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    hi i have the inequality:

    1+x / x^2 - 3x +2 >= 2

    i multiplied both sides by the bottom denominator and then multiplied the brackets out and then rearranged. then i factorised to give:

    (-2x + 1)(x - 3)

    giving me x >= 1/2 and x >= 3

    but wen i try x values bigger than 3, then it doesnt satisfy the equation. this happens when i try values lower as well.

    wen i try values bigger and smaller than 1/2, then this also does not satisfy the equation.... the only x values that satisfy the inequality are 1/2 and 3, which are the values i calculated. does this mean my answer is just x=1/2 and x=3???
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    BAHAHA
    i came here thinking you were being treated unequally!! XD
    Shows what i know about maths!! lol
    :P

    my bad...
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    Could you please put brackets round your inequality?

    Do you mean (1+x)/x^2 - 3x + 2

    or 1 + x/(x^2 -3x +2)

    or ??
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    (Original post by ian.slater)
    Could you please put brackets round your inequality?

    Do you mean (1+x)/x^2 - 3x + 2

    or 1 + x/(x^2 -3x +2)

    or ??
    sorry, i meant

    (1+x)/(x^2 -3x +2) >= 2
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    (Original post by sarah_vickers)
    hi i have the inequality:

    1+x / x^2 - 3x +2 >= 2

    i multiplied both sides by the bottom denominator and then multiplied the brackets out and then rearranged. then i factorised to give:

    (-2x + 1)(x - 3)

    giving me x >= 1/2 and x >= 3

    but wen i try x values bigger than 3, then it doesnt satisfy the equation. this happens when i try values lower as well.

    wen i try values bigger and smaller than 1/2, then this also does not satisfy the equation.... the only x values that satisfy the inequality are 1/2 and 3, which are the values i calculated. does this mean my answer is just x=1/2 and x=3???
    So the question is

    \displaystyle \frac{1+x}{x^2 - 3x + 2}\geq 2

    Remember that you're dealing with an inequality. Multiplying through by the denominator to get rid of the fraction is okay...so long as you know it's positive. Otherwise (if it's negative) then the inequality will flip over, because that's how inequalities work (2 > 1, but -2 < -1).

    In this question, then, you'll want to express

    \displaystyle 2 = \frac{x^2 - 3x + 2}{x^2 - 3x + 2}

    Then you can rearrange to get *algebra* \geq zero.
    And from there...
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    Well, you can't just multiply through by (x^2 -3x +2) because sometimes this is positive and sometimes negative. If you multiply by a negative number you must reverse the direction of the inequality. So you need a strategy.

    There are three ways:

    (1) break the values of x into two. Those where (x^2 -3x +2) is positive, and those negative. Work separately.

    (2) multiply all through by (x^2 -3x +2)^2 which is never negative, and deal with the messy algebra you get.

    (3) take the 2 to the LHS. Put everything over a common denominator. Build a table of 'critical values' and work from there.
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    Note that the "messy algebra" isn't actually that messy. In general, you have an obvious common factor, and the rest should factor reasonably easily.
 
 
 
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