If $a> 1$, roots of the equation $\displaystyle (1-a)x^{2}+3ax-1=0$ are
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one positive and one negative
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both negative
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both positive
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both nonreal complex
The equation is (1-a)x^2 + 3ax - 1 = 0. Since a > 1, the coefficient of x^2 is negative. The product of roots is c/a = -1 / (1-a) = 1 / (a-1), which is positive. The sum of roots is -b/a = -3a / (1-a) = 3a / (a-1), which is also positive. Since both sum and product are positive, both roots must be positive.
Using the discriminant test for the equation, we calculate $D = (3a)^2 - 4(1-a)(-1)$. This simplifies to $9a^2 + 4 - 4a$, which factors to $(3a - 2)^2$, and since it is a perfect square, $D$ is always non-negative and the roots are real. Next, we examine the signs of the roots using Vieta's formulas. The product of the roots is $c/a = -1 / (1-a)$, and for $a > 1$, the denominator is negative, making the product positive. The sum of the roots is $-b/a = -3a / (1-a)$, and for $a > 1$, this also yields a positive value, ensuring both roots are positive.