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Math Help - absolute values and quad. formulae

  1. #1
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    absolute values and quad. formulae

    absolute values and quad. formulae
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  2. #2
    GAMMA Mathematics
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    How about you solve for x from the original quadratic?

    x = \frac{(m+3) \pm \sqrt{(m+3)^2 - 4(2-m)}}{2}

    Therefore, a = \frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} and b = \frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}, since we let a > b

    |a|=|b+1|

    \frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} = \left|\frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}\right| + 1

    You can now solve for m.
    Last edited by colby2152; April 16th 2008 at 06:55 AM.
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by colby2152 View Post
    since a > b
    So, as |-2|=|1+1|,\,-2>1 ?
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  4. #4
    GAMMA Mathematics
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    Quote Originally Posted by flyingsquirrel View Post
    So, as |-2|=|1+1|,\,-2>1 ?
    Okay, we let a > b
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    Quote Originally Posted by colby2152 View Post
    How about you solve for x from the original quadratic?

    x = \frac{(m+3) \pm \sqrt{(m+3)^2 - 4(2-m)}}{2}

    Therefore, a = \frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} and b = \frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}, since we let a > b

    |a|=|b+1|

    \frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} = \left|\frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}\right| + 1

    You can now solve for m.
    i still don't understand. Why
    |a|=|b+1|
    \frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} = \left|\frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}\right| + 1
    shouldn't squaring the both sides is the only way to do it?
    a^2=(b+1)^2
    thanks for help!
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  6. #6
    GAMMA Mathematics
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    Quote Originally Posted by afeasfaerw23231233 View Post
    i still don't understand. Why
    |a|=|b+1|
    \frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} = \left|\frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}\right| + 1
    shouldn't squaring the both sides is the only way to do it?
    a^2=(b+1)^2
    thanks for help!
    No, keep the absolute value sign on the right side because there could be a negative.
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  7. #7
    Lord of certain Rings
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    Quote Originally Posted by afeasfaerw23231233 View Post
    i still don't understand. Why
    |a|=|b+1|
    \frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} = \left|\frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}\right| + 1
    shouldn't squaring the both sides is the only way to do it?
    a^2=(b+1)^2
    thanks for help!
    Denoting it by

    x + y = \left|x - y \right| + 1
    Consider two cases x > y and x <= y and then reduce it to y = \frac12 and x = \frac12 respectively.

    Now consider each case separately, and solve for m and check if the assumed condition (x > y or x <= y) satisfies.
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