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Math Help - inequality

  1. #1
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    inequality

    Challenge problem:



    Given:

    1) x^2-2ax+b=0 with b\neq 0

    2) |x_{2}|< |x_{1}|,where x_{1},x_{2} are the roots of the above equation

    Prove:

    2(2a^2-b)-|b|< x_{1}^2<2(2a^2-b)

    Moderator approved, CB
    Last edited by alexandros; June 6th 2010 at 04:23 PM. Reason: correction
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  2. #2
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    Quote Originally Posted by alexandros View Post
    Given:

    1) x^2-2ax+b=0 with b\neq 0

    2) |x_{2}|< |x_{1}|,where x_{1},x_{2} are the roots of the above equation

    Prove:

    2(2a^2-b)-|b|< x_{1}^2<2(2a^2-b)

    Moderator approved, CB
    In order to even consider the < relation on x_1^2, shouldn't we require that a^2 \geq b? (That is, assuming a,b \in \mathbb{R})
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    In order to even consider the < relation on x_1^2, shouldn't we require that a^2 \geq b? (That is, assuming a,b \in \mathbb{R})
    in fact  2a^2>b
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  4. #4
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    Quote Originally Posted by alexandros View Post
    in fact  2a^2>b
    This still doesn't give us that the solutions are real. Take a=2, ~ b=3 to get the equation x^2-4x+6=0 which has no real solutions, although 2a^2=2 \cdot 2^2 = 8 > 6 = b... So are you requiring that a^2>b or not?
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    This still doesn't give us that the solutions are real. Take a=2, ~ b=3 to get the equation x^2-4x+6=0 which has no real solutions, although 2a^2=2 \cdot 2^2 = 8 > 6 = b... So are you requiring that a^2>b or not?
    No the problem does require  2a^2>b

    In your example :  |x_{1}| = |x_{2}| ,in our problem we have :

    |x_{2}|<|x_{1}|
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  6. #6
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    Here is what I did.
    By Vieta's formulas, we have x_1+x_2=2a , and x_1\cdot x_2=b . Then {x_1}^2+{x_2}^2+2x_1x_2={x_1}^2+{x_2}^2+2b=4a^2. And then, {x_1}^2+{x_2}^2=4a^2-2a=2(2a^2+b) .

    Now, we have |x_1\cdot x_2|=|x_2|\cdot|x_1|=|b| , and since |x_{2}|< |x_{1}| we arrive at (*) {x_2}^2<|b|.

    Eventually, {x_1}^2=2(2a^2+b)-{x_2}^2. Notice that x_1, x_2 \neq 0 because 0^2-2a0+b=b\neq 0.
    It follows that {x_1}^2=2(2a^2+b)-{x_2}^2<2(2a^2+b) ( {x_2}^2 is positive).
    Also from (*), -{x_2}^2>-|b| and it follows that {x_1}^2=2(2a^2+b)-{x_2}^2>2(2a^2+b)-|b|.

    From all this we get the inequality: 2(2a^2-b)-|b|< x_{1}^2<2(2a^2-b).
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  7. #7
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    There are a few typos in there...

    (1) x^2-2ax+b=0,\ b\ne0

    Roots of f(x)=0 are x_1,\ x_2

    (2) |x_2|<|x_1|

    Prove the following...

    2\left(2a^2-b\right)-|b|<\left(x_1\right)^2<2\left(2a^2-b\right)


    Proof

    \left(x-x_1\right)\left(x-x_2\right)=x^2-2\left(x_1+x_2\right)+x_1x_2

    or

    x^2-2ax+b=x^2-(sum\ of\ roots)x+product\ of\ roots

    Therefore

    x_1+x_2=2a,\ x_1x_2=b

    \left(x_1+x_2\right)^2=\left(2a\right)^2\ \Rightarrow\ \left(x_1\right)^2+2x_1x_2+\left(x_2\right)^2=4a^2  =\left(x_1\right)^2+\left(x_2\right)^2+2b

    \left(x_1\right)^2+\left(x_2\right)^2=4a^2-2b=2\left(2a^2-b\right)

    Hence...since |x_2|<|x_1|,

    |x_1x_2|>\left(x_2\right)^2\ \Rightarrow\ \left(x_1\right)^2+\left(x_2\right)^2-|x_1x_2|=\left(x_1\right)^2-|k|

    This is <\left(x_1\right)^2

    it's already been established that 2\left(2a^2-b\right)=\left(x_1\right)^2+\left(x_2\right)^2

    which is >\left(x_1\right)^2
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  8. #8
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    Here is an easy double implication proof:

    2(2a^2-b)-|b|<(x_{1})^2< 2(2a^2-b)\Longleftrightarrow -|b|<(x_{1})^2-4a^2+2b<0\Longleftrightarrow -|x_{1}x_{2}|<(x_{1})^2-(x_{1}+x_{2})^2+2x_{1}x_{2}<0\Longleftrightarrow

    ...................(since x_{1}+x_{2}=2a and  x_{1}x_{2}=b)


    \Longleftrightarrow |x_{1}x_{2}|>|x_{2}|^2>0\Longleftrightarrow |x_{2}|<|x_{1}|

    ............(since b\neq 0\Longrightarrow |x_{1}|\neq 0\wedge|x_{2}|\neq 0)
    Last edited by alexandros; July 25th 2010 at 07:07 PM.
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