56. DES CARTES' RULE FOR BIQUADRATIC EQUATIONS'. RULE 1. Take away the second term from the given equation, (Art. 37.) and it will be reduced to this form, x++ax2 + bx +c=0; wherein the coefficients a, b, and c, may represent any quantities whatever, either positive or negative. II. Assume the product x2+px+q.x2+rx+s equal to the transformed equation x1 +ax2 + bx+c=o, and let the two factors be actually multiplied together; then will the product • Lewis Ferrari, the friend and pupil of the celebrated Cardan, was the first who discovered a rule for the solution of biquadratics; namely, about the year 1540. His rule, which is called the Italian method, was first published by Cardan with a demonstration, and likewise its application to a great variety of suitable examples: it proceeds on a very general principle, completing one side of the equation up to a square by the help of multiples, or parts of its own terms, and an assumed unknown quantity; the other side is then made a square, by assuming the product of its first and third terms equal to the square of half the second: then by means of a cubic equation, and other circumstances, the management of which greatly depends on the skill and judgment of the operator, the root is found. The rule we have given above was invented by that eminent French philosopher and mathematician, René Des Cartes, whose name it bears; and was first published in his Geometry, lib. 3. in 1637, but without any investigation: like Ferrari's method, it requires the intervention of a cubic and two quadratics; both methods are sufficiently laborious, but that of Des Cartes has in some respects the preference. The reason of the rule is extremely obvious; for it is plain that any biquadratic may be considered as the product of two quadratics; and if the coefficients of the terms of these latter can be found in terms of a, b, c, &c. the coefficients of the transformed biquadratic, (as we have shewn they can by means of a cubic, &c.) then those quadratics being solved, their roots will evidently be those of the transformed biquadratic, from whence the roots of the given equation will be known. All the roots of a complete biquadratic equation will be real and unequal. First, when of the square of the coefficient of the second term is greater than the product of the coefficients of the first and third terms. Secondly, when the square of the coefficient of the fourth term is greater than the product of the coefficients of the third and fifth terms. Thirdly, when the square of the coefficient of the third term is greater than the product of the coefficients of the second and fourth terms: in all other cases besides these three, the complete biquadratic equation will have imaginary roots. up2 +p1 b2 p2 of the last but one, subtract t or (since qsc) 4c which equation reduced, is po +2 ap1 + 3 = ........ _4c.p2=b2, from the solution of which (by Cardan's rule or otherwise) the value of p will be found. V. Having discovered p, the value of that of α p2 b it the root of will likewise be thence determined; that is, (since r-p,) all the quantities in the two assumed factors x2+px+q.x2+rx+s, except the value of x, are known. -32 VI. Next, find the roots of the two assumed quadratics x2 + px+q=0, and x2+rx+s=o, and we shall have, from the for-944-39 roots of the transformed biquadratic equation x2+ax2 + bx + c -s. Wherefore the four and -2- --q; the roots of the proposed equation. EXAMPLES.-1. To find the four roots of the biquadratic z1-4z3-8z+32=0. e the N To find the root of this equation by Cardan's rule, (Art. Wherefore the two quadratics to be solved, viz. x2+px+y =0, and x2+rx+s=0, (by substituting the above values of p, q, 7, and s,) become x2+4x=-7, and x2. 4x=- -3; of the former of these are x=−2±√−3; and of the latter, x=3, the two roots We have the solution of both these quadratics (or rather their answers) which the values of p, q, and s, being substituted, the roots of the transformed III. Make the coefficients of the same power of ≈ on each side this equation equal to each other, in order to find the values of the assumed coefficients p, q, r, and s; then will p+r=o, s+q+pr=a, ps+qr=b, and qs=c; from the first of these we get r-p, from the second s+q=(a—pr=since r= -p) a+p2, and from the third s =음. IV. From the square of the last but one, subtract the square of the last, and 4 qs=a2+2 ap2 +p1. or (since qs=c) 4 c b2 which equation reduced, is p° +2 ap1 + a2-4 c.p2=b', from the solution of which (by Cardan's rule or otherwise) the value of p will be found. α p2 b V. Having discovered p, the value of s=- + and 2 2 2p that of will likewise be thence determined; that α p2 b is, (since r-p,) all the quantities in the two assumed factors x2+px+q.x2+rx+s, except the value of x, are known. VI. Next, find the roots of the two assumed quadratics x2+ px+q=0, and x2+rx+s=o, and we shall have, from the forq; and from the latter, x=(-7± mer, x=-. Р p2 + -s, or since r=—p=)? ± -s. Wherefore the four 2 4 roots of the transformed biquadratic equation x+ax + bx+c EXAMPLES.-1. To find the four roots of the biquadratic 2-4 z3-8z+32=0. First, to take away the second term, (Art. 37.) let z=x+ 1; then will z*=x+4x3 + 6x2+ 4x+1 Here, putting a=—6, b=−16, and c=+21, the assumed cubic (po +2 ap1+a2 —4c.p2=b2) becomes by substitution po. 12 p*-48 p2=256. From this, let the second term be taken away, by putting p2=y+4; then will To find the root of this equation by Cardan's rule, (Art. Wherefore the two quadratics to be solved 1, viz. x2 +px+q =0, and x2+rx+s=0, (by substituting the above values of p, q, , and s,) become x2+4x=-7, and x2-4x=— -3; the two roots of the former of these are x=−2± √−3; and of the latter, x=3, We have the solution of both these quadratics (or rather their answers) which the values of p, q, and s, being substituted, the roots of the transformed |