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Discriminant

If x2+a-bx+...

Question

If $x^{2} + (a - b) x + (1 - a - b) = 0,$ where $a, b \in R,$ the value of $a$ such that the equation has distinct real roots for all value of $b$ are

a > 1

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a < 1

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a > 2

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a < 2

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Solution

The correct option is A $a > 1$
For real and distinct roots $Δ > 0$
$\Rightarrow$ $(a - b)^{2} - 4 (1 - a - b) > 0$
$\Rightarrow$ $(a - b - 2)^{2} + 8 a - 8 > 0$
$\Rightarrow$ $8 a - 8 > 0$
$a > 1$

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