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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 '14 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 '14 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 '14 at 19:55
Rather than saying that your inequality is "basic", it is better to say what it actually is. – ᴡᴏʀᴅs Aug 16 '14 at 21:23
up vote 7 down vote accepted

Another (perhaps more systematic?) approach:

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

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@Jason: This method is not correct as it does not verify that the solutions found are really solutions. By the same logic we could write $x^2 > 10 \Rightarrow x^2 > 0 \Rightarrow x \ne 0$! Only if we replace the implications with equivalences is it correct. – user21820 Dec 4 '14 at 4:46
@user21820: I'm not sure I understand what is wrong with saying your statement of $x^2 > 10 \implies x^2 > 0 \implies x \neq 0$... the statement is perfectly correct. – Mehrdad Dec 4 '14 at 5:54
Exactly, the statement is perfectly correct but does not solve for $x$ in $x^2 > 10$! Likewise this solution as written does not solve the question. – user21820 Dec 4 '14 at 5:59
@user21820: I don't understand what you're expecting to see, but I suggest writing your own answer the way you expect it to be. – Mehrdad Dec 4 '14 at 6:01
Hmm then I'm afraid you do not understand proofs and the meaning of implication. When you learn mathematical logic, then you will understand why this presentation is improper, even if you give a correct interpretation of it, which most students surely don't. – user21820 Dec 4 '14 at 7:32

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:

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

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

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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. – Kari Aug 15 '14 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 '14 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:

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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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