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Math Help - How to solve this equation by completing the square

  1. #1
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    How to solve this equation by completing the square

    Hello

    I can solve this question by using the quadratic formula but want to see how to solve by completing the square.

    Equation is \frac{-1}{5}x^2 + 2x - \frac{9}{2} = 0

    So my working is:

    \frac{-1}{5}(x^2 - 10x) - \frac{9}{2} = 0

    \frac{-1}{5}(x^2 - 10x) = \frac{9}{2}

    \frac{-1}{5}(x - 5)^2 - 25 = \frac{9}{2}

    \frac{-1}{5}(x - 5)^2 = \frac{50}{2} + \frac{9}{2} = \frac{59}{2}

    But how do I get rid of the \frac{-1}{5}. And is the above correct so far.

    My problem is that multiplying both sides by -5 means number on rhs will be negative and so no solution. But I know the curve does cross the x axis - so there is a solution. Can anyone show me the final steps to solve this using completing the square?

    Angus
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  2. #2
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    Re: How to solve this equation by completing the square

    You can multiply both sides by -5, yielding x^2 - 10x + \frac{45}{2} = 0. Then:

    x^2 - 10x + 25 + \frac{45}(2} = 25

    (x - 5)^2 = 25 - \frac{45}{2}

    (x - 5)^2 = \frac{5}{2}

    (Sorry, my LaTex isn't compiling)
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: How to solve this equation by completing the square

    When you add the 25 within the parentheses, you are really subtracting 5 because of the coefficient of -1/5, so you need to add 5 outside.

    What you have done is subtract 30 from the left side, while leaving the right side unchanged. You are correct up to:

    -\frac{1}{5}\left(x^2-10x\right)=\frac{9}{2}

    Your next step should be:

    -\frac{1}{5}\left(x^2-10x+25\right)=\frac{9}{2}-\frac{1}{5}\cdot25

    Can you finish from there?
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  4. #4
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    e^(i*pi)'s Avatar
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    Re: How to solve this equation by completing the square

    Quote Originally Posted by angypangy View Post
    Hello

    I can solve this question by using the quadratic formula but want to see how to solve by completing the square.

    Equation is \frac{-1}{5}x^2 + 2x - \frac{9}{2} = 0

    So my working is:

    \frac{-1}{5}(x^2 - 10x) - \frac{9}{2} = 0

    \frac{-1}{5}(x^2 - 10x) = \frac{9}{2}

    \frac{-1}{5}(x - 5)^2 - 25 = \frac{9}{2}

    \frac{-1}{5}(x - 5)^2 = \frac{50}{2} + \frac{9}{2} = \frac{59}{2}

    But how do I get rid of the \frac{-1}{5}. And is the above correct so far.

    My problem is that multiplying both sides by -5 means number on rhs will be negative and so no solution. But I know the curve does cross the x axis - so there is a solution. Can anyone show me the final steps to solve this using completing the square?

    Angus
    It's easier to multiply through by -5 initially to clear that fraction.

    -\dfrac{1}{5}x^2 + 2x - \dfrac{9}{2} = 0 \Leftrightarrow x^2 - 10x + \dfrac{45}{2} = 0

    (x-5)^2 - 25 + \frac{45}{2} = 0

    (x-5)^2 - \dfrac{5}{2} = 0

    (x-5)^2 = \dfrac{5}{2}

    This should now give two real solution
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  5. #5
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    Re: How to solve this equation by completing the square

    Quote Originally Posted by e^(i*pi) View Post
    It's easier to multiply through by -5 initially to clear that fraction.

    -\dfrac{1}{5}x^2 + 2x - \dfrac{9}{2} = 0 \Leftrightarrow x^2 - 10x + \dfrac{45}{2} = 0

    (x-5)^2 - 25 + \frac{45}{2} = 0

    (x-5)^2 - \dfrac{5}{2} = 0

    (x-5)^2 = \dfrac{5}{2}

    This should now give two real solution
    Ah, yes that is easier. thanks.
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