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Coefficients of a quadratic (Tomato subjective 73)

problem: Consider the equation \({\displaystyle{x^3 + Gx + H = 0}} \), where G and H are complex numbers. Suppose that this equation has a pair of complex conjugate roots. Show that both G and H are real.

solution: Let three roots of the equation \({\displaystyle{x^3 + Gx + H = 0}} \)

are \({\displaystyle{\alpha, \beta, \gamma}} \)                        [ Let \({\displaystyle{\alpha, \beta}} \) are complex conjugates]

Now \({\displaystyle{\alpha \beta \gamma}} \) = – H … (i)
\({\displaystyle{\alpha + \beta + \gamma}} \) = 0 … (ii)
\({\displaystyle{\alpha \beta + \beta \gamma + \gamma \alpha}} \) = G … (iii)

From (ii) we get
\({\displaystyle{\alpha + \beta + \gamma}} \) = 0 [ \({\displaystyle{\alpha, \beta}} \) are complex conjugates so they are real]
\({\Rightarrow} \) \({\gamma} \) = real

Now as \({\gamma} \) = real

\({\displaystyle{\beta \gamma + \gamma \alpha}} \) = \({\displaystyle{\gamma (\beta + \alpha)}} \)
= \({real \times real} \)
= \({real} \) … (iv)
\({\alpha, \beta} \) are complex conjugates so \({\alpha \beta = real} \) … (v)
From (iv) & (v) we get \({\displaystyle{\alpha \beta + \beta \gamma + \gamma \alpha}} \) = \({real + real = real} \)
\({\Rightarrow} \) G = real [from (iii)]

Now \({\alpha, \beta} \) is real and \({\alpha, \beta} \) is real
so \({\displaystyle{\alpha \beta \gamma}} \) = real
\({\Rightarrow} \) H = real.

August 12, 2015
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