If x is any real number, then prove that $\displaystyle x^2+2x+4\geq0$.

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- Aug 12th 2012, 06:37 AMKalodaPROVING an INEQUALITY
If x is any real number, then prove that $\displaystyle x^2+2x+4\geq0$.

- Aug 12th 2012, 06:41 AMProve ItRe: PROVING an INEQUALITY
- Aug 12th 2012, 06:43 AMKalodaRe: PROVING an INEQUALITY
Hahahahahahahaha! Now I know how dumb I am. Btw, thanks.

- Aug 12th 2012, 09:33 AMDevenoRe: PROVING an INEQUALITY
alternatively, you could look at the discriminant: b

^{2}- 4ac of the quadratic x^{2}+2x + 4, which is:

4 - 4(1)(4) = -12 < 0.

this tells you that the parabola:

y = x^{2}+2x + 4 never crosses the x-axis, and since it "opens up" (the x^{2}term is positive), it must lie entirely above the x-axis

(not only is the inequality true, but we can also replace "≥" with ">". in fact, the value of x^{2}+2x + 4 never gets smaller than 3, which happens at x = -1).