If the roots of the equatior 1-q-p2-2 x2-p1-q x-qq-1-p2-2-0 are equal- thenA- p2-8 qB- p 2-2 qC- p2-4 qD- p 2- q

The correct option is C p^2=4qAs roots are equal, so Δ=0 ⇒ p^2(1+q)^2=4 ( 1 q+p^22 ) ( q(q 1)+p^22 ) ⇒ p^2(1+q)^2 = (4(1 q)+2p^2) ( q(q 1)+p^22 ) ⇒ p^2(1+q)^2= ...

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Byju's Answer

Standard XIII

Mathematics

Nature of Roots

If the roots ...

Question

If the roots of the equation
$(1 - q + \frac{p^{2}}{2}) x^{2} + p (1 + q) x + q (q - 1) + \frac{p^{2}}{2} = 0$ are equal, then

p^{2} = 8 q

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p^{2} = 2 q

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p^{2} = 4 q

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p^{2} = q

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Solution

The correct option is C $p^{2} = 4 q$
As roots are equal, so $Δ = 0$
$\Rightarrow p^{2} (1 + q)^{2} = 4 (1 - q + \frac{p^{2}}{2}) (q (q - 1) + \frac{p^{2}}{2})$
$\Rightarrow p^{2} (1 + q)^{2} = (4 (1 - q) + 2 p^{2}) (q (q - 1) + \frac{p^{2}}{2})$
$\Rightarrow p^{2} (1 + q)^{2} = - 4 q (1 - q)^{2} + 2 p^{2} q (q - 1) + 2 p^{2} (1 - q) + p^{4}$
$\Rightarrow p^{2} (1 + q)^{2} - 2 p^{2} q (q - 1) - 2 p^{2} (1 - q) - p^{4} = - 4 q (1 - q)^{2}$
$\Rightarrow p^{2} [(1 + q)^{2} - 2 q^{2} + 4 q - 2 - p^{2}] = - 4 q (1 - q)^{2}$
$\Rightarrow p^{2} [- (1 - q)^{2} + (4 q - p^{2})] = 4 q (1 - q)^{2}$
$\Rightarrow - p^{2} (1 - q)^{2} + p^{2} (4 q - p^{2}) = - 4 q (1 - q)^{2}$
$\Rightarrow (1 - q^{2}) (- p^{2} + 4 q) + p^{2} (4 q - p^{2}) = 0$
$\Rightarrow (- p^{2} + 4 q) [(1 - q)^{2} + p^{2}] = 0$
As $(1 - q)^{2} + p^{2} \neq 0$ ,
$- p^{2} + 4 q = 0$
$∴ p^{2} = 4 q$

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