Grade 12Algebra
Solve 6x^{6}-25x^{5}+31x^{4}-31x^{2}+25x-6=0 of theory of equations
1 Answer
Arun
Dear student
6x^6 - 25x^5 + 31x^4 - 31x^2 +25x - 6 = 0
6x^6-6 -25x^5+25x+31x^4-31x^2=0
6(x^6-1)-25x(x^4-1)+31x^2(x^2-1)=0
6(x^2-1)(x^4+x^2+1)-25x(x^2-1)(x^2+1)+...
(x^2-1){6x^4+6x^2+6-25x^3-25x+31x^2}=0
(x^2-1){6x^4-25x^3+37x^2-25x+6}=0
i.e, either, x^2-1=0, 0r, (6x^4-25x^3+37x^2-25x+6)=0
x^2=1 gives roots x=1,-1 the other roots can be found out by the other part equating to zero, which is likely to return imaginary roots
6x^6-6 -25x^5+25x+31x^4-31x^2=0
6(x^6-1)-25x(x^4-1)+31x^2(x^2-1)=0
6(x^2-1)(x^4+x^2+1)-25x(x^2-1)(x^2+1)+...
(x^2-1){6x^4+6x^2+6-25x^3-25x+31x^2}=0
(x^2-1){6x^4-25x^3+37x^2-25x+6}=0
i.e, either, x^2-1=0, 0r, (6x^4-25x^3+37x^2-25x+6)=0
x^2=1 gives roots x=1,-1 the other roots can be found out by the other part equating to zero, which is likely to return imaginary roots
Regards
Arun (askIITians forum expert)
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