If $a > 1$ , roots of the equation $(1 - a) x^{2} + 3 a x - 1 = 0$ are

o n e o f p o s i t i v e o n e n e g a t i v e

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Solution

The correct option is D $b o t h p o s i t i v e$
$Let α, β are the roots of (1 - a) x^{2} + 3 a x - 1 =$
$s u m of roots= α + β = \frac{3 a}{1 - a}$
$p r o d u c t o f r o o t s = α β = \frac{- 1}{1 - a} = \frac{1}{a - 1}$
$i f a > 1$
$α + β > 0$
$α β > 0$
$i . e α > 0, β > 0$
$H e n c e b o t h a r e p o s i t i v e$

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