If (1 - p) is a root of quadratic equation $x^{2} + p x + (1 - p) = 0$ then its roots are

-1, 2

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-1, 1

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0, -1

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0, 1

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Solution

The correct option is C 0, -1
Let the second root be $α$ ,
Then $α + (1 - p) = - p \Rightarrow α = - 1$
Also $α . (1 - p) = 1 - p \Rightarrow (α - 1) (1 - p) = 0 \Rightarrow p = 1 [∵ α = - 1]$
$∴$ Roots are $α = - 1 a n d p - 1 = 0$

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