2
$\begingroup$

I'm reading Calculus with Analytic Geometry and in a problem in the first chapter (page 6)

Solve the following inequalities
$x^2 + 2x + 4 > 0$

Apparently, $x^2 + 2x + 4$ has no solution in real numbers for $x$ when the expression is equal to 0. Am I missing something here?

$\endgroup$
2
  • $\begingroup$ If it has no real solutions, since the strongest coefficient is positive, the parabola is always above the X axis, hence the inequality is true for every value of $x$. $\endgroup$
    – barak manos
    Feb 7, 2015 at 13:19
  • 1
    $\begingroup$ Since the expression on the left is never zero, it must be either negative or positive. It can't assume both negative and positive values without also assuming the value $0$ (by continuity--the image must be connected), so it must either be always negative or always positive. Testing a single point provides the result. $\endgroup$
    – MPW
    Feb 7, 2015 at 13:23

3 Answers 3

Reset to default
3
$\begingroup$

$$(x+1)^2+3>0$$

You know the rest!

$\endgroup$
2
$\begingroup$

Since the discriminant ($b^2-4ac$) is negative the graph won't touch the x axis. Now, because $a>0$, its always positive.

By the way if $a$ is troubling you then you can substitute any value you want in the equation to check whether it's always positive or negative.

$\endgroup$
1
$\begingroup$

If you compete the square, you see that $$x^2+2x+4=(x+1)^2+3$$. So its graph is an upward opening parabola with vertex at $(-1,3)$. Therefore, the minimum value of $x^2+2x+4$ is 3. Hence, $x^2+2x+4>0$ is true for all real numbers.

$\endgroup$

You must log in to answer this question.