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?
On
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:
$$x<-\sqrt{10} \vee x>\sqrt{10}$$
On
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
On
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
On
Another (perhaps more systematic?) approach:
$$x^2 > 10 \Leftrightarrow |x| > \sqrt{10} \Leftrightarrow x > +\sqrt{10}\ \lor\ x < -\sqrt{10}$$
On
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.
On
$x^2>10$
$\sqrt{x^2} > \sqrt{10}$
$|x| > \sqrt{10}$
$ \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, $x<-\sqrt{10}$
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