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} = a ^{4} + b^{4} + c^{4}+ 6a ^{2}b^{2} + 6a^{2}c^{2} + 6b^{2}c^{2}+ 4a ^{3}b + 4a^{3}c + 4b^{3}a + 4b^{3}c + 4c^{3}a + 4c^{3}b+ 12a ^{2}bc + 12ab^{2}c + 12abc^{2}= 2(a ^{2} + b^{2} + c^{2}) (a + b + c)^{2} + 8abc(a + b + c) −[a^{4} + b^{4} + c^{4} − 2(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})]-------(i)The fourth degree equation look like this a'x ^{4} + b'x^{3} + c'x^{2} + d'x + e' = 0which can be reduced to y ^{4} + py^{2} + qy + r = 0This can be written as y ^{4} = −py^{2} − qy − r As described in the previous post we can assume the solution of y ^{4} + py^{2} + qy + r = 0to be (a + b + c). On comparing it with (i) we get 2(a ^{2} + b^{2} + c^{2}) = −p -----(ii)8abc = −q -----(iii) −[a ^{4} + b^{4} + c^{4} − 2(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})] = −r -----(iv)Solving (iv) by substitution from (ii) [(a ^{2} + b^{2} + c^{2})^{2} − 4(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})] = r(a ^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2}) = [(a^{2} + b^{2} + c^{2})^{2} − r]/4(a ^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2}) = [p^{2}/4 − r]/4(a ^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2}) = [p^{2} − 4r]/16So we get the final three conditions as (a ^{2} + b^{2} + c^{2}) = −p/2 -----(v)abc = −q/8 -----(vi) (a ^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2}) = [p^{2} − 4r]/16 -----(vii)Dividing (vii) by square of (vi) we get (a ^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})/(a^{2}b^{2}c^{2})= 4[p^{2} − 4r]/q^{2}(1/c ^{2} + 1/a^{2} + 1/b^{2}) = 4[p^{2} − 4r]/q^{2} -----(viii)[(b ^{2}+c^{2})/ b^{2}c^{2} + 1/a^{2}]= 4[p^{2} − 4r]/q^{2} Substituting value of b ^{2}c^{2}= q^{2}/64a^{2} and (b^{2}+c^{2}) = −p/2 − a^{2}[(−p/2 − a ^{2})/(q^{2}/64a^{2}) + 1/a^{2}]= 4[p^{2} − 4r]/q^{2} [64a ^{2}(−p/2 − a^{2})/q^{2} + 1/a^{2}]= 4[p^{2} − 4r]/q^{2} [64a ^{4}(−p/2 − a^{2}) + q^{2} ]/q^{2}a^{2}= 4[p^{2} − 4r]/q^{2} [−32pa ^{4} − 64a^{6} + q^{2} ]= 4[p^{2} − 4r]a^{2} 64a ^{6} + 32pa^{4} + 4[p^{2} − 4r]a^{2 }− q^{2} = 0Solve 64a ^{6} + 32pa^{4} + 4[p^{2} − 4r]a^{2 }− q^{2} = 0and substitute in -b'/4a' + a + b + c to get the solution such that they satisfy the required conditions. 64a ^{6} + 32pa^{4} + 4[p^{2} − 4r]a^{2 } − q^{2} = 0a ^{6} + (1/2)pa^{4} + (1/16)[p^{2} − 4r]a^{2 } − (1/64)q^{2} = 0Express a ^{6} + (1/2)pa^{4} + (1/16)[p^{2} − 4r]a^{2 } − (1/64)q^{2} = 0 as y^{3} + ky + l = 0where y = a ^{2} + p/6Then its solutions is a ^{2} = −p/6 + [−l/2 + √(l^{2}/4 + k^{3}/27)]^{1/3} + [−l/2 − √(l^{2}/4 + k^{3}/27)]^{1/3}b ^{2} = −p/6 + ω[−l/2 + √(l^{2}/4 + k^{3}/27)]^{1/3} + ω^{2}[−l/2 − √(l^{2}/4 + k^{3}/27)]^{1/3}c ^{2} = −p/6 + ω^{2}[−l/2 + √(l^{2}/4 + k^{3}/27)]^{1/3} + ω[−l/2 − √(l^{2}/4 + k^{3}/27)]^{1/3}
a'x
k = (1/16)[p^{3} + b'x^{2} + c'x + d' = 0 is equivalent to y ^{3} + [c'/a' − b'^{2} /3a'^{2}]y + [(27a'^{2}d' +2b'^{3} −9a'b'c')/(27a'^{3})] = 0 wherey = x + b'/3a' or y ^{3} + cy + d = 0where c = (c'/a' − b' ^{2}/3a'^{2}) andd = (27a' ^{2}d' +2b'^{3} −9a'b'c')/(27a'^{3})
^{2} − 4r] − p^{2}/12 = −(p^{2} + 12r)/48l = −q ^{2}/64 + p^{3}/108 − p(p^{2} − 4r)/96The above solutions satisfy (v), (vi) and (vii).
Take the positive square roots of a
^{2}, b^{2}and c^{2} as a,b and c and find the solutions of the fourth degree equation asif(q>=0) then R _{1} = -b'/4a'−(a+b+c)R _{2} = -b'/4a'+(a+b−c)R _{3} = -b'/4a'+(a−b+c)R _{4} = -b'/4a'+(−a +b+c)if(q<0) then R _{1} = -b'/4a' -a+b-cR _{2} = -b'/4a' - a-b+cR _{3} = -b'/4a'+a−b-cR _{4} = -b'/4a'+a +b+c
a'x
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.
^{4} + b'x^{3} + c'x^{2} + d'x + e' = 0 is equivalent to y ^{4} + [c'/a' − 3b'^{2} /8a'^{2}]y^{2} + [(16a'^{2}d' +2b'^{3} −8a'b'c')/(16a'^{3})]y + [(16a'b'^{2}c' − 3b'^{4} − 64a'^{2}b'd' + 256e'a'^{3})/(256a'^{4})] = 0 wherey = x + b'/4a' or y ^{3} + py^{2} + qy + r = 0where p = [c'/a' − 3b' ^{2} /8a'^{2}]q = [(16a' ^{2}d' +2b'^{3} −8a'b'c')/(16a'^{3})] and r = [(16a'b' ^{2}c' − 3b'^{4} − 64a'^{2}b'd' + 256e'a'^{3})/(256a'^{4})] |

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