Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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;

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?

share|cite|improve this question
up vote 1 down vote accepted

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

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

share|cite|improve this answer

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 $a$.

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

share|cite|improve this answer
Wrong. The quadratic $3x^2 + bx + 7 = 0$ has no real solution. Hence the discriminant of the quadratic is strictly negative. – Deepak May 31 '15 at 9:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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