Question

If both the roots of the equation \( x^2 - 6ax + 2 - 2a + 9a^2 = 0 \) exceed 3, then:

• A. \( a < \frac{3}{2} \)
• B. \( a > \frac{3}{2} \)
• C. \( a < \frac{5}{2} \)
• D. \( a > \frac{11}{9} \)

Hint:

When both roots of a quadratic equation must be greater than a certain value, use the sum and product conditions along with the quadratic formula.

Answer 1

The Correct Option is D

Given the quadratic equation: \[ x^2 - 6ax + 2 - 2a + 9a^2 = 0 \] Step 1: Identify the Conditions for the Roots Let the roots be \( \alpha \) and \( \beta \). Since both roots must exceed 3, we need: \[ \alpha > 3 \quad \text{and} \quad \beta > 3 \] Step 2: Apply Vieta's Formulas By Vieta's relations: \[ \alpha + \beta = 6a \] \[ \alpha \beta = 2 - 2a + 9a^2 \] Step 3: Derive the Conditions For both roots to exceed 3: 1. Condition 1: Sum of roots condition \[ \alpha + \beta > 6 \] From \( \alpha + \beta = 6a \), this implies: \[ 6a > 6 \quad \Rightarrow \quad a > 1 \] 2. Condition 2: Product of roots condition \[ \alpha \beta > 3^2 = 9 \] From \( \alpha \beta = 2 - 2a + 9a^2 \), \[ 2 - 2a + 9a^2 > 9 \] Rearranging: \[ 9a^2 - 2a - 7 > 0 \] Using the quadratic formula to solve this inequality: \[ a = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(9)(-7)}}{2(9)} \] \[ a = \frac{2 \pm \sqrt{4 + 252}}{18} \] \[ a = \frac{2 \pm \sqrt{256}}{18} \] \[ a = \frac{2 \pm 16}{18} \] The two solutions are: \[ a = 1 \quad \text{or} \quad a = \frac{11}{9} \] Step 4: Identify the Correct Inequality Since the inequality \( 9a^2 - 2a - 7 > 0 \) holds true for: \[ a > \frac{11}{9} \quad \text{or} \quad a < -\frac{7}{9} \] Given that \( a > 1 \) from the sum condition, the valid solution is: \[ a > \frac{11}{9} \] Step 5: Final Answer 

\[Correct Answer: (4) \ a > \frac{11}{9}\]