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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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  • $\begingroup$ Try $x=-3$ and $x=-4$ and check your inequality for the negative square root. $\endgroup$ Commented Aug 15, 2014 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
    Commented Aug 15, 2014 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
    Commented Aug 16, 2014 at 19:55
  • $\begingroup$ Rather than saying that your inequality is "basic", it is better to say what it actually is. $\endgroup$
    – user147263
    Commented Aug 16, 2014 at 21:23

9 Answers 9

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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 $y=10$.

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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  • $\begingroup$ Very visual, very nice. $\endgroup$ Commented Aug 15, 2014 at 21:05
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    $\begingroup$ What software do you use to plot this function? $\endgroup$
    – matrix42
    Commented Aug 16, 2014 at 2:02
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    $\begingroup$ It's a free online tool: Desmos. $\endgroup$
    – rae306
    Commented Aug 16, 2014 at 7:27
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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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  • $\begingroup$ 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. $\endgroup$
    – Khallil
    Commented Aug 15, 2014 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
    Commented Aug 16, 2014 at 1:16
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Another (perhaps more systematic?) approach:

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

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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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    $\begingroup$ I like this. Very neat :) $\endgroup$
    – Mr Pie
    Commented Apr 7, 2018 at 10:06
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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 $x > \sqrt {10}$


Conversely,

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:

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

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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$$x>+\surd10\quad\text{or}\quad x<−\surd10.$$

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$x^2>10$

$\sqrt{x^2} > \sqrt{10}$

$|x| > \sqrt{10}$

$ \left\{ \begin{array}{l} x > \sqrt{10} &\mbox{, if $x \geq 0$}\\ -x>\sqrt{10} &\mbox{, if $x < 0$} \end{array} \right. \ $

$ \left\{ \begin{array}{l} x > \sqrt{10} &\mbox{, if $x \geq 0$}\\ x<-\sqrt{10} &\mbox{, if $x < 0$} \end{array} \right. \ $

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

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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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