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Math Help - Values such that the system of equations has only one solution

  1. #1
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    Values such that the system of equations has only one solution

    How many different values of a for which the system of equations
      y^2 + y + x^2 = 2 and x + y = a has only one solution?
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    Re: Values such that the system of equations has only one solution

    Quote Originally Posted by shiny718 View Post
    How many different values of a for which the system of equations
      y^2 + y + x^2 = 2 and x + y = a has only one solution?
    From the second equation, we have \displaystyle \begin{align*} x = a - y \end{align*}. Substituting into the first equation gives

    \displaystyle \begin{align*} y^2 + y + \left(a - y\right)^2 &= 2 \\ y^2 + y + a^2 - 2a\,y + y^2 &= 2 \\ 2y^2 + \left(1 - 2a\right)y + a^2 - 2 &= 0 \end{align*}

    For there to only be one solution, the discriminant must be 0, so

    \displaystyle \begin{align*} \left(1 - 2a\right)^2 - 4(2)\left(a^2 - 2\right) &= 0 \\ 1 - 4a + 4a^2 - 8a^2 + 16 &= 0 \\ -4a^2 - 4a + 17 &= 0 \end{align*}

    Checking the discriminant of this quadratic tells us how many values of a satisfy this equation, and therefore tell us how many values of a will give one solution to the original set of equations...

    \displaystyle \begin{align*} (-4)^2 - 4(-4)(17) &= 16 + 272 \\ &= 288 \\ &> 0 \end{align*}

    Since this is positive, there are two values of a which will satisfy the set of equations.
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    Re: Values such that the system of equations has only one solution

    Another method. The first equation is that of a circle, which you can verify by representing it in the form (y - y₀) + x = r. How many lines with slope -1 touch this circle?
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    Re: Values such that the system of equations has only one solution

    Thanks a lot! Thank you for making the solution really clear and simple.
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