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?
$\endgroup$ 49 Answers
$\begingroup$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:
$$x<-\sqrt{10} \vee x>\sqrt{10}$$
$\endgroup$ 3 $\begingroup$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
$\endgroup$ 2 $\begingroup$Another (perhaps more systematic?) approach:
$$x^2 > 10 \Leftrightarrow |x| > \sqrt{10} \Leftrightarrow x > +\sqrt{10}\ \lor\ x < -\sqrt{10}$$
$\endgroup$ 0 $\begingroup$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
$\endgroup$ 1 $\begingroup$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:
$-\sqrt {10} < x < \sqrt {10}$
$\endgroup$ $\begingroup$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$$x>+\surd10\quad\text{or}\quad x<−\surd10.$$
$\endgroup$ $\begingroup$$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, $x<-\sqrt{10}$
$\endgroup$ $\begingroup$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:
$\endgroup$ $\begingroup$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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