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Math Help - Completing the square for turning point when a != 1

  1. #1
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    Completing the square for turning point when a != 1

    Say if we have:

    x^2 + 12x + 5
    (x^2 + 6) - 36 + 5
    (x^2 + 6) - 31
    x = -6, y = -31

    How how could I find the TP if a != 1?
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    Re: Completing the square for turning point when a != 1

    Quote Originally Posted by SplashDamage View Post
    x^2 + 12x + 5
    (x^2 + 6) - 36 + 5
    (x^2 + 6) - 31
    That should be (x+6)^2 - 31.

    Anyway, to complete the square when there is a leading coefficient, just factor it out of the first two terms:

    2x^2 - 12x + 5

    =2\left(x^2 - 6x\right) + 5

    =2\left(x^2 - 6x + 9 - 9\right)+5

    =2\left(x^2 - 6x + 9\right)-2(9)+5

    =2(x - 3)^2-13
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    Re: Completing the square for turning point when a != 1

    The 9 is from halving the 6 and doing 3^2 isnt it?

    However I do not know where you get the -2 from. You lost me on that step.
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    Re: Completing the square for turning point when a != 1

    Do you know that a(b+ c)= ab+ ac? He had 2(x^2- 6x+ 9- 9) and wrote that as 2(x^2- 6x+ 9)+ 2(-9).
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    Re: Completing the square for turning point when a != 1

    Ah that makes much more sense now that you put it that way. Thanks.
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    Re: Completing the square for turning point when a != 1

    By the way. How would I go about solving this with logs? 2^x + 2^(x+3) = 36

    I know it can be solved simply by breaking 2^(x + 3) into 2^x and 2^3. Giving you 2^x * 8, and then just adding 1 and making it 9.

    But there is a way to do it give logarithms is there not?
    Last edited by SplashDamage; May 30th 2012 at 03:26 PM.
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    Re: Completing the square for turning point when a != 1

    Quote Originally Posted by SplashDamage View Post
    By the way. How would I go about solving this with logs? 2^x + 2^(x+3) = 36

    I know it can be solved simply by breaking 2^(x + 3) into 2^x and 2^3. Giving you 2^x * 8, and then just adding 1 and making it 9.

    But there is a way to do it give logarithms is there not?
    Right, after you do that you have

    9\cdot2^x=36

    \Rightarrow2^x=4

    Now, it's probably obvious that the answer is 2. But we could use logarithms to make our work more explicit:

    2^x=4\quad\Rightarrow\quad\log_22^x=\log_24 \quad\Rightarrow\quad x=\log_24=2.
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