of cubic equations as being reduced, by the preceding pro cess, to the solution of an equation of the form w3. 2 27 w 2 = 3 27' (a), since the right members of (1′′), (3′′), (5′′), etc., as their number is increased, continually approximate to as their limit; and it is clear, when their number is infinitely great, that their right members will differ from by a difference 2 27 which will be less than any given difference. Now, although (1''), (3′), (5′′), etc., can not in practice be continued to infinity, yet they can easily be continued so far as to enable us to find a to any required degree of exactness; since, if they are continued until an equation is found whose right member agrees with to n decimal places, it is clear that we 2 2 27 may put the greatest root of (a) for the greatest root of the 3 equation thus obtained, and that we can thence find a correctly to at least n-1 decimal places; and it is clear that the value of a thus found will (generally) be the numerically greatest root of (1′). If r stands for the root of (1'), thus found, then, reducing (1) to the form a ax-b0, we shall, by division, get 23 ax b = x2 + rx + p2 — a + ly, since ar b= 0, to n decimal places, if we put x2 + rx + p2 — a equal to 0, we shall get the quadratic equation + rx + r2 — a = 0, whose solution gives a = √4a-3r 32 2 2 x (b), which gives the two remaining roots of (1) when the values of a and r are substituted for them in the equation, these roots being thus correctly found to at least n 1 decimal places. To illustrate what has been done, take the following CUBIC EQUATIONS. 10.70710678, and (1′′) and (2′′) give V3 2 27 27 27 IX IX V V it is clear that we may put =2; consequently, from ⇓ = 2 3 VIII 1.99999442 by putting for V, we get 3V = 1.99999721. VIII 3 Similarly 3V = gives 3V = 1.99999023, and so VII 3V 1 which is one of the roots of the given equation, correctly found to five (and very nearly to six) decimal places. Substituting 6 for a and 2.601679 for r in (b), we easily get – 2.261802 and x = — 0.339876 for the remaining roots of the given equation; these roots being correctly found to five decimal places. (See Vol. I., p. 155 of Bonnycastle's Algebra, Question 4.) 2. To find the roots of 2-3x=1 to three places of have 1.879 for one of the roots of the given equation; correctly found to three decimal places. Putting 3 for a and 1.879 for r in (b), we easily get x=1.532 and x = — 0.347 for the first figures of the two remaining roots of the given equation. 3. To find the roots of x3 7ас figures. Here we have a = 7, 1.96396049; and hence V3 — 1.96396049 = 7 to four places of V V = 7 3 27 x= 3.048 for one of the roots of the given equation, which is correct in all its figures. It is hence easy, from (b), to BIQUADRATIC EQUATIONS. find 1.692 and 1.356 for the first figures of the two remaining roots of the given equation. (3.) Hence it follows that we have completed the solution of cubic equations. For those which do not fall under the irreducible case can be solved by the formulæ of Cardan, and those which fall under the irreducible case can be solved to any degree of exactness by the preceding method. SOLUTION OF BIQUADRATIC EQUATIONS. (1.) In accordance with the definitions that have been given, we may clearly take y* + Ay3 + By2 + Cy = D, (1), to represent a biquadratic equation. A If we put y = x it is easy to perceive that (1) will be reduced to the more simple form a + ax2 + bx + c = 0, (2). (2.) Putting x2 = mx + n, (3), and substituting this value of a2 in (2), it will become (mx + n)2 + a(mx + n) + bx + c = m2x2 + 2mnx + n2 + amx + an + bx + c = m2x2 + (2mn + am + b)x + n2 + an + c = 0; and if in this we put mx + n for a2, it will become m2(mx + n) + (2mn + am + b)x + n2 + an + c = (m3 + 2mn + am + b)x + n2 + m2n + an + c = 0; in which m and n are undetermined. If we determine m and n on the supposition that the coefficient of x is 0 (so as to leave a arbitrary), we shall get the equations m3 + 2mn + am +b=0 and n2 + m3n + an + c = 0. From the first of these equations we easily get_n= m3 + am + b which being put for n in the second equation, 2m reduces it to (m2 + a)(— +c= ・m3 + am + b [m3 (b + ma + m3 2m 2m m3 + am + b' (m3 + am + 2m (m2 + a)] ]x[ + (m3 • m23 + am + b' 2m 2m +c= (b − (ma + m2)) + c = 0, or we shall have 2m b2 — (ma + m3)2 + 4cm2 = 0, which is easily reduced to the equation m + 2am1 + (a2 — 4c)m2 = b2. If we put m2 = 2, (4), the preceding equation becomes ≈3 + 2az2 + (a2 — 4c)z = b2, (5), which agrees with the result obtained in a very different manner by Euler. (See Bonnycastle's Algebra, Vol. I., p. 161.) (3.) If we substitute the value of n in (3), it will become Hence, the solution of the biquadratic equation (2) is reduced to that of the cubic equation (5); for having found z from (5), we get m from (4), and then a can be found from (6); consequently, since a cubic equation can either be exactly solved, or solved to any required degree of exactness, it follows that we have completed the solution of biquadratic equations. (4.) To illustrate what has been done, take the following EXAMPLES. 1. To find the roots of a ·27x2 +14x + 120 = 0. Since a, b, and c are represented by 27, 14, and 120, (5) becomes 23 54z2 + 2492 = 196, one of whose roots is easily seen to be 1. z Substituting 1 for 2 in (4), we get m2 = 1, whose roots give m= 1 and m = - 1. Putting 1 and 1 separately for m in (6), we easily get 3, 2, 4, and 5 for the roots of the proposed equation. Here a, b, and c being expressed by 17, 20, and −6, (5) becomes 23-342 + 3132 3423132 = 400; and it is easy to see that 16 is one of the roots of this equation. Putting 16 for z in (4), we get m = 4 and m = 4 for the corresponding values of m. Substituting 4 and -4 separately for m in (6), we get 2+ √7, 2−√7, −2+12, and 2-12 for the roots of the proposed equation. (See Question 4, p. 162, Vol I., of Bonny castle's Algebra. 3. To find the roots of y8y' + 20y3 - 19y + 12 = 0. Putting y=x+ 2 in the given equation, it becomes a1 — 4x2 - 3x+6= 0, which wants its second term. |