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Mathematics Question on Complex Numbers and Quadratic Equations

If α,β,γ\alpha , \beta ,\gamma are the roots of x3−2x+1=0x^3-2x+1=0, then the value of (∑1α+β−γ(\sum \frac {1} {\alpha +\beta -\gamma}) is

A

−12\frac {-1}{2}

B

−1-1

C

00

D

12\frac {1}{2}

Answer

−1-1

Explanation

Solution

Given cubic equation is, x3−2x+1=0,(α,β,γ)x^{3}-2 x+1=0,(\alpha, \beta, \gamma) are roots of this equation.
Then, sum of roots Σα=0\Sigma \alpha=0
⇒α+β+γ=0\Rightarrow \alpha+\beta+\gamma=0
Σαβ=−2,αβγ=−1\Sigma \alpha \beta=-2, \alpha \beta \gamma=-1
Now, we have
Σ1α+β−γ=Σ1−γ−γ=−12Σ1γ\Sigma \frac{1}{\alpha+\beta-\gamma} =\Sigma \frac{1}{-\gamma-\gamma}=-\frac{1}{2} \Sigma \frac{1}{\gamma}
=−12(1α+1β+1γ)=-\frac{1}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)
=−12(αβ+βγ+αγαβγ)=-\frac{1}{2}\left(\frac{\alpha \beta+\beta \gamma+\alpha \gamma}{\alpha \beta \gamma}\right)
=−12⋅Σαβαβγ=−12⋅(−2)(−1)=−1=-\frac{1}{2} \cdot \frac{\Sigma \alpha \beta}{\alpha \beta \gamma}=-\frac{1}{2} \cdot \frac{(-2)}{(-1)} =-1