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Homework Help: X^2 + x + 1 > 2

  1. Apr 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve for X

    2. Relevant equations

    x^2 + x + 1 > 2


    3. The attempt at a solution

    x2 + x - 1 > 0

    (x + 2)(x - 1/2) > 0

    x > 2
    x > 1/2

    Wolfram Alpha said that the solutions for X are:

    x>1/2 (sqrt(5)-1)

    and

    x<1/2 (-1-sqrt(5))

    what did I do wrong?
     
    Last edited: Apr 5, 2012
  2. jcsd
  3. Apr 5, 2012 #2

    Curious3141

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    Homework Helper

    Well, for one thing, your factorisation of x2 + x - 1 to (x + 2)(x - 1/2) is wrong.

    You can see that by multiplying those factors together. They won't equal the original quadratic.

    What's the discriminant of the quadratic?

    The other error is in supposing that, if AB > 0, then A > 0 and B > 0 is the only solution. Don't forget that A < 0 and B < 0 is also a valid solution. Remember, if you multiply two negative numbers together, you get a positive.

    Another mistake was going from x + 2 > 0 to x > 2. Shouldn't that be x > -2?

    The last mistake was in the way you expressed your solution. Even though you considered the A > 0 and B > 0 case, you wrote the solution as x > 2, x > 1/2. Remember the "AND" condition. If x > 2 AND x > 1/2, you should simply "collapse" that solution to x > 2 (the bigger value) since everything that's bigger than 2 is also bigger than 1/2. Of course, 2 and 1/2 are wrong values for this question (as I mentioned earlier), but I'm using them to demonstrate the concept for you.
     
    Last edited: Apr 5, 2012
  4. Apr 5, 2012 #3

    D H

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    Staff Emeritus
    Science Advisor

    Here is what you did wrong:
    (x+2)(x-1/2) = x2 + 1.5x - 1, not x2 + x - 1.
     
  5. Apr 5, 2012 #4
    Why would you go trying to factor when all you have to do is bring the 2 to the left side and solve the equation (x^2+x+?=0) by finding the discriminant Δ and you'll get 2 real roots. Then go on from there.
     
  6. Apr 6, 2012 #5

    verty

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    Homework Helper

    Another method is to complete the square, getting p^2 > ...
     
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