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Math Help - quadratic equation

  1. #1
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    quadratic equation

    Challenge problem:

    Given that with the equation having two unequal real roots,

    Prove:

    moderator approved CB
    Last edited by alexandros; June 6th 2010 at 05:22 PM.
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  2. #2
    Super Member Deadstar's Avatar
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    I've probably done something wrong but here goes...

    Using standard formula we get roots to be

    \frac{2a \pm \sqrt{4a^2 - 4b}}{2} = a \pm \sqrt{a^2 - b} .

    Since roots are real and unequal, a^2 > b

    Hence...

    16a^4 + a^2 b^2 - 16a^2b + 2ab + 4b^2 + 3 > 16b^2 + a^2 b^2 - 16b^2 + 2ab + 4b^2 + 3.

    Canceling we get

    = a^2 b^2 + 2ab + 4b^2 + 3 = (ab + 1)^2 - 1 + 3 + 4b^2 = (ab+1)^2 + 4b^2 + 2 \geq 2...

    Hence result.
    Last edited by Deadstar; June 6th 2010 at 05:38 AM.
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  3. #3
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    There are rules for posting in this particular subforum. Read them.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    There are rules for posting in this particular subforum. Read them.
    I have received the solution and said to post. But when I checked it had not been posted so I did not add the imprimatur.

    Just shows the point of sending the solution to all the moderators.

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    I have received the solution and said to post. But when I checked it had not been posted so I did not add the imprimatur.

    Just shows the point of sending the solution to all the moderators. Mr F says: That is, to do what it says to do in the rules (clearly stated in the sticky) for posting in this subforum.

    CB
    ..
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  6. #6
    Super Member Deadstar's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Wow Deadstar what a Fantastic solution!
    Quote Originally Posted by CaptainBlack View Post
    Yes Mr F I agree! I believe Deadstar should be promoted to a mod!

    CB
    Quote Originally Posted by mr fantastic View Post
    I don't think a mod is good enough... Administrator would be better.
    Cheers guys!
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  7. #7
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    Quote Originally Posted by Deadstar View Post
    I've probably done something wrong but here goes...

    Using standard formula we get roots to be

    \frac{2a \pm \sqrt{4a^2 - 4b}}{2} = a \pm \sqrt{a^2 - b} .

    Since roots are real and unequal, a^2 > b

    Hence...

    16a^4 + a^2 b^2 - 16a^2b + 2ab + 4b^2 + 3 > 16b^2 + a^2 b^2 - 16b^2 + 2ab + 4b^2 + 3.

    Canceling we get

    = a^2 b^2 + 2ab + 4b^2 + 3 = (ab + 1)^2 - 1 + 3 + 4b^2 = (ab+1)^2 + 4b^2 + 2 \geq 2...

    Hence result.
    If i understand you correctly :from  a^2>b you get 16a^4>16b^2 ,by raising 1st the inequality to the power of 2 and then multplying by 16.

    But when the inequality is raised to 2 must we not have that b>0 ,otherwise the inequality may change if b<0.The problem assumes only b\neq 0.

    Example:take b=-3 and a=1 :we have b<a^2 ,but b^2>a^4
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  8. #8
    Super Member Deadstar's Avatar
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    Quote Originally Posted by alexandros View Post
    If i understand you correctly :from  a^2>b you get 16a^4>16b^2 ,by raising 1st the inequality to the power of 2 and then multplying by 16.

    But when the inequality is raised to 2 must we not have that b>0 ,otherwise the inequality may change if b<0.The problem assumes only b\neq 0.

    Example:take b=-3 and a=1 :we have b<a^2 ,but b^2>a^4
    Yeah you're right.

    Split it up into two cases then.

    a^2 > b we know.

    So let case 1 be

    |a| > \sqrt{|b|} which was my first post and hence a^4 > b^2.

    Case 2 is

    |a| < \sqrt{|b|} and hence a^4 < b^2.

    From which we get that...

    16a^4 + a^2 b^2 - 16a^2b + 2ab + 4b^2 + 3 = (ab + 1)^2 + 2 + 16a^4 - 16a^2b + 4b^2

    > (ab+1)^2 + 2 + 16a^4 - 16b^2 + 4b^2

    > (ab+1)^2 + 2 + 16a^4 - 16a^4 + 4b^2 \geq 2


    Although I haven't checked I think some of the the > can be replaced with \geq for one of the cases to include a^2 > b with a^4 = b^2. (This will only happen when b is negative since if b is +ve then a^2 - b = 0)
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  9. #9
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    Quote Originally Posted by Deadstar View Post
    Yeah you're right.

    Split it up into two cases then.

    a^2 > b we know.

    So let case 1 be

    |a| > \sqrt{|b|} which was my first post and hence a^4 > b^2.

    Case 2 is

    |a| < \sqrt{|b|} and hence a^4 < b^2.

    From which we get that...

    16a^4 + a^2 b^2 - 16a^2b + 2ab + 4b^2 + 3 = (ab + 1)^2 + 2 + 16a^4 - 16a^2b + 4b^2

    > (ab+1)^2 + 2 + 16a^4 - 16b^2 + 4b^2

    > (ab+1)^2 + 2 + 16a^4 - 16a^4 + 4b^2 \geq 2


    Although I haven't checked I think some of the the > can be replaced with \geq for one of the cases to include a^2 > b with a^4 = b^2. (This will only happen when b is negative since if b is +ve then a^2 - b = 0)

    Forming the inequality : 16a^4 + a^2 b^2 - 16a^2b + 2ab + 4b^2 + 3 = (ab + 1)^2 + 2 + 16a^4  - 16a^2b + 4b^2

    > (ab+1)^2 + 2 + 16a^4 - 16b^2 + 4b^2

    Do you assume : -16a^2b>-16b^2?
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