III. Solution of Biquadratic Equations. Let the proposed biquadratic be x2 + 2px3 =qx2 + rx + s. Now (22+ px+n)2 = x2 + 2px3 + (p2+2n)x2 + 2pnx+n2: if therefore (p2 + 2n) x2 + 2pnx + n2 be added to both sides of the proposed biquadratic, the first will become a complete square (x2 + px + n)2, and the latter part (p2 + 2n + q)x* + (2pn + r)x + n2 + s, is a complete square if 4(p2 + 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 r2 = 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 + n2+s; then extracting the square root on both sides x2+px+n=° {√(p2+2n+q)x+√√(n2+s) when 2pn+r is positive; or.x2+px+n={√(p2+2n+q)x(n2+s) { when 2pm+r is negative. 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 x2 + px. 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 pro. posed 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 root of the last term, multiply the sum by r2, from the product take the third term, and add the remainder to both sides of the biquadratics. * This rule, for solving biquadratics, by conceiving each to be the difference of twosquares, is frequently ascribed to Dr. Waring; but its original inventor was Mr. Thomas Simpson, formerly Professor of Mathematics in the Royal Military Academy. root 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. Let x-ax2-bx - c = 0, be the given biquadratic equation wanting the second term. Take f = a, g = a2 + 4c, and h = b2, or √h = b; with which values of f, g, h, form the cubic equation, z3 - fz2 + 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. 4. x = -√p-q + √r.x=-√P-√q-√r. 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 x + ax + 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 4ab + 8c, 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. 7 Er. 1. Find the roots of the equation x2+x by tables of sines and tangents. Here p = 449 7 1695 12716 88 44 1695 12716 and the equation agrees with the 1695 1st form. Also tan A = 12716 In logarithms thus : 1695 ,and x = tan A = 12716 Log 1695 = 3.2291697 Arith. com. log 12716 = 5.8956495 sum + 10 = 19.1248192 half sum = 9.5624096 log 88 = 1.9444827 Arith. com. log 7 = 9.1549020 10 = log tan A = 10.6617943 = log tan 77°42′31′′; log tan A = 9-9061115 = log tan 38°51′15′′7; log q, as above sum - 10 = log x = 9.5624096 1.4685211 = log 2941176. 5 17 This value of x, viz 2941176, is nearly equal to To find 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 nume 1 rator is unity: thus is the fraction found to be exactly, and this is manifestly equal to. As to the other root of the equation, it is equal to 17 339 Ex. 2. Find the roots of the cubic equation the second term is negative, and 4p3 > 27q2: so that the example falls under the irreducible case. Hence, sin 3A = The logarithmic computation is subjoined. half sum = 0.0429026 const. log. Arith. com. const. log = 9.9570974 log 414 Arith. com. log 403 = 2.6170003 = 7.3946950 1. sum - 10 = log x= - 1.6320232=log 4285714=log. Log sin (60°-A) = 9.7810061 const. log. = 0.0429026 .. 2. sum - 10 = log x= - 1.8239087=log 6666666=log. Log sin (60°+A) = 9.9966060 const. log 3. sum - 10 = log - x = 0.0395086=log1.095238=log. So that the three roots are, 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 x 60x Here a = 25, b = - 60, and c = 36; therefore = 25 = 625 Consequently the cubic equation will be 25 769 The three roots of which are 2= 9 225 225 25 25x2 + = p, and z = 4 = q, and z = = r; 4 the square roots of these are p = 1⁄2, √ q = 2 ort, √r= Hence, as the value of tb is negative, the four roots are Ex. 4. Produce a quadratic equation whose roots shall be and. Ans. x2 -x+= 0. Ex. 5. Produce a cubic equation whose roots shall be 2, E.r. 6. roots 1, 4, Produce a biquadratic which shall have for the -5, and 6 respectively. 120 = 0. Ans. x4 6x3 21.x2 + 146x Ex. 7. Find x, when r2 + 347x = 22110. Ans. x = 55, x = -402. Ex. 8. Find the roots of the quadratic 2 Ex. 9. Solve the equation r2 55 325 6 65 12 139 25 Ex. 10. Given x2 - 24113x = - 481860, to find r. Ans. x = 20, x = 24093. Ex. 11. Find the roots of the equation x3 - 3x − 1 = 0. Ans. the roots are sin 70°, sin 50°, and sin 10°, to a radius = 2; or the roots are twice the sines of those arcs as given in the tables. Ex. 12. Find the real root of x3 x-6=0. Ans. 3 x sec 54° 44′20′′. Ex. 13. Find the real root of 25x3 + 75x Ex. 14. Given x4 8x3 find x by quadratics. 7. 46 = 0. Ans. 2 cot 74°27′ 48′′. 12x2 + 84x 63 = 0, to Ans. x = 2 + √7±√11+√٦٠ Ex. 15. Given r+ +36x3 - 400x2 - 3168x +7744=0, to find x, by quadratics. Ex. 16. Given r2 + 24x3 - 114x2 - 24x + 1 = 0, to find r. Ans. x = ± √197 – 14, x = 2 ± √5. Ans. x = 11+√209. Ex. 17. Find x, when x4 - 12x - 5 = 0. Ans. x = 1 ± √2, x = −1±2-1. Ex. 18. Find a', when x4 - 12x3 +47x2-72x+36=0 Ans. x = 1, or 2, or 3, or 6. Ex. 19. Given x5 - 5ax4 - 80a2x3-68a3x2+7a+x+a=0, to find r. a, x = 6a ± a√37, x = ±a/10-3a. CHAPTER |