First slide

Theory of equations

Question

$If α and β are the roots of x^{2} - p (x + 1) - c = 0 then the values of (α + 1) (β + 1) and \frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c} a r e$

Moderate

A

$1 + c, - 1$
B

$1 - c, - 1$
C

$1 - c, 1$
D

$1 - c, 0$

Solution

$The given equation is x^{2} - p x - (p + c) = 0$
$\begin{array}{l} ∴ α + β = p and α β = - (p + c) \\ So, (α + 1) (β + 1) = α β + (α + β) + 1 \end{array}$
$= - (p + c) + p + 1 = 1 - c$
$(α + 1) (β + 1) = 1 - c - - - - (1)$
$\begin{array}{l} \frac{(α + 1)^{2}}{(α + 1)^{2} - (1 - c)} + \frac{(β + 1)^{2}}{(β + 1)^{2} - (1 - c)} \\ \begin{array}{rc} = & \frac{(α + 1)^{2}}{(α + 1)^{2} - (α + 1) (β + 1)} + \frac{(β + 1)^{2}}{(β + 1)^{2} - (α + 1) (β + 1)} \end{array} using eq 1 \end{array}$
$\frac{{(α + 1)}^{2}}{(α + 1) (α + 1 - β - 1)} + \frac{{(β + 1)}^{2}}{(β + 1) (β + 1 - α - 1)}$
$\begin{matrix} = \frac{α + 1}{α - β} + \frac{β + 1}{β - α} = \frac{(α + 1) - (β + 1)}{α - β} = 1 \end{matrix}$

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