I am very confused whenever I encounter an inequality . Like suppose $ \frac{| \sqrt {1 -4x^{2}}|}{|x|} \le 1 $ Now I am confused whether to square or not . I am very well aware of the rule at we do not square if the sign of $x$ is not known . But what if the range of $x$ is given as $ \frac {-1}{2} \le x \le \frac 12 $ . Now here I am confused whether to square or not as $ x $ can take either negative value. If someone could give me a link to a set of rules of inequalities under such conditions , it would be appreciated .

  • 1
    $\begingroup$ Note: $|x|$ and $\sqrt{1-4x^2}$ are non-negative $\endgroup$ Aug 4, 2019 at 16:05
  • $\begingroup$ $ \frac{| \sqrt {1 -4x^{2}}|}{|x|} \le 1 \implies 1-4x^2\le x^2 \implies 1\le 5x^2\implies \dfrac15\le x^2\implies x\le\dfrac{-1}{\sqrt5}$ or $x\ge\dfrac1{\sqrt5}$ $\endgroup$ Aug 4, 2019 at 16:13

3 Answers 3


In general equalities, where you're not sure of the sign, you can still square, but just have to realize you may be introducing extraneous solutions since, for the example you stated, $(-x)^2 = x^2$. With inequalities, you have to be more careful. If they are both negative, you need to reverse the inequality, and you also need to do more checks if the sign(s) are different (e.g., as asked, and with several good answers, at MSE's Squaring both sides of an inequality of opposite signs). However, in your particular example, as J. W. Tanner's question comment indicated, both $|x|$ and $\sqrt{1-4x^2}$ are non-negative in your inequality of

$$\frac{\sqrt{1 - 4x^2}}{|x|} \le 1 \tag{1}\label{eq1}$$

As for possibly an extraneous solution being added, this can be handled by just confirming your solution(s) work with the original inequality, with this being a good idea in general to help check for any mistake you perhaps have made.

Regarding the range $\frac{-1}{2} \le x \le \frac{1}{2}$, note that for the denominator, you get $|x| \le \frac{1}{2}$. Also, the numerator already has $x^2$ in it, so the sign is not implicit in it. As such, squaring here and solving for the inequality should not add any extraneous solutions. As J. W. Tanner's question comment indicates, squaring both sides, getting $x^2$ by itself, dividing and taking the square roots gives the solution of $x \le \frac{-1}{\sqrt{5}}$ or $x \ge \frac{1}{\sqrt{5}}$. Combined with the conditions for the numerator to be real gives

$$\frac{-1}{2} \le x \le \frac{-1}{\sqrt{5}} \; \text{ or } \; \frac{1}{\sqrt{5}} \le x \le \frac{1}{2} \tag{2}\label{eq2}$$

You can also combine these $2$ inequalities, if you wish, to get the equivalent

$$\frac{1}{\sqrt{5}} \le |x| \le \frac{1}{2} \tag{3}\label{eq3}$$

To confirm this range works, note at $|x| = \frac{1}{\sqrt{5}}$ that the LHS of \eqref{eq1} is $1$. As $|x|$ increases to $\frac{1}{2}$, the numerator is decreasing towards $0$, while the denominator is increasing. Thus, the fraction is going towards $0$ and, thus, remains $\le 1$, so the inequality in \eqref{eq1} always holds.

Regarding links to a general set of rules for inequalities, doing an online search reveals a number of fairly good ones, such as Rules for solving inequalities. As for your specific concerns about squaring an inequality involving negative quantities (or, at least, possibly negative quantities), there are multiple online resources which you can use, with the MSE post I referenced above being a good one, as well as others like Should the sign be reversed if I square both sides of an inequality?.


Note: All of what follows is based on the axiom that if $a < b$ (regardless of the signs of either) and $c > 0$ then $ac < bc$ and likewise if $d < 0$ then $ad > bd$.

I'm assuming that is known and familiar.

You can do whatever you want. You can take them out to dinner for all I care. You just have to know what you are doing.

If you have $x < 3$ and $x$ is negative, say, then you can not say $x^2 < 9$ because: If we multiple both sides of the equation $x < 3$ by $x$ we flip the inequality to get $x^2 > 3x$ and if we multiply both sides of $x < 3$ by $3$ we get $3x < 9$. So we get $x^2 > 3x$ and $9> 3x$ but we have no way of comparing $x^2$ to $9$.

That's not wrong. It's just not what we'd expect if we were stupid and just monkeying around with rules we didn't understand and "squaring both sides" without thinking.

Anyway.... If we do know the signs say $0 < x < 3$ then we can square both sides because $x< 3$ and $x > 0$ means $x^2 < 3x$ and $x < 3$ and $3>0$ so $3x < 9$. So $0 < x^2 < 9$.

We can actually do if $-a < x < b$ where $-a < 0$ and $b < 0$ we can get.

If $x < 0$ then $-a < x \implies -ax > x^2; (-a)^2 = a^2 > -ax$ so $x^2 < a^2$. But if $x > 0$ we have $-ax < x^2$ and $(-a)^2 =a^2 > -ax$ and we have no way of compare $a^2$ to $x^2$.

And if $x < 0$ we get $x < b$ means $x^2 > bx$ and $bx < b^2$ but we have no way to compare $x^2$ to $b^2$. But if $x > 0$ we have $x^2 < b^2$.

So we have either $x^2 < a^2$ or $x^2 < b^2$ so we do have $x^2 < \max(a^2,b^2)$.

And note. If $a = b$ and you have $-a < x < a$ we have $0 \le x^2 < a^2$.


But to your problem.....

Well, yours is simple. $|K| \ge 0$ so you have

$0 \le \frac {|1-4x^2|}{|x|} \le 1$

So we know the signs and we can square both sides.

$0 \le \frac {1-4x^2}{x^2} \le 1$ so

$0 \le 1-4x^2 \le x^2$ so

$4x^2 \le 1 \le 5x^2$ so

$0 \le 4x^2 \le 1 \le 5x^2$ so

$0 \le 2|x| \le 1 \le \sqrt 5|x|$. So $|x| \le \frac 12$ and $\frac 1{\sqrt 5} \le |x|$ so $\frac 1{\sqrt {5} }\le |x| \le \frac 12$.

We don't know the sign of $x$ so

Either $-\frac 12 \le x \le -\frac 1{\sqrt 5}$ or $\frac 1{\sqrt {5} }\le x \le \frac 12$

  • $\begingroup$ Very much thank you to all of the people who have replied here this just solved all my doubts regarding inequalities . $\endgroup$ Aug 5, 2019 at 15:36

As a general rule, when you have to solve an inequality in the form $P(x)\geq,\leq,>,<Q(x)$, where $P(x)$ and $Q(x)$ are polynomials in $x$, you have to resolve this inequality system: $$\left\{\begin{matrix} P(x)\geq0 \\Q(x)\geq0 \\ P(x)^2\geq,\leq,>,<Q(x)^2 \end{matrix}\right.$$

In your case $P(x)=\sqrt{1-4x^2}$ and $Q(x)=|x|$ (I can multiply both sides by $|x|$ since it's always positive); so:

$$\left\{\begin{matrix} \sqrt{1-4x^2}\geq0 \\|x|\geq0 \\ \sqrt{1-4x^2}\leq|x| \end{matrix}\right.$$

In other words:

$$\left\{\begin{matrix} \forall x\in R \\1-4x^2\leq x^2 \end{matrix}\right.$$

The solution is $x\leq -\frac{1}{\sqrt{5}} \vee x\geq \frac{1}{\sqrt{5}}$.

I hope it has helped you...


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