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Find the ordered pair (a,b) of real numbers for which x^2+ax+b has a non-real root whose cube is 1.

 Apr 30, 2022
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If x^3 = 1, then x^3 - 1 = 0.

Factorizing gives (x - 1)(x^2 + x + 1) = 0.

 

If the root x is non-real and its cube is 1, then x^2 + x + 1 = 0.

Comparing coefficients gives a = 1, b = 1.

 May 2, 2022

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