Prove or Disprove that there exists a real number x such that x^2+x+1 is less than or equal to zero.

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- Mar 27th 2008, 11:52 AM #1

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- Mar 27th 2008, 11:59 AM #2

- Mar 27th 2008, 12:02 PM #3

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- Mar 27th 2008, 12:07 PM #5

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- Mar 27th 2008, 12:14 PM #6

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From the quadratic equation, you know there are no real roots, meaning the function is never 0 (either always negative, or always positive). With that, you can note that letting x=0 gives you a value of 1, so the function is always positive.

Not quite the most rigorous proof and probably not what your professor is looking for, but it is valid.

Edit: Also, I forgot to add that this works because the function is continuous over all real numbers.

- Mar 27th 2008, 12:15 PM #7
Well the quadratic formula tells you where you have a value of x such that

$\displaystyle x^2 + x + 1 = 0$

which is one of the criterion of your proof. So the existence of x proves that the function can be less than or equal to 0.

Likely the best method is the intermediate value theorem, if you can do it that way.

-Dan