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Math Help - Completing the Square

  1. #1
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    Completing the Square

    Can someone help me out and go through the steps for how to complete the square for this problem? I can never quite figure it out.

    The problem is: x^2-6x-1=0 Then I know I would want to switch it to: x^2-6x=1 Normally I know I would divide everything by what is in front of the x, but in that case it's nothing (1), and dividing by 1 obviously wouldn't do anything. So what else do I need to do?




    thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by deathtolife04 View Post
    Can someone help me out and go through the steps for how to complete the square for this problem? I can never quite figure it out.

    The problem is: x^2-6x-1=0 Then I know I would want to switch it to: x^2-6x=1 Normally I know I would divide everything by what is in front of the x, but in that case it's nothing (1), and dividing by 1 obviously wouldn't do anything. So what else do I need to do?




    thanks
    add and subtract the square of (1/2) the coefficient of x (that way, we're adding zero and hence not changing anything.

    x^2 - 6x - 1 = 0

    \Rightarrow x^2 - 6x ~{ \color {red} + (-3)^2 - (-3)^2} - 1 = 0

    now, contract the first three terms as you have been taught

    \Rightarrow (x - 3)^2 -(-3)^2 - 1 = 0

    simplify

    \Rightarrow (x - 3)^2 - 10 = 0

    the square has been completed
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    add and subtract the square of (1/2) the coefficient of x (that way, we're adding zero and hence not changing anything.

    x^2 - 6x - 1 = 0

    \Rightarrow x^2 - 6x ~{ \color {red} + (-3)^2 - (-3)^2} - 1 = 0

    now, contract the first three terms as you have been taught

    \Rightarrow (x - 3)^2 -(-3)^2 - 1 = 0

    simplify

    \Rightarrow (x - 3)^2 - 10 = 0

    the square has been completed

    ok thanks!


    so how would i find out what x equals?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by deathtolife04 View Post
    ok thanks!


    so how would i find out what x equals?
    solve for x the old-fashion way. try it
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    solve for x the old-fashion way. try it
    what would I do after I get to x^2-6x=1?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by deathtolife04 View Post
    what would I do after I get to x^2-6x=1?
    why are you going back there? did you let me complete the square for nothing? start from (x - 3)^2 - 10 = 0

    remember, when we are solving for x, it means we want to get x by itself on one side of the equation, so do it
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    why are you going back there? did you let me complete the square for nothing? start from (x - 3)^2 - 10 = 0

    remember, when we are solving for x, it means we want to get x by itself on one side of the equation, so do it

    ok so (x-3)^2=10

    where would i go from there to get x by itself? the () and ^2 confuse me.

    Could it just be x=the square root of 10+3?
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    add and subtract the square of (1/2) the coefficient of x (that way, we're adding zero and hence not changing anything.

    x^2 - 6x - 1 = 0

    \Rightarrow x^2 - 6x ~{ \color {red} + (-3)^2 - (-3)^2} - 1 = 0

    now, contract the first three terms as you have been taught

    \Rightarrow (x - 3)^2 -(-3)^2 - 1 = 0

    simplify

    \Rightarrow (x - 3)^2 - 10 = 0

    the square has been completed
    Quote Originally Posted by deathtolife04 View Post
    ok thanks!


    so how would i find out what x equals?
    To continue:
    (x - 3)^2 - 10 = 0

    (x - 3)^2 = 10

    x - 3 = \pm \sqrt{10}

    x = 3  \pm \sqrt{10}

    -Dan
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