Question

If $a>1$ roots of the equation $(1-a)x²+3ax-1=0$ are

• A. one positive and one negative
• B. both negative
• C. both positive
• D. both non-real complex

Hint:

If $a>1$ roots of the equation $(1-a)x+3ax-1=0$ are

Answer 1

The Correct Option is C

Step 1: Concept
For the equation $(1-a)x^2+3ax-1=0$, let the roots be $\alpha$ and $\beta$.
Step 2: Analysis
The sum of roots $\alpha+\beta = \frac{3a}{a-1}$ and the product of roots $\alpha\beta = \frac{-1}{1-a} = \frac{1}{a-1}$.
Step 3: Evaluation
Since $a>1$, it follows that $\alpha+\beta>0$ and $\alpha\beta>0$. The discriminant $D = 9a^2 + 4(1-a) = 9(a^2 - \frac{4}{9}a + \frac{4}{9})$, which is greater than 0 for $a>1$.
Step 4: Conclusion
Because the sum, product, and discriminant are all positive, the equation has real and positive roots.
Final Answer: (c)