# Thread: Dividing by variable and taking square root of inequalities

1. ## Dividing by variable and taking square root of inequalities

How does one solve $\displaystyle 4x \le x^3$?

This is actually a two part question. First, you can't really divide both sides by $\displaystyle x$, since it could be negative (or even zero).

I figure we can do this problem by separating it into three cases. First, let $\displaystyle x=0$ and see if it is a solution (it is). Then let $\displaystyle x>0$ and $\displaystyle x<0$.

When $\displaystyle x>0$, then $\displaystyle 4 \le x^2$. Then we can take the square root of both sides and get $\displaystyle \pm 2 \le x$. If we interpret that as $\displaystyle 2 \le \left| x \right|$ and then write it as $\displaystyle x \le -2$ or $\displaystyle x \ge 2$, we have the correct full answer. But why is that the correct interpretation? How does someone explain that to his daughter?

Last case is to let $\displaystyle x<0$. Then $\displaystyle 4 \ge x^2$ and $\displaystyle \pm 2 \ge x$. Using the same logic as last paragraph, that would indicate that $\displaystyle 2 \ge \left| x \right|$, or that $\displaystyle -2 \le x \le 2$, which is wrong.

What's the correct way to solve $\displaystyle 4x \le x^3$?

2. ## Re: Dividing by variable and taking square root of inequalities

How does one solve $\displaystyle 4x \le x^3$?

This is actually a two part question. First, you can't really divide both sides by $\displaystyle x$, since it could be negative (or even zero).

I figure we can do this problem by separating it into three cases. First, let $\displaystyle x=0$ and see if it is a solution (it is). Then let $\displaystyle x>0$ and $\displaystyle x>0$.
You are right about separating it in three cases (the last one should say x < 0). For each of those cases you need to solve the inequality. In each case, you'll get a set of solutions. Then you need to take the intersection with the corresponding set of x's: take only positive solutions in the second case and only negative solutions in the third case. Written in terms of set operations, the final set of solutions is $\displaystyle \{0\}\cup (A\cap\{x\mid x>0\})\cup (B\cap\{x\mid x<0\})$ where A and B are sets of solutions in the second and third case, respectively. Equivalently, x is a solution to the original equation iff (1) x = 0, or (2) x > 0 and x is a solution in the second case, or (3) x < 0 and x is a solution in the third case. Note the distribution of "and"s and "or"s.

In practice, I would write the three cases separately and clearly note the side condition (x = 0, x > 0, x < 0) for each case. After solving each case, it is important not to forget and combine the solution with the side conditions using "and," and then combine the solutions for different cases using "or."

When $\displaystyle x>0$, then $\displaystyle 4 \le x^2$. Then we can take the square root of both sides and get $\displaystyle \pm 2 \le x$.
The last formula is ambiguous: does it mean $\displaystyle -2 \leq x\text{ and }2 \leq x$ or $\displaystyle -2 \leq x\text{ or }2 \leq x$? And both of these interpretations are wrong. In fact, who said that it is allowed to convert $\displaystyle a^2\leq b^2$ into $\displaystyle a\le b$? I would just memorize that

$\displaystyle a^2\le b^2$ is equivalent to $\displaystyle |a|\le|b|$

In this case, it can be verified in two ways. First,

\displaystyle \begin{align*}4 \le x^2&\iff x^2-4\ge0\\&\iff (x-2)(x+2)\ge0\\& \iff (x-2\ge0\text{ and }x+2\ge0)\text{ or }(x-2\le0\text{ and }x+2\le0)\\& \iff x\ge2\text{ or }x\le-2\\&\iff |x|\ge 2\end{align*}

Alternatively, you can move a point along the x-axis from 0 to the right or to the left and see where the inequality becomes true.

Remember that you need to take only the positive solutions from the second case.

Last case is to let $\displaystyle x<0$. Then $\displaystyle 4 \ge x^2$ and $\displaystyle \pm 2 \ge x$. Using the same logic as last paragraph, that would indicate that $\displaystyle 2 \ge \left| x \right|$, or that $\displaystyle -2 \le x \le 2$, which is wrong.
No, it is not wrong. You need to take the negative x satisfying $\displaystyle -2 \le x \le 2$.

3. ## Re: Dividing by variable and taking square root of inequalities

Originally Posted by emakarov
You are right about separating it in three cases (the last one should say x < 0).
Edited my OP.

Good explanation. Just the way I needed it.

For the second case, $\displaystyle x \ge 0$, you get to
$\displaystyle x\ge2\text{ or }x\le-2$.

Because this is the [tex]x>0[\tex] case (and by your earlier comment), I think you forgot to discard $\displaystyle x\le-2$, leaving you with $\displaystyle x \ge 2$? (There's so many parts, it's hard to get everything exactly right, while at the same time typing it up in latex).

For the last case, $\displaystyle x<0$, using the same method you outlined for the second case, we started with $\displaystyle x^2 \le 4$. I end up with $\displaystyle -2 \le x \le 2$, which really means $\displaystyle -2 \le x <0$

So the final solution becomes $\displaystyle [-2,0] \cup [2, \infty)$.

Checks (since I always encourage them, I figure I should write them up, too!):
$\displaystyle 4x \le x^3$
$\displaystyle 4(-3)=-12 \not\leq -27 = (-3)^3$
$\displaystyle 4(-2)= -8 \leq -8 = (-2)^3$
$\displaystyle 4(-1)=-4 \leq -1 = (-1)^3$
$\displaystyle 4(0)=0 \leq 0 = (0)^3$
$\displaystyle 4(1)=4 \not\leq 1 = (1)^3$
$\displaystyle 4(2)=-8 \leq 8 = (2)^3$
$\displaystyle 4(3)=12 \leq 27 = (3)^3$

4. ## Re: Dividing by variable and taking square root of inequalities

Because this is the [tex]x>0[\tex] case (and by your earlier comment), I think you forgot to discard $\displaystyle x\le-2$, leaving you with $\displaystyle x \ge 2$?
No, I wrote, "Remember that you need to take only positive solutions from the second case."

For the last case, $\displaystyle x<0$, using the same method you outlined for the second case, we started with $\displaystyle x^2 \ge 4$. I end up with $\displaystyle -2 \le x \le 2$, which really means $\displaystyle -2 \le x \le 0$
This should say $\displaystyle x^2 \le 4$. Also, strictly speaking, the solution to the last case is $\displaystyle -2 \le x < 0$, but with x = 0 from the first case this indeed makes $\displaystyle -2 \le x \le 0$.

So the final solution becomes $\displaystyle [-2,0] \cup [2, \infty]$.
Yes. Usually ∞ is followed by ) instead of ] to indicate that infinity is not included in the final answer since it is not a real number.

5. ## Re: Dividing by variable and taking square root of inequalities

Originally Posted by emakarov
No, I wrote, "Remember that you need to take only positive solutions from the second case."

This should say $\displaystyle x^2 \le 4$. Also, strictly speaking, the solution to the last case is $\displaystyle -2 \le x < 0$, but with x = 0 from the first case this indeed makes $\displaystyle -2 \le x \le 0$.

Yes. Usually ∞ is followed by ) instead of ] to indicate that infinity is not included in the final answer since it is not a real number.
Bunch of typos. I fixed them, too.