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Math Help - Completing the square

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    Completing the square

    I do not understand this at all. Click image to enlarge Completing the square-imposibru.jpg. I keep getting the answer 5(t-8/5)^2+1/5, which means that the vertex of this equation is NOT (8/5,-64/25). I even used h: 8/5 to plug it into the entire equation 5t^2-16t+13.
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    Re: Completing the square

    Your completing the square work is correct. (The answer in the "Click to enlarge" panel, that is.)

    -Dan
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    Re: Completing the square

    Why? Why is the k value not 1/5?
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    Re: Completing the square

    Can you show me the steps for the solution so I can look at the mechanics of it?
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    Re: Completing the square

    Quote Originally Posted by Eraser147 View Post
    I do not understand this at all. Click image to enlarge Click image for larger version. 

Name:	imposibru.JPG 
Views:	5 
Size:	3.6 KB 
ID:	25278. I keep getting the answer 5(t-8/5)^2+1/5, which means that the vertex of this equation is NOT (8/5,-64/25). I even used h: 8/5 to plug it into the entire equation 5t^2-16t+13.
    \displaystyle \begin{align*} 5 \left[ \left( t - \frac{8}{5} \right) ^2 - \frac{64}{25} \right] &= 5 \cdot \left( t - \frac{8}{5} \right)^2 + 5 \cdot \left( -\frac{64}{25} \right) \\ &= 5\left( t - \frac{8}{5} \right)^2 - \frac{64}{5} \end{align*}

    So your turning point should be \displaystyle \begin{align*} \left( \frac{8}{5}, -\frac{64}{5} \right) \end{align*}.
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    Re: Completing the square

    It wouldn't make sense because if you plugin 8/5 into the original equation, 1/5 IS the answer for the k value.
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    Re: Completing the square

    Quote Originally Posted by Eraser147 View Post
    It wouldn't make sense because if you plugin 8/5 into the original equation, 1/5 IS the answer for the k value.
    No amount of arguing is going to make your answer correct. You correctly completed the square, you did NOT do the expansion correctly.
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    Re: Completing the square

    No no no no, just no. That image is what the answer the solution manual has. THAT's NOT my answer. My answer is with the 1/5. I know how to expand, but no matter how I calculate it, it won't show the k value to have 1/5. Again, it is easily testable by jut plugging in k into the original polynomial above.
    Last edited by Eraser147; October 19th 2012 at 07:16 AM.
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