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Thread: inequality

  1. #1
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    inequality

    Challenge problem:



    Given:

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

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

    Prove:

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

    Moderator approved, CB
    Last edited by alexandros; Jun 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) $\displaystyle x^2-2ax+b=0$ with $\displaystyle b\neq 0$

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

    Prove:

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

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

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

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

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

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

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

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

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

    (2) $\displaystyle |x_2|<|x_1|$

    Prove the following...

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


    Proof

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

    or

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

    Therefore

    $\displaystyle x_1+x_2=2a,\ x_1x_2=b$

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

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

    Hence...since $\displaystyle |x_2|<|x_1|,$

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

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

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

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

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

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


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

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