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Inequalities- tricky question

  1. Aug 15, 2011 #1
    Inequalities- tricky question!!

    hiii,
    i was wondering if anybody knew how to help me with this one tricky homework question. i can do most of the inequaliies ive come across, but how do you solve an inequality if you cant factor it???
    the question is: (x^2-4x+7)/(x^2+x-6)
    i know that the denominator is easily factorable ((x+3)(x-2)) but the numerator isnt? help plzzzz ive been stuck on this for ages!!!:(:(
     
  2. jcsd
  3. Aug 15, 2011 #2

    disregardthat

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    Re: Inequalities- tricky question!!

    Rewrite (x^2-4x+7) to (x-2)^2+3. What can you say about this?
     
  4. Aug 15, 2011 #3

    Mentallic

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    Re: Inequalities- tricky question!!

    Ok so I'm guessing by inequalities you mean something like [tex]\frac{x^2-4x+7}{x^2+x-6}>1[/tex] for example?

    Try multiplying both sides by [tex](x^2+x-6)^2[/tex] since you know this has to be a non-negative number, and don't expand! Use your knowledge of factorizing to solve it.
     
  5. Aug 15, 2011 #4
    Re: Inequalities- tricky question!!

    ohh sorry i didnt even put the rest of the question in!! it was (x^2-4x+7)/(x^2+x-6)≤ 0
    i just wasnt sure if i was allowed to automatically multiply by the denominator because i didnt know if it was positive or negative. if i rephrase it as (x-2)^2 + 3, i still cant cancel out any brackets because its not completely factored. sorry:( just havent done maths in ages and i suck at it..
     
  6. Aug 15, 2011 #5

    disregardthat

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    Re: Inequalities- tricky question!!

    Can you tell when (x-2)^2+3 is positive and negative though? Notice the squared term.
     
  7. Aug 29, 2011 #6
    Re: Inequalities- tricky question!!

    oohhh i got it now!!!:) yay took ages tho lol but thanks every1:)
     
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