, viz 0·5314379, and consider it as the logarithm of the denominator of a fraction whose numerator is unity: thus is the fraction found to be exactly,

5

3.4

and this is manifestly equal to. As to the other root of

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Ex. 2. Find the roots of the cubic equation

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4p3 > 27q2: so that the example falls under the irreducible case.

403

46

=

441

1479

the second term is negative, and

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The logarithmic computation is subjoined.
Log 1612 3.2073650

Arith. com. log 1323 = 6·8784402

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half sum = 0·0429026 const. log.

Arith. com. const. log =9-9570974

log 414 ..

Arith. com. log 403

log sin 3A.

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= 2.6170003

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x=

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Log sin (60°+A)
const. log.

3. sum

1.8239087=log 6666666=log.

9.9966060

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= 0.0429026

10= log -x=0·0395086= =log1·095238=log}}.

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So that the three roots are 4, 3, and ; of which the first two are together equal to the third with its sign changed, as they ought to be.

Ex. 3. Find the roots of the biquadratic 4 — 25x2 + 360, by Euler's rule.

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9

2= = p, and z = 4 = 9, and z = =r;

4

25

4

the square roots of these are p = 3, √ q = 2 or 4, √√r=3. Hence, as the value of +b is negative, the four roots are

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Produce a quadratic equation whose roots shall be

Ans. x + 3 = 0.
x2
+3=0.

Produce a cubic equation whose roots shall be 2,

11x+30= 0.

Ans. x3.

4x2

Er. 6. roots 1, 4,

Ex. 7.

Produce a biquadratic which shall have for the 5, and 6 respectively.

Ans. x+ 6x3 21x146x 120 0. Find x, when x2 + 347x = 22110.

Ex. 8.

Ans. x 55, x= - 402. Find the roots of the quadratic 2

55

325

=

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sin 50°, and

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Ex. 11. Find the roots of the equation x3- 3.x — 1=0. Ans. the roots are sin 70°, sin 10°, to a radius 2; or the roots are twice the sines of those arcs as

given in the tables.

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Ex. 13. Find the real root of 25x3+ 75x

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Ans. 2 cot 74° 27′ 48′′.

12x2 + 84x 63 = 0, to

Ans. x2 + √7± √11+√7.

Ex. 15. Given xa +36x3 — 400x2 — 3168x + 7744≈0,

to find x, by quadratics.

Ans. 11+√209.

Given 4+24x3- 114x2 - 24x+10, to

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Er. 19. Given xs-5ax4-80a2x3-68a3x2+7a*x+a3=0, to find x.

Ans. xa, x = 6a ±a√√37, x = ± a√/10-3a.

CHAPTER

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