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Math Help - Solving for B

  1. #1
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    Solving for B

    I need help solving for B in this equation:

    (NOTE: 5b^2 is 5b squared)

    5b^2 - 5b +1 = 0


    I started by subtracting -1 from both sides getting:

    5b^2 - 5b = -1

    Then I pulled out 5b from the left side and got:

    5b (b - 1) = -1

    And now I have no idea what to do. I think I am doing the entire thing wrong. Can someone help me out?
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  2. #2
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    Re: Solving for B

    oohh wait a second.. do i use Quadratic equation?
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  3. #3
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    Re: Solving for B

    Quote Originally Posted by gurrry View Post
    I need help solving for B in this equation:

    (NOTE: 5b^2 is 5b squared)

    5b^2 - 5b +1 = 0


    I started by subtracting -1 from both sides getting:

    5b^2 - 5b = -1

    Then I pulled out 5b from the left side and got:

    5b (b - 1) = -1

    And now I have no idea what to do. I think I am doing the entire thing wrong. Can someone help me out?
    yes, you are.

    5b^2 - 5b + 1 = 0

    the trinomial in not factorable, so use quadratic formula ...

    b = \frac{5 \pm \sqrt{(-5)^2 - 4(5)(1)}}{2(5)}
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  4. #4
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    Re: Solving for B

    Heh.. I guess I shouldve thought harder before posting. It didnt occur to me to use QF. I guess im still rubbing off the summer "rust". Lol.

    thanks for the help! ill more than likely be back sometime today...
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  5. #5
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    Re: Solving for B

    just to be sure.. is the answer:

    5 +/- sq root of 5 over 10?

    (obviously it can be reduced)
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  6. #6
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    Re: Solving for B

    Another way- which I personally prefer to just plugging into a formula: complete the square:
    5b^2- 5b= -1 as you say. Divide through by 5:
    b^2- b= -1/5.

    (b- a)^2= b^2- 2ab+ a^2 will match that is 2a= 1 so a= -1/2 and a^2= 1/4.

    We can make b^2- b into a "perfect square by adding 1/4:
    b^2- b+ 1/4= -1/5+ 1/4= -4/20+ 5/20= 1/20
    (b- 1/2)^2= 1/20= 1/(4(5))
    b- 1/2= \pm 1/(2(sqrt(5))

    b= 1/2\pm sqrt(5)/10 as you say.
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  7. #7
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    Re: Solving for B

    Sorry to bump this again, but I have a question reguarding the reducing.

    Is the fully reduced answer 1 +/- Sqrt of 1 over 2? Or can whats inside the sq. root not be reduced?
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  8. #8
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    Re: Solving for B

    \frac{5 \pm \sqrt{5}}{10}

    or

    \frac{1}{2} \pm \frac{\sqrt{5}}{10}

    the first one is considered "reduced" as is ... the second is just a splitting of the fractions.
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