Finding a for x^10 + ax + 1 = 0 with reciprocal solutions

[galois427]

I need some help on how to solve this question. It asks me to find all real numbers a with the property that the polynomial equation x^10 + a*x +1 = 0 has a real solution r such that 1/r is also a solution. I tried plugging in r and 1/r and equating the 2 equations, but that got me nowhere.

 

[TenaliRaman]

rearrange the equation as,
a = (-x^10-1)/x
put x=r and call it eqn 1
then put x=1/r and call it eqn 2
shouldn't RHS of both 1 and 2 be same?

-- AI

 

[Hurkyl]

Well, could you show what you got when you plugged in r and 1/r?

 

[galois427]

rearrange the equation as,
a = (-x^10-1)/x
put x=r and call it eqn 1
then put x=1/r and call it eqn 2
shouldn't RHS of both 1 and 2 be same?

-- AI

i euqated that equation and found out that r= + or - 1 so a = + or - 2.
well, that yielded 2 equations. x^10 - 2x + 1 and x^10 + 2x +1. r = 1 is a zeo, but r=-1 is not. what am i doing wrong?

 

[Rocko]

Do you know the answer?

 

[TenaliRaman]

for r = 1, a=-2
so x^10 - 2x + 1 = 0
put r = 1 , it is zero ... put 1/r = 1 .. again it is zero.

for r=-1, a = 2
so x^10 + 2x + 1 = 0
put r = -1 , it is zero ... put 1/r = -1 .. again it is zero.

so our conditions are satisfied...

-- AI