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Quadratic inequality

  1. Oct 18, 2015 #1
    • Member warned about posting with little or no prior research
    Can someone explain to be in detail what is quadratic inequality? It's rather confusing. Thank you
     
  2. jcsd
  3. Oct 18, 2015 #2

    jedishrfu

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    There are some video tutorials on it that may help:

     
  4. Oct 18, 2015 #3
    Try to find the 2 roots of the quadratic equation and discuss the intervals made of these numbers.
     
  5. Oct 18, 2015 #4

    symbolipoint

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    What is a quadratic inequality?
    [tex]ax^2+bx+c<0[/tex]or[tex]ax^2+bx+c<=0[/tex]or[tex]ax^2+bx+c>0[/tex]or[tex]ax^2+bx+c>=0[/tex]
     
  6. Oct 19, 2015 #5

    symbolipoint

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    To be clearer, do not let the relation to 0 fool you. The relation can be with a term or a value other than zero on one side. Merely adding the additive inverse to both sides can bring the inequality to relate a quadratic expression to zero. Also, if the quadratic is factorable, you may be able to have something like (ax+b)(cx+d) (relation-symbol)(0).
     
  7. Oct 19, 2015 #6
    ImageUploadedByPhysics Forums1445234898.698360.jpg
    For visualisation's sake, is it something like that? The space between the two intersection in the graph is equivalent to the space between he two lines on the number line?
     
  8. Oct 19, 2015 #7

    symbolipoint

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    Lym Y K,
    The Mathispower4u (which jedishru posted) video you should find very helpful in understanding what to do with solving a quadratic inequality. The roots of the quadratic expression form the x-number line into three intervals, and any value in each interval can be chosen to test the truth or falsity for the interval.
     
  9. Oct 19, 2015 #8

    symbolipoint

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    The example in the paper in your included photograph shows [itex]-4x^2-4x-1[/itex] and we must assume it's meant as related versus 0. You can/should DIVIDE both sides by NEGATIVE 4, and this MUST reverse the direction of the inequality symbol. Why? because multiplication or division by a negative VALUE.
    That step now gives you [itex]x^2+x+1/4[/itex] versus 0, as said, with relation reversed from what it was originally. This quadratic is factorable giving you exactly ONE critical x value.

    [itex](x+1/2)^2[/itex]. versus 0. (You did not show on your paper the inequality symbol relating). The critical value is at [itex]x=-1/2[/itex], just one single value, cutting the x-number line into just two intervals. Now, you test each interval, and maybe also you need to test that critical x value of [itex]-1/2[/itex].
     
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