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Math Help - Express quadratic equation into perfect square form

  1. #1
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    Express quadratic equation into perfect square form

    Express y = 2 - 5x - 3x^2 in the form y = a - b(x + c)^2. Hence find the maximum value of y and the corresponding value of x.
    This is what I did but I can't seem to get the answer y = \frac{49}{12} - 3(x + \frac{5}{6})^2

    y = 2 - 5x - 3x^2
    y = -3 (x^2 + \frac{5}{3}x - \frac{2}{3})
    y = -3 [(x + \frac{5}{6})^2 - \frac{2}{3} + \frac{23}{56})]
    y = -\frac{1}{12} - 3(x + \frac{5}{6})^2

    Is there a careless mistake somewhere that I can't seem to spot? I've been redoing this for at least four times and I can't seem to see where I've gone wrong ):
    Any help would be greatly appreciated and thank you very much in advance.
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  2. #2
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    Quote Originally Posted by caramelcake View Post
    This is what I did but I can't seem to get the answer y = \frac{49}{12} - 3(x + \frac{5}{6})^2

    y = 2 - 5x - 3x^2
    y = -3 (x^2 + \frac{5}{3}x - \frac{2}{3})
    y = -3 [(x + \frac{5}{6})^2 - \frac{2}{3} + \frac{23}{56})]
    This line is wrong.
    y = -\frac{1}{12} - 3(x + \frac{5}{6})^2

    Is there a careless mistake somewhere that I can't seem to spot? I've been redoing this for at least four times and I can't seem to see where I've gone wrong ):
    Any help would be greatly appreciated and thank you very much in advance.
    Your equation reads:

    y = 2-3\left(x^2+\frac53 x\right)

    y = 2-3\left(x^2+\frac53 x + \frac{25}{36} - \frac{25}{36} \right)
    Here I added the missing square and subtracted it at once so the equation didn't change.
    y = 2-3\left(x+\frac56 \right)^2 + \frac{25}{12}

    y = \frac{49}{12}-3\left(x+\frac56 \right)^2
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  3. #3
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    You could also expand the RHS, and compare the coefficients:

    2-5x-3x^2 = a-b(x+c)^2 = (a-bc^2)-(2bc)x-bx^2.

    Thus b = 3, and 2bc = 5 hence b = \frac{5}{6}, and a-bc^2 = 2, so a = \frac{49}{12}.
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