Solving quadratic inequality $x^2 > 0$

Question

This particular inequality has me puzzled: $$ x^2 > 0 $$

Graphically it is simple to solve, as you can see the curve only touches $0$ but stretches off in both negative and positive directions as shown here:

However I cannot express in a calculation how I can come up with $$x > 0 \lor x < 0. $$

For instance if you attempt to solve the inequality by finding the square root on either side, you end up with: $$ x > \pm\sqrt{0} $$ Since zero is neither positive or negative, this doesn't really make much sense to me. Or is there a different way to interpret this?

Thanks

Answer 1

Caution,

$$a^2>b$$ does not imply $$a>\pm\sqrt b.$$

But $$\pm a>\sqrt b$$ is correct. (With a somewhat sloppy notation.)

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More rigorously

$$a^2>b\\\iff (a-\sqrt b)(a+\sqrt b)>0\\\iff (a>\sqrt b\land a>-\sqrt b)\lor (a<\sqrt b\land a<-\sqrt b)\\\iff a>\sqrt b\lor a<-\sqrt b.$$

Answer 2

We have that

• for $x=0 \implies x^2=0$

• for $x \neq 0 \implies x^2>0$

and the proof is complete by exhaustion.

Following your idea, using that $\sqrt{x^2}= |x|$, we can take the square root both sides to obtain

$$x^2>0 \iff \sqrt{x^2}>\sqrt 0 \iff |x|>0$$

which is always true for $x\neq 0$.

Answer 3

When you applied the square root on each side you shouldn't have "cancelled out" the square root with the square power.

I think you are used to do something like this: $x^2=a$ then $x=\pm\sqrt a$, which is fine, but doesn't work with inequalities. Instead you should be thinking it more like this: if $x^2=a$ then $|x|=+\sqrt a$, which also works with inequalities.

For example: if $x^2=4$ usually we say straightforward $x=\pm2$, but you could do a middle step, which is $|x|=2$, and then think that the absolute value just gets rid of the minus sign, so your solutions can be $x=\pm2$.

This is the way you should do the inequality: if $x^2>0$ then $|x|>0$, so you can have either $x>0$ or $x<0$, since the absolute value will get rid of the minus sign.

Answer 4

$$x^2=0\iff x=0.$$ Hence

$$x^2>0\iff x\ne0.$$

Answer 5

As a general rule, I have discovered the following:

$$\mathrm{For\:}u^n\:>\:0\mathrm{,\:if\:}n\:\mathrm{is\:even}\mathrm{\:then\:}u\:<\:0\quad \mathrm{or}\quad \:u\:>\:0$$

Courtesy of Symbolab. https://www.symbolab.com/solver/inequalities-calculator/x%5E%7B2%7D%3E0