First slide

Theory of equations

Question

$Let α, β be the roots of x^{2} + b x + 1 = 0 . Then the equation whose roots are - (α + \frac{1}{β}) and - (β + \frac{1}{α}) is$

Moderate

A

$x^{2} - 2 b x + 4 = 0$
B

$x^{2} - b x + 1 = 0$
C

$x^{2} = 0$
D

$x^{2} + 2 b x + 4 = 0$

Solution

$\begin{aligned} Since α, β are roots of x^{2} + b x + 1 = 0, α + β = - b, α β = 1 \\ We have (- α - \frac{1}{β}) + (- β - \frac{1}{α}) = - (α + β) - (\frac{1}{β} + \frac{1}{α}) \end{aligned}$
$\begin{array}{l} = - (α + β) - \frac{(α + β)}{α β} \\ = b + b = 2 b \end{array}$
$\begin{array}{l} and, (- α - \frac{1}{β}) (- β - \frac{1}{α}) = α β + 2 + \frac{1}{α β} = 1 + 2 + 1 = 4 \\ Thus, the equation whose roots are \end{array}$
$- α - \frac{1}{β} and - β - \frac{1}{α} is x^{2} - x (2 b) + 4 = 0$

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