hey_thr67 wrote:Find all the values of 'a', so that 6 lies between the roots of the equation x^2 + 2(a-3)x + 9 =0

The easiest and methodical solution to this problem requires some advanced understanding of quadratic equations. For any quadratic equation ax² + bx + c = 0, the curve of the graph f(x) = ax² + bx + c is a upward (or downward) facing parabola if a > 0 (or a < 0). See the following diagram for better understanding.


Now we can conclude that the graph of f(x) = x² + 2(a - 3)x + 9 will be an upward facing parabola. Hence, if 6 lies between the roots of x² + 2(a - 3)x + 9 = 0, f(6) must be less than zero.

So, f(6) = 6² + 2(a - 3)*6 + 9 < 0
--> [36 + 12a - 36 + 9] < 0
--> (12a + 9) < 0
--> a < -9/12 = -3/4

The correct answer is A.