# How to solve the inequality $x^2>10$ using square roots?

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?

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

Another (perhaps more systematic?) approach:

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

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

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

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

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Very visual, very nice. –  André Nicolas Aug 15 at 21:05
What software do you use to plot this function? –  mathe Aug 16 at 2:02
It's a free online tool: Desmos. –  Rainier van Es Aug 16 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

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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. –  Khallil Aug 15 at 21:30
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. –  SimonT Aug 16 at 1:16

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

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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

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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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