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Solving double inequalities

  1. Dec 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Instructions: For the following inequalities, determine if 0 is a number in the solution set.
    1) 3x [tex]\leq[/tex] x+1 [tex]\leq[/tex] x-1
    2) x(2x-1) [tex]\leq[/tex] x+7

    2. Relevant equations



    3. The attempt at a solution
    1)
    LS:
    [tex]3x \leq x+1 [/tex]
    [tex]3x-1 \leq x [/tex]
    [tex]-1 \leq -3x [/tex]
    [tex]-1/-2 \geq -2x/-2[/tex]
    [tex]1/2 \geq x [/tex]

    RS:
    [tex]x+1 \leq x-1[/tex]
    [tex]0 \leq -2[/tex]

    I said for the right side that there are no solutions, since the inequality is not true.

    So I said that 0 is not included in the solution set in this inequality, based from the inequality from the left side.

    2)
    [tex] x(2x-1) \leq x+7 [/tex]
    [tex] 2x^2 - x \leq x+7 [/tex]
    [tex] 2x^2 - 2x - 7 \leq 0 [/tex]

    After that I tried factoring the equation. It would work, just not with even numbers though...have some decimals on it.

    This is what I tried so far for both questions.

    Note: I looked at the wrong part of the question, that's why I changed the question, but its still the same equations.
     
    Last edited: Dec 13, 2009
  2. jcsd
  3. Dec 13, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi francis21! Welcome to PF! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Dec 13, 2009 #3
    tiny-tim, I just changed by first post.

    I hope this helps.
     
  5. Dec 13, 2009 #4

    tiny-tim

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    1) looks ok, and for 2), use the standard formula for a quadratic equation …

    except why not just substitute x = 0 into the inequalities, and see if they're true?
     
  6. Dec 13, 2009 #5
    That would actually make sense. But I guess, for this particular homework, they want to train us on how to solve linear inequalities. That's why I was solving for the inequalities.

    For number 1, I tried placing zero on the inequality, and does not actually include zero as its solution in the solution set.

    For number 2, isn't written in standard form, [tex] 2x^2-2x-7 \leq 0 [/tex] ([tex] ax^2+bx +c=0[/tex]), but just expressed as an inequality?

    But let's say if I factor the quadratic equation, the x values would be the interval right (x= some value or x= some other value)?
     
  7. Dec 13, 2009 #6
    I don't think graphing is suggested yet which is easiest way to approach this problem IMO
     
  8. Dec 13, 2009 #7

    tiny-tim

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    Yes, it's a continuous function, so if you find the zeros, you automatically (well, with a little care :wink:) also find the positive and negative ranges.
     
  9. Dec 14, 2009 #8

    HallsofIvy

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    Notice that this does NOT ask you to solve the inequality. It only asks whether "0 is a number in the solution set"- that is, whether or not 0 satisfies the inequality. Just replace x by 0 in each and see if they are true.

    1) [itex]3(0)\le 0+1\le 0-1[/itex]
    Is that true?

    2) [itex]0(2(0)-1)\le 0+ 7[/itex]
    Is that true?
     
  10. Dec 14, 2009 #9

    Mentallic

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    For 2) try drawing the parabola. It will make more sense this way.
    Now, factorize (or use the quadratic formula, whichever suits your fancy) to find the x-intercepts, and now try to see where the parabola is less than or equal to 0. In other words, for what range of x-values is [itex]y\leq 0[/itex]? If the x-value of 0 is in this range, then 0 is in the solution set of the inequality.

    p.s. quadratic inequalities aren't linear :cool:
     
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