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Math Help - Quadratic and Completing square

  1. #1
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    Quadratic and Completing square

    Having some difficulty attempting this problem...




    1) (x-a)^{2} - b^{2} = 0

    Solving:
    (x-a)(x-a) - b^{2} = 0

    x^{2} - ax - ax - a^{2} - b^{2} = 0

    x^{2} - 2ax - a^{2} - b^{2} = 0



    ... stuck.*




    Completing the square


    2) x^{2} + 2bx + c = 0


    .... stuck
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  2. #2
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    Quote Originally Posted by pychon View Post
    Having some difficulty attempting this problem...




    1) (x-a)^{2} - b^{2} = 0

    Solving:
    (x-a)(x-a) - b^{2} = 0

    x^{2} - ax - ax - a^{2} - b^{2} = 0

    x^{2} - 2ax - a^{2} - b^{2} = 0



    ... stuck.*




    Completing the square


    2) x^{2} + 2bx + c = 0


    .... stuck
    For the first problem, try adding b^2 to both sides of the equation.

    For the second problem, take 1/2 of the coefficient of x, square it and add to both sides of the equation.

    Can you go on to solve?
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  3. #3
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    Ha.. you got to it before I could edit. The first problem is just a quadratic equation, the second is completing the square.
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  4. #4
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    Quote Originally Posted by pychon View Post
    Having some difficulty attempting this problem...




    1) (x-a)^{2} - b^{2} = 0

    Solving:
    (x-a)(x-a) - b^{2} = 0

    x^{2} - ax - ax - a^{2} - b^{2} = 0 you should have a +a^2, not a -a^2

    x^{2} - 2ax - a^{2} - b^{2} = 0

    shouldn't you be going

    (x-a)^2=b^2\;\;?

    x-a= \pm b

    x=a \pm b



    ... stuck.*




    Completing the square


    2) x^{2} + 2bx + c = 0


    .... stuck
    .
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    .
    Nice catch.... yes, a typo should be -a^2 and yes... should have used the square root instead of factoring the entire thing out (tedious). Correct?

    so...

    \sqrt{(x-a)^{2}} = \sqrt{b^{2}}

    x-a = b

    x = a \pm b



    Just a quick check if someone wouldn't mind (different problem):

    9x^{2}-6x-2=0

    9x^{2}-6x=2

    x^{2}-\frac{2}{3}+\frac{1}{9}=\frac{2}{9}+\frac{1}{9}

    \sqrt{(x-\frac{1}{3})^{2}} = \sqrt{\frac{3}{9}}

    x=\frac{1}{3}\pm\frac{\sqrt{5}}{5}
    thanks!



    as for completing the square... still trying to burn that into my brain with no luck.
    Last edited by pychon; October 25th 2010 at 04:17 PM.
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  6. #6
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    Quote Originally Posted by pychon View Post
    Nice catch.... yes, a typo should be +a^2 and yes... should have used the square root instead of factoring the entire thing out (tedious). Correct?

    so...

    \sqrt{(x-a)^{2}} = \sqrt{b^{2}}

    x-a = b take the positive and negative square roots of.. b^2 here

    x = a \pm b

    thanks!



    as for completing the square... still trying to burn that into my brain with no luck.
    The completing the square idea works according to

    (x+q)(x+q)=x(x+q)+q(x+q)=x^2+xq+xq+q^2=x^2+2xq+q^2

    To work backwards from that, you examine the multiplier of x and take half of it...

    x^2+2bx+c=0

    x is multiplied by 2b, so half of that is b, hence we need to introduce b^2 by adding to both sides

    x^2+2bx+b^2+c=b^2\Rightarrow\ x^2+2bx+b^2=b^2-c

    Now the left side is a "multiplied out" square

    (x+b)^2=x^2+2bx+b^2

    so you can continue by taking the square root of both sides as before
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  7. #7
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    Quote Originally Posted by pychon View Post
    Just a quick check if someone wouldn't mind (different problem):

    9x^{2}-6x-2=0

    9x^{2}-6x=2

    x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{2}{9}+\frac{1}{9}

    \sqrt{(x-\frac{1}{3})^{2}} = \sqrt{\frac{3}{9}}

    x=\frac{1}{3}\pm\frac{\sqrt{5}}{5} incorrect, maybe a typo, should have 3 in there

    thanks!
    \displaystyle\ x-\frac{1}{3}=\pm\sqrt{\frac{1}{3}}=\pm\frac{1}{\sqr  t{3}}\Rightarrow\ x=\frac{1}{3}\ \pm \frac{1}{\sqrt{3}}=\frac{1}{3}\ \pm\frac{\sqrt{3}}{3}=\frac{1\ \pm\sqrt{3}}{3}

    Alternatively...

    9x^2-6x=2\Rightarrow\ (3x)^2-(2)3x+1=2+1

    (3x-1)(3x-1)=3\Rightarrow\ 3x-1=\pm\sqrt{3}

    3x=1 \pm \sqrt{3}\Rightarrow\ x=\displaystyle\frac{1\pm\ \sqrt{3}}{3}
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