I have a problem that says:
Suppose $3x^2+bx+7 > 0$ for every number $x$, Show that $|b|<2\sqrt21$.
Since the quadratic is greater than 0, I assume that there are no real solutions since
$y = 3x^2+bx+7$, and $3x^2+bx+7 > 0$, $y > 0$
since $y>0$ there are no x-intercepts. I would use the discriminant $b^2-4ac<0$.
I now have $b^2-4(3)(7)<0$
Now how do I change $b$ to $|b|$? Can I take the absolute value of both sides of the equation or is there a proper way to do this?