Help with solving a polynomial

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To solve the polynomial equation x^10 + a*x + 1 = 0 with the condition that if r is a solution, then 1/r is also a solution, the value of a can be determined. By substituting r and 1/r into the rearranged equation a = (-x^10 - 1)/x, two cases arise: for r = 1, a = -2, leading to the equation x^10 - 2x + 1 = 0, which has r = 1 as a root. For r = -1, a = 2 results in the equation x^10 + 2x + 1 = 0, which also satisfies the condition with r = -1. Both values of a ensure that the original conditions are met, confirming that the solutions are valid. Thus, the values of a that satisfy the polynomial equation are -2 and 2.
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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.
 
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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
 
Well, could you show what you got when you plugged in r and 1/r?
 
TenaliRaman said:
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?
 
Do you know the answer?
 
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
 
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