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Math Help - Proving quadratics

  1. #1
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    Proving quadratics

    How do you prove that x^2y^2-5xy+7>0
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    Re: Proving quadratics

    Proving quadratics-inequality.png
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  3. #3
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    Re: Proving quadratics

    Quote Originally Posted by RuyHayabusa View Post
    How do you prove that x^2y^2-5xy+7>0
    \displaystyle \begin{align*} x^2y^2 - 5\,x\,y + 7 &= \left( x\,y \right)^2 - 5\,x\,y + 7 \\ &= \left( x\, y \right)^2 - 5\,x\,y + \left( -\frac{5}{2} \right)^2 - \left( -\frac{5}{2} \right)^2 + 7 \\ &= \left( x\,y - \frac{5}{2} \right)^2 - \frac{25}{4} + \frac{28}{4} \\ &= \left( x\,y - \frac{5}{2} \right)^2 + \frac{3}{4}  \end{align*}

    Since \displaystyle \begin{align*} \left( x\,y - \frac{5}{2} \right)^2 \geq 0 \end{align*} for all \displaystyle \begin{align*} x, y \end{align*}, that means \displaystyle \begin{align*} \left( x\,y - \frac{5}{2} \right)^2 + \frac{3}{4} \geq \frac{3}{4} > 0 \end{align*} for all \displaystyle \begin{align*} x,y \end{align*}.
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    Re: Proving quadratics

    Thanks everyone!
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