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$23 Answers
$\begingroup$$$(x+1)^2+3>0$$
You know the rest!
$\endgroup$$\begingroup$Since the discriminant ($b^2-4ac$) is negative the graph won't touch the x axis. Now, because $a>0$, its always positive.
By the way if $a$ is troubling you then you can substitute any value you want in the equation to check whether it's always positive or negative.
$\endgroup$$\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.
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