Multiple choice

The value of 'a' for which both roots of the equation x2-6ax+2-2a+9a2 = 0 exceeds 3 are

  1. a<1

  2. a>1

  3. a<11/9

  4. a>11/9

  5. 0
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For roots to exceed 3: 1) Discriminant D >= 0, 2) f(3) > 0, 3) -b/2a > 3. D = (6a)^2 - 4(9a^2 - 2a + 2) = 36a^2 - 36a^2 + 8a - 8 = 8a - 8. D >= 0 -> a >= 1. f(3) = 9 - 18a + 9a^2 - 2a + 2 = 9a^2 - 20a + 11 > 0. Roots of 9a^2 - 20a + 11 are 1 and 11/9. For >0, a < 1 or a > 11/9. Vertex -b/2a = 6a/2 = 3a > 3 -> a > 1. Intersection of all conditions: a > 11/9.

AI explanation

For the quadratic equation x2 - 6ax + 2 - 2a + 9a2 = 0, the discriminant D equals (-6a)2 - 4(1)(2 - 2a + 9a2), which simplifies to 8a - 8. For both roots to be real, the discriminant must be non-negative, requiring 8a - 8 >= 0 and thus a >= 1. The sum of the roots is 6a, so for both to exceed 3, their sum must be strictly greater than 6, meaning 6a > 6 and a > 1. The vertex of the parabola is at x = 3a, which is strictly greater than 3 when a > 1, and the function evaluates to a negative value for any a > 1, proving both roots exceed 3. The result is a>11/9.

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