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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?

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Your second guess is correct: absolute value (by definition) can not be negative.

Please note that the restriction says $-2\sqrt{21}<b<2\sqrt{21}$

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Let $x\ge 0$. Then, by definition, $\sqrt{x}$ is the non-negative number whose square is $x$.

The function $f(x)=\sqrt{x}$ is an increasing function. Thus, if $p$ and $q$ are non-negative, then $p\lt q$ iff $\sqrt{p}\lt \sqrt{q}$.

Remark: Regrettably, it is not uncommon in the schools for teachers, and texts, to write, for example, $\sqrt{9}=\pm 3$. The convention when functions are studied is that $\sqrt{x}$ is non-negative. This is done so that $\sqrt{x}$ will be a function of $x$. Recall that for $f$ to be a function, we cannot have two different values of $f$ at a number $a$.

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Mmm...be careful. The question supposes the inequation exists for all real $x$. So your first inequality in your solution should be $b^2 - 4ac \geq 0$.

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  • $\begingroup$ Wrong. The quadratic $3x^2 + bx + 7 = 0$ has no real solution. Hence the discriminant of the quadratic is strictly negative. $\endgroup$ – Deepak May 31 '15 at 9:22

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