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Math Help - Solve for x-- the inequality of quadratic

  1. #1
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    Solve for x-- the inequality of quadratic

    Solve \frac{2x}{x^2-9}\le\frac{1}{x+2}

    ------

    x^2-9\not=0
    .'. x\in R-\{-3,3\}
    and

    x+2\not=0
    .'. x\in R-\{-2\}

    then converting the original inequality to
    (2x)(x+2)\le(x^2-9)
    (2x^2+4x)(-x^2+9)\le 0
    (x^2+4x+9)\le 0
    As Discriminant <0 it has no real roots
    so how to do further...




    My assumptions:-
    x^2-9should not be zero
    x+2should also not be zero

    Are my assumptions right?
    Any help will be highly appreciated.
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  2. #2
    Member anonimnystefy's Avatar
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    Re: Solve for x-- the inequality of quadratic

    hi sumedh

    your "conversion" isn't correct.you cannot multiply an inequality by something you don't know the sign of.

    \frac{2x}{x^2-9}\leq\frac{1}{x+2}

    \frac{2x}{(x-3)(x+3)}-\frac{1}{x+2}\leq 0

    \frac{2x(x+2)-(x-3)(x+3)}{(x-3)(x+2)(x+3)}\leq 0

    \frac{2x^2+4x-(x^2-9)}{(x-3)(x+2)(x+3)}\leq 0

    \frac{2x^2+4x-x^2+9}{(x-3)(x+2)(x+3)}\leq 0

    \frac{x^2+4x+9}{(x-3)(x+2)(x+3)}\leq 0

    can you do it from here?
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  3. #3
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    Re: Solve for x-- the inequality of quadratic

    i am really very thankful to you
    it will be your greatness if you tell few next steps .
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Solve for x-- the inequality of quadratic

    If you know:
    \frac{x^2+4x+9}{(x-3)\cdot (x+3)\cdot (x+2)}\leq 0
    That means:
    x^2+4x+9 \leq 0
    Now solve x^2+4x+9=0.
    What do you notice?

    After this you only have to make the sign-table to determine where f(x)\leq 0
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  5. #5
    Member anonimnystefy's Avatar
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    Re: Solve for x-- the inequality of quadratic

    hi Siron

    actually he should make a sign table for:
    \frac{x^2+4x+9}{(x-3)(x+2)(x+3)}

    because (x-3)(x+2)(x+3) might be negative and thus change the sign of the inequality.so we mustn't multiply by (x-3)(x+2)(x+3).
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Solve for x-- the inequality of quadratic

    Yes, I ment that but maybe I had better said:
    f(x)=\frac{x^2+4x+9}{(x-3)(x+2)(x+3)}\leq 0
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  7. #7
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    Re: Solve for x-- the inequality of quadratic

    x^2+4x+9=0 has imaginary roots.
    so it will be always positive
    and we will have to consider the denominator now, right??
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: Solve for x-- the inequality of quadratic

    Indeed  x^2+4x+9>0, \forall x \in \mathbb{R}, an easy way to determine this statement is by using completing the square:
    x^2+4x+9>0 \Leftrightarrow (x^2+4x+4)+5>0 \Leftrightarrow (x+2)^2+5>0 \Leftrightarrow (x+2)^2>-5
    Which is always true!

    Go on with the denominator.
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  9. #9
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    Re: Solve for x-- the inequality of quadratic

    thanks i got it
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: Solve for x-- the inequality of quadratic

    Okay, you're welcome and good work!
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