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Show x² + x + 1 > 0

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data

    Show x² + x + 1 > 0
    well this is basically asking to prove that any value of x you put in will give a result bigger than zero, but how do i prove it

    2. Relevant equations
    According to the markscheme you should use complete the sqare


    3. The attempt at a solution

    (X + 1/2) + 3/4

    but how does that help

    Thanks :)
     
  2. jcsd
  3. Sep 13, 2008 #2

    statdad

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    Can you determine whether this has any real zeros? What if it doesn't?
     
  4. Sep 13, 2008 #3
    err dont see what you mean. there should be nothing to do with complex numbers in this if that's where your heading?

    Thanks :)
     
  5. Sep 13, 2008 #4
    (X + 1/2)2 + 3/4

    Meaning that the graph's min pt is 3/4. Therefore it is >0.
     
  6. Sep 13, 2008 #5
    You'll also have to show that the function is increasing.
     
  7. Sep 13, 2008 #6
    to show it's increasing you'll have to have somthing > 3/4 right?


    (X + 1/2)² + 3/4 = 0
    x + 1/2 = sqrt(-3/4)

    you cant root a minus :eek:

    Is that what you have to do?
     
  8. Sep 13, 2008 #7

    statdad

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    The solution has nothing to do with showing this is increasing - it can't, because [tex] x^2 + x + 1 [/tex] is not an increasing function of [tex] x [/tex].


    There are only a few possibilities for a quadratic:

    1. It does not have any real zeros - in this case it is always positive or always negative

    2. It has one real zero - in this case it is either positive and zero in one spot, or it is negative and zero in one spot

    3. It has exactly two real zeros - in this case it is positive for some values of x, negative for some, and zero for 2.

    So, if you can determine how many real zeros this has, and if you can determine at its sign at a value of [tex] x [/tex], you are done. Do not waste your time worrying about whether it is increasing or decreasing - that is the wrong track.
     
  9. Sep 13, 2008 #8

    Dick

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    You've shown x^2+x+1=(x+1/2)^2+3/4. (x+1/2)^2 is nonnegative (it's a SQUARE). 3/4 is positive. Hence the sum is positive.
     
  10. Sep 13, 2008 #9

    Redbelly98

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    Yes, absolutely. At this point, the problem is solved. The squared quantity can be zero at minimum, therefore the whole expression has a minimum of 3/4. In other words, it's always positive.

    End of discussion?
     
  11. Sep 13, 2008 #10
    yes thankyou very much :)
     
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