Solve the inequality: $$x^2>10$$

How am I supposed to do this? It doesn't make sense when I take into account that if $x^2=10$ then $x=+\sqrt{10}$ and $x=-\sqrt{10}$

But how am I supposed to apply this to an inequality, I would get $x>\sqrt{10}$ and $x>-\sqrt{10}$

But for some reason this just doesn't make sense to me. Can someone explain it to me mathematically, instead of just having to memorize these kinds of things?

  • $\begingroup$ Try $x=-3$ and $x=-4$ and check your inequality for the negative square root. $\endgroup$ – Mark Bennet Aug 15 '14 at 20:44
  • $\begingroup$ It would be $x > \sqrt{x}$ or $x < - \sqrt{10}$ i.e. $x\in\mathbb{R} - [-\sqrt{10},\sqrt{10}]$ $\endgroup$ – Darth Geek Aug 15 '14 at 20:44
  • $\begingroup$ $x=+\sqrt{10}\text{ and }x=-\sqrt{10}$ is a contradiction. Your solution of $x^2=10$ should instead be $x=+\sqrt{10}\text{ or }x=-\sqrt{10}$. $\endgroup$ – Ruslan Aug 16 '14 at 19:55
  • $\begingroup$ Rather than saying that your inequality is "basic", it is better to say what it actually is. $\endgroup$ – user147263 Aug 16 '14 at 21:23

The only problem you have here lies in understanding the difference between and and or.

Concerning the equation $x^2=10$, your statement $$x=+\surd10\quad\text{ and }\quad x=−\surd10$$ is self-contradictory and false. Correct is:$$x=+\surd10\quad\text{or}\quad x=−\surd10.$$Similarly, as you already point out regarding the inequality $x^2>10$, the statement $$x>+\surd10\quad\text{ and }\quad x<−\surd10$$makes little sense. The correct version is$$x>+\surd10\quad\text{or}\quad x<−\surd10.$$


Sketch the graph of $x^2$ (it's a parabola opening upwards with vertex in $(0,0)$) and sketch the line $y=10$.

enter image description here

They intersect in $x=-\sqrt{10}$ and $x=\sqrt{10}$, and the sketch immediately gives the solution to the inequality:

$$x<-\sqrt{10} \vee x>\sqrt{10}$$

  • $\begingroup$ Very visual, very nice. $\endgroup$ – André Nicolas Aug 15 '14 at 21:05
  • 2
    $\begingroup$ What software do you use to plot this function? $\endgroup$ – mathe Aug 16 '14 at 2:02
  • 1
    $\begingroup$ It's a free online tool: Desmos. $\endgroup$ – rae306 Aug 16 '14 at 7:27

Using $a^2 - b^2 = (a+b)(a-b)$, we get $(x-\sqrt{10})(x+\sqrt{10}) > 0$, which mean $x+\sqrt{10}$ and $x-\sqrt{10}$ have the same sign

  • $\begingroup$ Nice answer! This is how I like to do it. Just using the basic properties of numbers is more appealing to me than using the graphical approach. $\endgroup$ – Kari Aug 15 '14 at 21:30
  • $\begingroup$ When solving polynomial inequalities, it's often best to collect and factor. Then sorting each factor from least to greatest (when all of the factors are of degree one) will lead you to find the correct intervals. $\endgroup$ – SimonT Aug 16 '14 at 1:16

Another (perhaps more systematic?) approach:

$$x^2 > 10 \Leftrightarrow |x| > \sqrt{10} \Leftrightarrow x > +\sqrt{10}\ \lor\ x < -\sqrt{10}$$


Another way to see it algebraicaly/analyticaly is this:

$(-x)^2 = x^2 > 10$ then you have 2 conditions:

a) $-x > \sqrt{10} \implies x < -\sqrt{10}$

b) $x > \sqrt{10}$

which both provide solutions

  • $\begingroup$ I like this. Very neat :) $\endgroup$ – Feeds Apr 7 '18 at 10:06

Here is how I think about problems like this:

If $x^2 > 10$, then

$x^2 - 10 > 0$

$(x + \sqrt {10})(x - \sqrt {10}) > 0 $

$(x + \sqrt {10})$ and $(x - \sqrt {10})$ are either both positive or both negative.

Thus, when they are both positive and we divide both sides by one of the terms:

$x + \sqrt {10} > 0$ and $x - \sqrt {10} > 0$

This simplifies to:

$x > \sqrt {10}$

When they are both negative and we divide both sides by one of the terms:

$x + \sqrt {10} < 0$ and $x - \sqrt {10} < 0$

This simplifies to:

$x < -\sqrt {10}$

The final solution is:

$x < -\sqrt {10}$ or $x > \sqrt {10}$


if $x^2 < 10$, then

$x^2 - 10 < 0$

$(x - \sqrt{10})(x + \sqrt{10}) < 0$

We know that one of the terms, $(x - \sqrt{10})$ or $(x + \sqrt{10})$, is negative.

Since $(x - \sqrt{10})$ is always smaller, we know that it is the negative term.

Thus, when we divide both sides by $(x - \sqrt{10})$ we get:

$x + \sqrt{10} > 0$

This simplifies to:

$x > -\sqrt{10}$

when we divide both sides by $(x + \sqrt{10})$ we get:

$x - \sqrt{10} < 0$

This simplifies to:

$x < \sqrt{10}$

The final solution is:

$-\sqrt {10} < x < \sqrt {10}$


Expanding upon Petite Etincelle's answer, but for LHS < 0 inequalities. This is to solve $x^2 < 10$ instead.

First factorize the inequality so that the RHS is 0. This gives $(x-\sqrt{10})(x+\sqrt{10}) < 0$.

Since $(x-\sqrt{10})(x+\sqrt{10})$ is less than 0 (therefore a negative number), it follows that $x-\sqrt{10}$ and $x+\sqrt{10}$ must have different signs. (Since $-a \cdot b < 0$ and $a \cdot -b < 0$). To find $x$, we can write two statements:

1) $x-\sqrt{10}<0$ and $x+\sqrt{10}>0$ (Therefore, $x<\sqrt{10}$ and $x>-\sqrt{10}$)
2) $x-\sqrt{10}>0$ and $x+\sqrt{10}<0$ (Therefore, $x>\sqrt{10}$ and $x<-\sqrt{10}$)

From there, we can see that statement 2 is clearly wrong. $x$ cannot be larger than $\sqrt{10}$ and be smaller than $-\sqrt{10}$ at the same time. Statement 1 is clearly the correct solution.


One way to think about this is as a graph. What happens if you plot $y= x^2$? You get a parabola. Now, for which values of $x$ is $y > 10$? The answer is $x>\sqrt{10}$ and $x<\sqrt{10}$.

You can see a graph like this here: http://www.wolframalpha.com/input/?i=x%5E2+%3D+10


A quadratic equation usually has two solutions (except x2=0 etc.). Consequently, a quadratic inequality such at this one has two sets of solutions, in this case one positive and one negative.


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