PROBLEM VI. To find the roots of biquadratic equations, according to the method of DES CARTES. RULE. 1. Take away the second term of the equation by problem 2, and it will be reduced to the form x4+qx2+ rx+8=0. 2. From the cubic equation y3+2qy2+(q2—48)y-r2 =0 take the second term, and find the value of y by the last problem. e2 3. Let e be assumed =√y,ƒ=}7+} 2——, and g= 2e' 4. Find the roots of the two quadratic equations x2+ ex+f=0, and x2. ex+g=0, and they will be the four roots of the biquadratic required. Investigation of the rule. Let the given equation x4+qx2+rx +so be equal to the product of the two quadratic equations x2+ •x+f=0, and x2—ex+8=0=x1+(ƒ+,8—e2)x2+(eg—ef )x+ƒ8 =0. Then, by equating the homologous terms, we shall have ƒ+8—e2=q, eg—ef=r, and fg=s; and therefore f=17+}e2. 19+1e2+ and s=fxg=1q2+ £qe2+ !, e4— 2e r2 4e2* r 8 And from this last equation we shall have e6+2qe1+(92-4s) e2r2 to a cubic equation, in which the value of e may be found, as in the last problem. 1 EXAMPLES. 1. Let z4-423-8z+32-0 be the equation proposed, in which it is required to find the value of z. 1. To take away the second term, let x+1=z; then 24=x4+4x3 + 6x2+ 4x+ 1 x46x216x+21 =0, or y3-12y2-48y-256-0, for the cubic equation. 2. To take away the second term from this equation, let +4=y; then 2e But ƒ (=19+1e2——) and g (=19+ ! e2 + Ze) 8 (=19+1e2+2) are also known; and therefore the roots of the quadratic equations x2+ex+f=0, and x2—ex+8=o, may be determined, and are the four roots of the biquadratic equation required. Q. E. I. Note. The co-efficient of x is put equal to e, in both the equa tions, because, when the second term is wanting, the sum of the positive roots is always equal to the sum of the negative ones, with a contrary sign. This rule has sometimes been ascribed to Des Cartes, and sometimes to Bombelli, an Italian; but the original inventor of it was Louis Ferrari. 3. To find the value of p, by CARDAN's rule for cubic 16 16 8 12; and therefore y=16, or y=4, f= +==7, and g= f=7, and g=3. + 2 2 In the method of Des Cartes, above explained, all biquadratic equations are supposed to be generated from the multiplication of two quadratic ones: but, according to the following way, which is taken from Simpson's Algebra, every such equation is conceived to arise by taking the difference of two complete squares. Here, the general equation x4+ax3+bx2+cx+d=o being pro. posed, we are to assume (x2+ax+A)2—(BX+C)2=x4+ax3+cx +d; in which A, B, and C, represent unknown quantities to be determined. Then x2+a2+A, and вx+c being actually involved, will give x4+ax3+2Ax2 +‡a2x2+a^x+a2С=xa+ax3+bx2 +cx+d: from whence, by equating the homologous terms, we shall Let now the first and last of these equations be multiplied together, and the product will, evidently, be equal to 4 of the square of the second; that is, 2A3+(4a2—b)×â2—Qdî—({ a2—b)×d=(B2c2) Xa2A2-2αca+c2. M 4. To find the roots of the two quadratic equations x2+ex+f=0, and x2-ex+g=0. x2+ex+f=x2+4x+7=0. x2 -ex+g=x2—4x+3=0. In the first of these x=−2+√—3, or —2—√√/—3. And in the second x=3, and 1. Therefore, 3, 1, −2+√—3, or —2——3, are the four roots of the equation x4—6x2—16x+21=0. And if unity be added to each of them, we shall have 4, 2, −1+√—3, and —1——3, for the roots of z4-4z3—8z+32=0, the equation proposed; the two last of which are impossible. 2. Given x4+2x3-7x2-8x+12=0, to find the values of x. Ans. x=1, 2, -3, and -2. Whence, by denoting the given quantities ac—d, and {c2+d× (4a2-b) by k and l, respectively, there will arise this cubic equation, A3— ba2+ka—=0; by means of which the value of a may be determined; and therefore, from the preceding equations, both B and c will also be known; в being found from thence ✔✅✔✅(2A+ 4a2-6), and c 2B The several values of A, B, and c, being thus found, that of x will also be readily obtained: for (x2+}ax+A)2—(BX+C)2 being universally, in all circumstances of x, equal to x4+ax3+bx2+ca +d, it is evident, that, when the value of x is taken such that the latter of these expressions becomes equal to nothing, the former must likewise be equal to nothing; and consequently (x2+(ax+A)2= (Bx+c)2. And therefore, by extracting the square root of both sides of the equation, we shall have x2+ax+A+BX±C; oг x=±в— a±[ ( a + 3B)2 ±c—A]2±±вα±(1α2±¡αB+\в2±C—A)?, which exhibits all the different roots of the given equation, according to the variation of the signs. 3. Given x4-25 x2+60x-36=0, to find the values of x. Ans. 1, 2, 3, and -6. of 4. Given y4-8y3+14y2+4y—8=0, to find the values y. Ans. y=3+√5, 3−−√5, 1+√3, and 1−√√3. 5. Given x4+12x-17=0, to find the values of x. Ans, x=}√2±√(−3√2—§), and also —§√2± √(3√2—1). 4x3-3x2-4x+1=0, to find the va 7. Given x4-10x3 +35x2—50x +24=0, to find the values of x. Ans. 1, 2, 3, and 4. This method will be found to have many advantages over that given above. In the first place, there is no necessity for being at the trouble of exterminating the second term of the equation, in order to prepare it for a solution: secondly, the equation a3—¿ba2+ A-0, here brought out, is of a more simple form than that derived from the former method: and, thirdly, the value of a in this equation will always be commensurate and rational; not only when all the roots of the given equation are commensurate, but also when they are irrational, and even impossible. EXAMPLE. Let there be given x4+12x—17=0, to find the value of x. Here, by comparing this with the general equation x4+ax3 + bx +cx+do, we shall have ao, bo, c=12, and d-17; and therefore kac—d—17, l=c2+d×(4a2—b)=36, and 13— 1ba2+ka-1=A3+17A—18=0. And, from this equation, a will be found equal to 1; and there fore B =(2A+‡a2—b)3=✔✅✓/2, c= aА-c -12 2B 2/2 -3/2, and |