### Quartic Equations (Solution 1)

 When I found the solution of third degree equation I moved to fourth degree equation. The solution is present on sites.google.com/site/rahulsmaths I originally posted it on rashmaths.blogspot.com on 4 November 2014. http://rashmaths.blogspot.in/2014/11/solution-of-fourth-degree-equation.html but it took me about 15 days to verify. I made a program to solve it by my method. It gives the correct solution. In a fourth degree equation we have four constants. We can reduce them to equation with three constants. When we need to solve such kind of equations, we can solve them by finding three equations each representing the constants. So we can assume that the solution is the sum of three variables. Find the three equations and solve them. (a + b + c)4 = a4 + b4 + c4 + 6a2b2 + 6a2c2 + 6b2c2 + 4a3b + 4a3c + 4b3a + 4b3c + 4c3a + 4c3b + 12a2bc + 12ab2c + 12abc2 = 2(a2 + b2 + c2) (a + b + c)2 + 8abc(a + b + c) −[a4 + b4 + c4 − 2(a2b2 + b2c2 + c2a2)]-------(i) The fourth degree equation look like this a'x4 + b'x3 + c'x2 + d'x + e' = 0 which can be reduced to y4 + py2 + qy + r = 0 This can be written as y4 = −py2 − qy − r As described in the previous post we can assume the solution of y4 + py2 + qy + r = 0 to be (a + b + c). On comparing it with (i) we get 2(a2 + b2 + c2) = −p -----(ii) 8abc = −q -----(iii) −[a4 + b4 + c4 − 2(a2b2 + b2c2 + c2a2)] = −r -----(iv) Solving (iv) by substitution from (ii) [(a2 + b2 + c2)2 − 4(a2b2 + b2c2 + c2a2)] = r (a2b2 + b2c2 + c2a2) = [(a2 + b2 + c2)2 − r]/4 (a2b2 + b2c2 + c2a2) = [p2/4 − r]/4 (a2b2 + b2c2 + c2a2) = [p2 − 4r]/16 So we get the final three conditions as (a2 + b2 + c2) = −p/2 -----(v) abc = −q/8 -----(vi) (a2b2 + b2c2 + c2a2) = [p2 − 4r]/16 -----(vii) Dividing (vii) by square of  (vi) we get (a2b2 + b2c2 + c2a2)/(a2b2c2)= 4[p2 − 4r]/q2 (1/c2 + 1/a2 + 1/b2) = 4[p2 − 4r]/q2  -----(viii) [(b2+c2)/ b2c2 + 1/a2]= 4[p2 − 4r]/q2 Substituting value of b2c2= q2/64a2 and (b2+c2) = −p/2 − a2 [(−p/2 − a2)/(q2/64a2) + 1/a2]= 4[p2 − 4r]/q2 [64a2(−p/2 − a2)/q2 + 1/a2]= 4[p2 − 4r]/q2 [64a4(−p/2 − a2) + q2 ]/q2a2= 4[p2 − 4r]/q2 [−32pa4 − 64a6 + q2 ]= 4[p2 − 4r]a2 64a6 + 32pa4 + 4[p2 − 4r]a2 − q2 = 0 Solve 64a6 + 32pa4 + 4[p2 − 4r]a2 − q2 = 0 and substitute in -b'/4a' + a + b + c to get the solution such that  they satisfy the required conditions. 64a6 + 32pa4 + 4[p2 − 4r]a2 − q2 = 0 a6 + (1/2)pa4 + (1/16)[p2 − 4r]a2 − (1/64)q2 = 0 Express a6 + (1/2)pa4 + (1/16)[p2 − 4r]a2 − (1/64)q2 = 0 as y3 + ky + l = 0 where y = a2 + p/6 Then its solutions is a2 = −p/6 + [−l/2 + √(l2/4 + k3/27)]1/3 + [−l/2 − √(l2/4 + k3/27)]1/3 b2 = −p/6 + ω[−l/2 + √(l2/4 + k3/27)]1/3 + ω2[−l/2 − √(l2/4 + k3/27)]1/3 c2 = −p/6 + ω2[−l/2 + √(l2/4 + k3/27)]1/3 + ω[−l/2 − √(l2/4 + k3/27)]1/3 a'x3 + b'x2 + c'x + d' = 0 is equivalent to y3 + [c'/a' − b'2 /3a'2]y + [(27a'2d' +2b'3 −9a'b'c')/(27a'3)] = 0 where y = x + b'/3a' or y3 + cy + d = 0 where c = (c'/a' − b'2/3a'2) and d = (27a'2d' +2b'3 −9a'b'c')/(27a'3) k = (1/16)[p2 − 4r] − p2/12 = −(p2 + 12r)/48 l = −q2/64 + p3/108 − p(p2 − 4r)/96 The above solutions satisfy (v), (vi) and (vii). Take the positive square roots of a2, b2and c2 as a,b and c and find the solutions of the fourth degree equation as if(q>=0) then R1 = -b'/4a'−(a+b+c) R2 = -b'/4a'+(a+b−c) R3 = -b'/4a'+(a−b+c) R4 = -b'/4a'+(−a +b+c) if(q<0) then R1 = -b'/4a' -a+b-c R2 = -b'/4a' - a-b+c R3 = -b'/4a'+a−b-c R4 = -b'/4a'+a +b+c a'x4 + b'x3 + c'x2 + d'x + e' = 0 is equivalent to y4 + [c'/a' − 3b'2 /8a'2]y2 + [(16a'2d' +2b'3 −8a'b'c')/(16a'3)]y + [(16a'b'2c' − 3b'4 − 64a'2b'd' + 256e'a'3)/(256a'4)] = 0 where y = x + b'/4a' or y3 + py2 + qy + r = 0 where p = [c'/a' − 3b'2 /8a'2] q = [(16a'2d' +2b'3 −8a'b'c')/(16a'3)] and r = [(16a'b'2c' − 3b'4 − 64a'2b'd' + 256e'a'3)/(256a'4)] After solving the fourth degree equation I moved to fifth degree equation but was not able to express it an a function of unknowns. Either the solution don't have some powers so I left the idea and took a break from it.