First slide

Theory of equations

Question

If $α, β, γ$ are the roots of $x^{3} - x^{2} - 1 = 0$ then the value of $\frac{1 + α}{1 - α} + \frac{1 + β}{1 - β} + \frac{1 + γ}{1 - γ}$ is equal to

Moderate

A

-5
B

-6
C

-7
D

2

Solution

$\begin{array}{l} Σα = 1, Σαβ = 0, αβγ = 1 \\ \begin{aligned} \sum \frac{1 + α}{1 - α} & = - Σ \frac{- α + 1 - 2}{1 - α} = Σ (\frac{2}{1 - α} - 1) \\ = 2 Σ \frac{1}{1 - α} - 3 \end{aligned} \end{array}$
Now, $x^{3} - x^{2} - 1 = (x - α) (x - β) (x - γ)$
$∴ \log (x - α) + \log (x - β) + \log (x - γ) = \log (x^{3} - x^{2} - 1)$
Differentiating, we get
$\begin{array}{l} \frac{1}{(x - α)} + \frac{1}{(x - β)} + \frac{1}{(x - γ)} = \frac{3 x^{2} - 2 x}{x^{3} - x^{2} - 1} \\ \Rightarrow \frac{1}{1 - α} + \frac{1}{1 - β} + \frac{1}{1 - γ} = \frac{3 - 2}{1 - 1 - 1} = - 1 \\ \Rightarrow \sum_{} \frac{1 + α}{1 - α} = - 5 \end{array}$

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