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Math Help - Completing the square

  1. #1
    Senior Member Mukilab's Avatar
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    Completing the square

    How would I complete the square with the following?

    4x^2-25

    I thought of (4x+\frac{0}{2}^2)-25+\frac{0}{2}^2 but that doesn't make any sense!

    Any help please?
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  2. #2
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    e^(i*pi)'s Avatar
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    Why would you want to complete the square? This is the difference of two squares and will easily factor

    4x^2-25 = (2x)^2 - 5^2 and the difference of two squares is a^2-b^2 = (a-b)(a+b)


    ====================================

    If you want/must to complete the square you have the right idea but the first step is always to make sure the coefficent of x^2 is 1

    4x^2-25 = x^2 - \dfrac{25}{4} = (x+0)^2 - \dfrac{25}{4} = (x+0)^2 - \left(\dfrac{5}{2}\right)^2
    Last edited by e^(i*pi); October 4th 2010 at 10:52 AM. Reason: math tag
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  3. #3
    Senior Member Mukilab's Avatar
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    need it to work out the minimum point of a graph. Thanks!
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  4. #4
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    e^(i*pi)'s Avatar
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    Ah right, in this case it might be easier to spot that 4x^2 \geq 0 for all real x.

    The mininum will therefore be when 4x^2 = 0 hence the minimum will be (0,-25)
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  5. #5
    MHF Contributor

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    You "complete the square" to get rid of a term of the form "ax", that is x to the 1 power, in a quadratic. In your example, 4x^2- 25, there is no "x" term- " 4x^2" is already a perfect square. The minimum value occurs when x= 0 and is -25. Since 4x^2- 25= (2x- 5)(2x+ 5) the graph of y= 4x^2- 25 is a parabola with vertex at (0, -25), vertical axis, opening upward, and with x-intercepts at (5/2, 0) and (-5/2, 0).
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