Abs Value Inequality with a Squared Term

opus

Gold Member
1. The problem statement, all variables and given/known data
$\left|\left(\frac{x}{2}\right)^2\right| < 1$

2. Relevant equations

3. The attempt at a solution
The absolute value situation is throwing me off for some reason. Would it be correct to split this into two equations?

$-\left(\frac{x^2}{4}\right) > -1$ and $\frac{x^2}{4} < 1$?
I dont think this is right as the first equation would yield $x^2 < -4$

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StoneTemplePython

Gold Member
The absolute value situation is throwing me off for some reason.

question
:
can you interchange the absolute value and the squaring? I.e. does

$\left|\left(\frac{x}{2}\right)^2\right| = \left|\left(\frac{x}{2}\right)\right|^2 = \left(\frac{x}{2}\right)^2$

?

• opus

andrewkirk

Homework Helper
Gold Member
If $x$ must be real then the problem is trivial, since the square of any real number is positive. So we can drop the original absolute value signs to get $x^2<4$. What interval on the real number line contains all numbers whose square is less than 4?

It becomes more interesting when $x$ can be complex, so that the $|\cdot |$ signs represent modulus rather than absolute value. Then $x$ is any number inside the circle of radius 2, centred at the origin of the complex plane.

• Delta2 and opus

SammyS

Staff Emeritus
Homework Helper
Gold Member
1. The problem statement, all variables and given/known data
$\left|\left(\frac{x}{2}\right)^2\right| < 1$
I suppose the problem statement is missing the fact that you are asked to solve the given inequality. Yes, it is an inequality, not an equation.
2. Relevant equations

3. The attempt at a solution
The absolute value situation is throwing me off for some reason. Would it be correct to split this into two equations?

$-\left(\frac{x^2}{4}\right) > -1$ and $\frac{x^2}{4} < 1$?
I don't think this is right as the first equation would yield $x^2 < -4$
You're right, it's not right.

That first inequality, $\ -\left(\frac{x^2}{4}\right) > -1\,,\$ simplifies to $\ \left(\frac{x^2}{4}\right) < 1 \,,\$ which is the second inequality.

The piece-wise definition of absolute value can be helpful in showing that @StoneTemplePython's simplification is valid.
$|u| = \begin{cases} u & \text{if } x \geq 0 \\ -u & \text{if } u < 0 \end{cases}$

Apply that to $\left|\left(\frac{x}{2}\right)^2\right|$ and ask yourself if $\left(\frac{x}{2}\right)^2$ can be negative.

The real chore is to solve $x^2 < 4$ for $x$ .

• opus

opus

Gold Member

question
:
can you interchange the absolute value and the squaring? I.e. does

$\left|\left(\frac{x}{2}\right)^2\right| = \left|\left(\frac{x}{2}\right)\right|^2 = \left(\frac{x}{2}\right)^2$

?
I want to say that if we have an even power, then yes we can.
If $x$ must be real then the problem is trivial, since the square of any real number is positive. So we can drop the original absolute value signs to get $x^2<4$. What interval on the real number line contains all numbers whose square is less than 4?

It becomes more interesting when $x$ can be complex, so that the $|\cdot |$ signs represent modulus rather than absolute value. Then $x$ is any number inside the circle of radius 2, centred at the origin of the complex plane.
(-2,2) would be the answer then. And your second part sounds interesting.
I suppose the problem statement is missing the fact that you are asked to solve the given inequality. Yes, it is an inequality, not an equation.

You're right, it's not right.

That first inequality, $\ -\left(\frac{x^2}{4}\right) > -1\,,\$ simplifies to $\ \left(\frac{x^2}{4}\right) < 1 \,,\$ which is the second inequality.

The piece-wise definition of absolute value can be helpful in showing that @StoneTemplePython's simplification is valid.
$|u| = \begin{cases} u & \text{if } x \geq 0 \\ -u & \text{if } u < 0 \end{cases}$

Apply that to $\left|\left(\frac{x}{2}\right)^2\right|$ and ask yourself if $\left(\frac{x}{2}\right)^2$ can be negative.

The real chore is to solve $x^2 < 4$ for $x$ .
Ah that piecewise comment helps. Thanks!

• Delta2

"Abs Value Inequality with a Squared Term"

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