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by hey_thr67 » Mon Jun 04, 2012 9:06 pm
Find all the values of 'a', so that 6 lies between the roots of the equation x^2 + 2(a-3)x + 9 =0

A: a< −3/4
B: a> 3/4
C: a<0 or a>6
D: a>6
E: a < −1/4

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by Anurag@Gurome » Mon Jun 04, 2012 10:05 pm
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.
Image

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.
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