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    Hey,

    Im stuck on this question, where am i going wrong because I cant get the answer to equal 0

    A demand curve for peanuts is represented by a section of the quadratic:

    P(Q) = -Q^2 - 8Q + 84

    Complete the square of this quadratic function. Use this result with explanation to (i) sketch the function (ii) suggest values for Q for which this might be a sensible demand curve.

    P(Q) = -Q^2 - 8Q + 84

    P(Q) = -Q^2 - 8Q + 84 = 0

    (-Q^2 - 8x + 16) + 84 - 16 = 0

    (-Q-4) (-Q-4) + 68 = 0

    (-Q-4)^2 = -68

    square root: (-Q-4)^2 = + or - square root: (-68)

    -Q-4 = + or - -68

    -Q-4= -68(+4) ______________-Q-4 = 68(-4)

    =-64 or 64

    thanks
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    It would be easier to complete the square on -\mbox{P}(Q). Your method for completing the square looks more complicated than it needs to be... the idea is that if you have x^2 + 2ax + b, then you halve the coefficient of x and put it inside the bracket which gets squared, and then remove the square of it; so x^2 + 2ax + b \equiv (x+a)^2 - a^2 + b. Applying this here gives -\mbox{P}(Q) = Q^2 + 8Q - 84 = (Q+4)^2 - 16 - 84. This gives you information about an upside-down version of the graph (i.e. a reflection in the x-axis), so you can multiply both sides by -1 to give you the right-way-up version.
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    sorry, but im lost in what you are saying.

    x -1 by both sides? Wouldn't that just give you the same answer?
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    What I mean by \times (-1) on both sides is that what you end up with by following my steps is -\mbox{P}(Q) = (Q+4)^2 - 100 when what you want is \mbox{P}(Q) = -(Q+4)^2 + 100.

    As for what you did, for some reason you have 84-16 = 68 when you should have 84+16 = 100. You somehow have also managed to get that (-Q)(-Q) = -Q^2, which isn't true, and you've made a few more arithmetical errors here and there... for example you square-rooted a negative, and after that somehow the square-root disappeared on the next line.

    The equation you should end up with is -(Q+4)^2 + 100 = 0 \Rightarrow (Q+4)^2 = 100, which is fairly easy to solve since 100 is a perfect square, and is positive :p:
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    Ohh i see thanks.

    So the final answer is when q =4, 100
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    (Original post by stem01)
    Ohh i see thanks.

    So the final answer is when q =4, 100
    Well that's not the answer really...

    The answer to part (i) is a plot of the graph. To do that you can use the information in the equation to find out where the vertex is, which in this case is at the point (-4, 100), and you can solve it to find out where it crosses the Q-axis, which is at the points (6, 0) and (-14, 0)... if you don't know how I got any of these answers ask me and I'll go through it.

    For part (ii) you have to say which region gives sensible answers. Well, I'm assuming that Q is the number of peanuts and P(Q) is the demand for them. If so, the number of peanuts can't be negative (because that'd be silly), so Q > 0. Also, the demand for peanuts can't be less than zero either, so P(Q) > 0, meaning that -14 < Q < 6 (because outside of this region the curve crosses the Q-axis and P(Q) becomes negative). Combining the two, we get just 0 \le Q \le 6. Again, if you don't know how I worked this out, ask me.
 
 
 
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