If - and - are the roots of x2-px-1-q-0- then

The correct options are A (α+1) (β + 1) = 1 q D α^2 + 2 α + 1/α^2 + 2 α + q + β^2 + 2 β + 1/β^2 + 2 β + q = 1 x^2 p(x+1) q = 0 x^2 px p q=0 α + β =p, αβ = ...

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Byju's Answer

Standard XII

Mathematics

Roots of a Quadratic Equation

If α and β ar...

Question

If $α$ and $β$ are the roots of $x^{2} - p (x + 1) - q = 0$ , then

$(α + 1) (β + 1) = 1 - q$

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$(α + 1) (β + 1) = 1 + q$

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$\frac{(α + 1)^{2}}{(α + 1)^{2} + q - 1} + \frac{(β + 1)^{2}}{(β + 1)^{2} + q - 1} = q$

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$\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + q} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + q} = 1$

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Solution

The correct options are
A
$(α + 1) (β + 1) = 1 - q$

D
$\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + q} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + q} = 1$

$x^{2} - p (x + 1) - q = 0 x^{2} - p x - p - q = 0 α + β = p, α β = - p - q α + β + α β = - q α + β + α β + 1 = 1 - q (1 + α) (1 + β) = 1 - q \dots \dots (1) \Rightarrow \frac{α^{2} + 2 α + 1}{α^{2} + 2 α + q} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + q}$
On putting the value of q - 1 from equation (1)
$\frac{(α + 1)^{2}}{(α + 1)^{2} - (1 + α) (1 + β)} + \frac{(β + 1)^{2}}{(β + 1)^{2} - (1 + α) (1 + β)} \frac{α + 1}{α - β} - \frac{β + 1}{α - β} \Rightarrow \frac{(α - β)}{(α - β)} = 1$

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If α and β are the roots of x2−p(x+1)−q=0, then

If $α$ and $β$ are the roots of $x^{2} - p (x + 1) - q = 0$ , then