JEE MAIN - Mathematics (2004 - No. 35)
If $$\left( {1 - p} \right)$$ is a root of quadratic equation $${x^2} + px + \left( {1 - p} \right) = 0$$ then its root are
$$ - 1,2$$
$$ - 1,1$$
$$ 0,-1$$
$$0,1$$
Explanation
Let the second root be $$\alpha .$$
Then $$\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$$
Also $$\alpha .\left( {1 - p} \right) = 1 - p$$
$$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$$
$$ \Rightarrow p = 1$$ [as $$\alpha = - 1$$]
$$\therefore$$ Roots are $$\alpha = - 1$$ and $$p-1=0$$
Then $$\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$$
Also $$\alpha .\left( {1 - p} \right) = 1 - p$$
$$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$$
$$ \Rightarrow p = 1$$ [as $$\alpha = - 1$$]
$$\therefore$$ Roots are $$\alpha = - 1$$ and $$p-1=0$$
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