How Do You Solve the Inequality ##x^2<4## for Negative Values of ##x##?

[RChristenk]

Homework Statement

Solve $x^2<4$

Relevant Equations

Inequalities

Just by inspection, the answer is obviously $-2<x<2$. But I tried calculating it step by step and couldn't get the negative portion of the inequality.

For $x>0$, $x^2<4 \Rightarrow x<2$ . Hence $0<x<2$.

For $x<0$, $x^2<4 \Rightarrow x>2$. I flipped the inequality because $x<0$. But then I get a contradiction: $x<0$ and $x>2$.

So how do I calculate for the negative part of the answer?

 

[fresh_42]

I would build $0>x^2-4=(x-2)(x+2)$ and examine the factors.

 

[FactChecker]

For $x<0$, $x^2<4 \Rightarrow x>2$. I flipped the inequality because $x<0$.

That is the rule for division by a negative number. You did not divide by $x$. If you did, you would have $x \gt 4/x$, which is correct.

 

[Mark44]

@RChristenk there are a couple rules for inequalities you should memorize. In the following it's assumed that $a \ge 0$.
$x^2 < a \Rightarrow -\sqrt a < x < \sqrt a$
$x^2 > a \Rightarrow x < -\sqrt a \text{ OR } x > \sqrt a$

As already mentioned, one can prove these by factoring $x^2 - a$ and investigating when the product of the factors is negative or when it is positive.

 

[PeroK]

Homework Statement: Solve $x^2<4$
Relevant Equations: Inequalities

Just by inspection, the answer is obviously $-2<x<2$. But I tried calculating it step by step and couldn't get the negative portion of the inequality.

For $x>0$, $x^2<4 \Rightarrow x<2$ . Hence $0<x<2$.

For $x<0$, $x^2<4 \Rightarrow x>2$. I flipped the inequality because $x<0$. But then I get a contradiction: $x<0$ and $x>2$.

So how do I calculate for the negative part of the answer?

There are six cases:
$$x < -2 \implies x^2 > 4$$$$x = -2 \implies x^2 = 4$$$$-2 < x \le 0 \implies x^2 < 4$$$$0 < x < 2 \implies x^2 < 4$$$$x = 2 \implies x^2 = 4$$$$x > 2 \implies x^2 > 4$$This can be seen by drawing the graph of $y = x^2$:

 

[docnet]

For $x<0$, $x^2<4 \Rightarrow x>2$. I flipped the inequality because $x<0$. But then I get a contradiction: $x<0$ and $x>2$.

So how do I calculate for the negative part of the answer?

You assumed $x<0$ and then contradicted your assumption when you stated $x>2$.

If $x<0$, you should have
\begin{align*}x^2<4 \Rightarrow & \,x<2 \Rightarrow -x>-2.\end{align*}
$-x$ is a positive quantity so the last inequality gives the lower bound of $(-2,2)$.

But it would be much clearer to keep $x$ the same: you have $f(x)=x^2=(-x)^2=f(-x)$ so that $-x<2$ and $x<2$ are the two possibilities. The first case gives $-2<x$, the second case gives $x<2$ and the two cases combine to give the interval $-2<x<2$.

 

[vela]

For $x>0$, $x^2<4 \Rightarrow x<2$ . Hence $0<x<2$.

For $x<0$, $x^2<4 \Rightarrow x>2$. I flipped the inequality because $x<0$. But then I get a contradiction: $x<0$ and $x>2$.

So how do I calculate for the negative part of the answer?

One thing you should learn is that $\sqrt {x^2} = \lvert x \rvert$. A common misconception is thinking $\sqrt{x^2}=x$, but that's only true if $x \ge 0$. Taking the square root of the original inequality, you should get $\lvert x \rvert < 2$. Now you break it into the two cases, $x\ge 0$ and $x<0$.