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Math Help - solving quadratic system

  1. #1
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    solving quadratic system

    Hi;
    I have the equations y = 2x^2 - 2x +3 and y = x^2 + 5x -7 therefor they are both equal so set them equal and solve I'm ok with that but...

    I thought I could set them both to zero and solve for x by elimination ie
    2x^2 - 2x + 3 = 0
    x^2 + 5x - 7 = 0

    giving
    2x^2 - 2x = -3
    x^2 + 5x = 7


    multiplt the second equation by 2

    2x^2 - 2x = -3
    2x^2 + 10x = 14

    I thought this would let me eliminate the x^2 terms and solve for x but I get the wrong solution and I don't know why?

    Thanks
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  2. #2
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    Re: solving quadratic system

    Since they are both equal to \displaystyle y then they are equal. This means

    \displaystyle \begin{align*} 2x^2 - 2x + 3 &= x^2 + 5x - 7 \\ x^2 - 7x + 10 &= 0 \\ (x - 2)(x - 5) &= 0 \\ x = 2\textrm{ or }x &= 5\end{align*}

    Now substitute each point into one of the equations to find the \displaystyle y values.
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  3. #3
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    Re: solving quadratic system

    Is this the method for all quadratic systems that are equal.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: solving quadratic system

    The problem with your method is you assume y=0, but that's not given so the method Prove It has given you is a general method to solve exercices like this one.
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  5. #5
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    Re: solving quadratic system

    Quote Originally Posted by anthonye View Post
    Is this the method for all quadratic systems that are equal.
    It works for any kinds of systems that are equal. If y= A and y= B then A= B.
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