What is the proper way to show numbers (x) that satisfy this inequality?

[The Student]

I am studying the first year of honors calculus. The basic inequality properties that the book teaches does not help me with this question x^2+x+1 > 2. I am at the beginning of the textbook, but and it does not seem to give an example on the proper way to do this question. So, here is my work. x^2+x+1 > 2; x^2+x+1/4 > 5/4; (x+1/2)^2 > 5/4; x+1/2 > 5^(1/2)/2; x > (-1+5^(1/2))/2 or (-1-5^(1/2))/2. I did this by tinkering with the question and just remembering what I learnt in high school math. Is there a more proper way that is expected for a first year honors calculus program?

 

[DrPhil]

I am studying the first year of honors calculus. The basic inequality properties that the book teaches does not help me with this question x^2+x+1 > 2. I am at the beginning of the textbook, but and it does not seem to give an example on the proper way to do this question. So, here is my work. x^2+x+1 > 2; x^2+x+1/4 > 5/4; (x+1/2)^2 > 5/4; x+1/2 > 5^(1/2)/2; x > (-1+5^(1/2))/2 or (-1-5^(1/2))/2. I did this by tinkering with the question and just remembering what I learnt in high school math. Is there a more proper way that is expected for a first year honors calculus program?

I think it is a little easier if you subtract 2 from both sides of the inequality:

\(\displaystyle \displaystyle x^2 + x - 1 > 0\)

Then you can find the two roots of the quadratic. Using the quadratic formula,

\(\displaystyle \displaystyle x = -\dfrac{1}{2} \pm \dfrac{\sqrt{5}}{2} \)

Your method - completing the square - is just as valid. A continuous function can change sign only where it crosses zero. The expression will be positive when \(\displaystyle x\) is less than the lower root or larger than the larger root.

\(\displaystyle \displaystyle x \in \left(-\infty, -\dfrac{1+\sqrt{5}}{2}
\right) + \left(-\frac{1- \sqrt{5}}{2}, +\infty \right) \)

 

[The Student]

I think it is a little easier if you subtract 2 from both sides of the inequality:

\(\displaystyle \displaystyle x^2 + x - 1 > 0\)

Then you can find the two roots of the quadratic. Using the quadratic formula,

\(\displaystyle \displaystyle x = -\dfrac{1}{2} \pm \dfrac{\sqrt{5}}{2} \)

Your method - completing the square - is just as valid. A continuous function can change sign only where it crosses zero. The expression will be positive when \(\displaystyle x\) is less than the lower root or larger than the larger root.

\(\displaystyle \displaystyle x \in \left(-\infty, -\dfrac{1+\sqrt{5}}{2}
\right) + \left(-\frac{1- \sqrt{5}}{2}, +\infty \right) \)

Oh yeah, thank-you very much!

 

[lookagain]

\(\displaystyle \displaystyle x \in \left(-\infty, \ -\dfrac{1+\sqrt{5}}{2}
\right) \ \cup \ \left(-\frac{1- \sqrt{5}}{2}, \ +\infty \right) \)

DrPhil, you can achieve the union symbol by typing "\cup" in the center where you have the "+."