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Math Help - solve for s

  1. #1
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    solve for s

    1/s+a + 1/s+b = 1/c

    solve for s
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by aeroflix View Post
    1/s+a + 1/s+b = 1/c

    solve for s
    Is it \dfrac{1}{s}+a+\dfrac{1}{s}+b=\dfrac{1}{c} or \dfrac{1}{s+a}+\dfrac{1}{s+b}=\dfrac{1}{c}? Depending on the interpretation, you will get completely different answers!!
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  3. #3
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    Quote Originally Posted by aeroflix View Post
    1/s+a + 1/s+b = 1/c
    solve for s
    IF: 1/(s+a) + 1/(s+b) = 1/c
    THEN:
    s = {2c - a - b +- SQRT[(a + b - 2c)^2 - 4(ab - ac - bc)]} / 2
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  4. #4
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    Quote Originally Posted by aeroflix View Post
    1/s+a + 1/s+b = 1/c

    solve for s
    \frac{1}{s + a} + \frac{1}{s + b} = \frac{1}{c}

    \frac{s + b}{(s + a)(s + b)} + \frac{s + a}{(s + a)(s + b)} = \frac{1}{c}

    \frac{s + b + s + a}{(s + a)(s + b)} = \frac{1}{c}

    \frac{2s + a + b}{s^2 + (a + b)s + ab} = \frac{1}{c}

    c(2s + a + b) = s^2 + (a + b)s + ab

    2cs + ac + bc = s^2 + (a + b)s + ab

    0 = s^2 + (a + b - 2c)s + ab - ac - bc.


    Now solve using the Quadratic Formula...
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  5. #5
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    whats the value of c ???

    value of a is 1 and b is what also?
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  6. #6
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    You need to solve for s in terms of a, b, and c. The values of a, b, and c can be taken to be arbitrary constants (ie, they don't change, but you also don't know what they are).
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  7. #7
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    ive came up with a quadratic but the part -4ac kinda freaks me out
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  8. #8
    Senior Member eumyang's Avatar
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    Quote Originally Posted by aeroflix View Post
    whats the value of c ???

    value of a is 1 and b is what also?
    Quote Originally Posted by Math Major View Post
    You need to solve for s in terms of a, b, and c. The values of a, b, and c can be taken to be arbitrary constants (ie, they don't change, but you also don't know what they are).
    No, the OP is asking what is a, b, and c in the quadratic formula. So in the equation
    s^2 + (a + b - 2c)s + ab - ac - bc = 0
    "a" is 1,
    "b" is a + b - 2c, and
    "c" is ab - ac - bc.

    So the setup would be
    s = \frac{-(a + b - 2c) \pm \sqrt{(a + b - 2c)^2 - 4(1)(ab - ac - bc)}}{2(1)}
    ...
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