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Inequations

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hey_thr67 Master | Next Rank: 500 Posts Default Avatar
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Post Mon Jun 04, 2012 9:06 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    Post 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.
    ttp://postimage.org/image/5dz252ilf/" target="_blank">

    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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