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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 \(1\).

 

\(\phantom{cube is 343}\)

 May 22, 2022
 #1
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Suppose that the non-real root is \(\omega\). Obviously, since it is non-real, it cannot be equal to 1.

Since its cube is 1, \(\omega^3 = 1\), which means \(\omega^3 - 1 = 0\). If we factorize the left-hand side of the equation, we have \((\omega - 1)(\omega^2 + \omega + 1) = 0\). Since \(\omega\) is not 1, we can cross out \((\omega - 1)\), so we're left with \(\omega^2 + \omega + 1 = 0\).

 

With the derivation above, it turns out that if \(\omega\) is a non-real root whose cube is 1, then \(x = \omega\) must be a root of the equation \(x^2 + x + 1 = 0\). Therefore, a = b = 1 by comparing coefficients.

 May 23, 2022
 #2
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I used this symbol for the non-real root for a reason. It is called the "cube roots of unity". Generally, any root of unity is commonly denoted by \(\omega\)

 

WIki page: https://en.wikipedia.org/wiki/Root_of_unity

MaxWong  May 23, 2022

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