# PROVING an INEQUALITY

Printable View

• Aug 12th 2012, 06:37 AM
Kaloda
PROVING an INEQUALITY
If x is any real number, then prove that $x^2+2x+4\geq0$.
• Aug 12th 2012, 06:41 AM
Prove It
Re: PROVING an INEQUALITY
Quote:

Originally Posted by Kaloda
If x is any real number, then prove that $x^2+2x+4\geq0$.

\displaystyle \begin{align*} x^2 + 2x + 4 &= x^2 + 2x + 1^2 - 1^2 + 4 \\ &= (x + 1)^2 + 3 \\ &\geq 0 \end{align*}
• Aug 12th 2012, 06:43 AM
Kaloda
Re: PROVING an INEQUALITY
Hahahahahahahaha! Now I know how dumb I am. Btw, thanks.
• Aug 12th 2012, 09:33 AM
Deveno
Re: PROVING an INEQUALITY
alternatively, you could look at the discriminant: b2 - 4ac of the quadratic x2+2x + 4, which is:

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

this tells you that the parabola:

y = x2+2x + 4 never crosses the x-axis, and since it "opens up" (the x2 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 x2+2x + 4 never gets smaller than 3, which happens at x = -1).