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?

$\endgroup$
  • $\begingroup$ 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 '15 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 '15 at 13:23
4
$\begingroup$

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.