I have a question regarding taking square roots in inequalities. I have a problem asking:
Suppose $3x^2+bx+7>0$ for every real number x. Show that $|b|<2\sqrt{21}$.
In an earlier question it was established that I couldn't take the absolute value of both sides since it is not an equation. However, I could use $\sqrt{b^2}=|b|$ supposing $b$ is a real number.
Here is my work for this problem:
$b^2-4ac<0$ This is because the quadratic has no solution;
$b^2-4(3)(7)<0$
$b^2-84<0$
$b^2<84$
$\sqrt{b^2}<\sqrt{84}$
$|b|<\sqrt{84}$
$|b|<2\sqrt{21}$
My question is, when you take the square root of both sides of the inequality, why does it stay positive as opposed to $\pm$ like in equations? Is this because of the restriction stated in the problem that $3x^2+bx+7>0$ or is it because the absolute value of a variable cannot be negative? Could someone clear this up for me?