modulus in Inequality

[mimie]

How to change [MATH]\mid 4x^2\mid<1[/MATH] to [MATH]\mid x\mid[/MATH] form?
Can someone show me how? Thanks.

 

[pka]

How to change [MATH]\mid 4x^2\mid<1[/MATH] to [MATH]\mid x\mid[/MATH] form?
Can someone show me how?

First a question: Why was this posted in the Beginning Algebra forum?
Is the solution \(\displaystyle |x|<\tfrac{1}{2}~?\) Why or why not?

 

[mimie]

First a question: Why was this posted in the Beginning Algebra forum?
Is the solution \(\displaystyle |x|<\tfrac{1}{2}~?\) Why or why not?

Sorry if I posted this in wrong place. I am not sure where to post this question.
Your answer is correct. My question is how to get this answer.
I square root both side, [MATH]\mid \sqrt{4x^2} \mid<\pm \sqrt{1}[/MATH]then [MATH]\mid 2x \mid<\pm 1[/MATH][MATH]\mid x \mid<\pm \tfrac{1}{2}~[/MATH]I get plus and minus sign, then how to eliminate the minus sign?
Please point out my mistake, thanks.

 

[Deleted member 4993]

Sorry if I posted this in wrong place. I am not sure where to post this question.
Your answer is correct. My question is how to get this answer.
I square root both side, [MATH]\mid \sqrt{4x^2} \mid<\pm \sqrt{1}[/MATH]then [MATH]\mid 2x \mid<\pm 1[/MATH][MATH]\mid x \mid<\pm \tfrac{1}{2}~[/MATH]I get plus and minus sign, then how to eliminate the minus sign?
Please point out my mistake, thanks.

Absolute value cannot be negative "by definition".

 

[pka]

Sorry if I posted this in wrong place. I am not sure where to post this question.

The point was to find out why were you given this question. What do you know about absolute value? What methods can you use?

My question is how to get this answer. I square root both side, [MATH]\mid \sqrt{4x^2} \mid<\pm \sqrt{1}[/MATH] then [MATH]\mid 2x \mid<\pm 1[/MATH] [MATH]\mid x \mid<\pm \tfrac{1}{2}~[/MATH]

Frankly I see from the above that you know almost nothing about absolute value.
You were given that \(\displaystyle |4x^2|<1\) from which you should know that implies\(\displaystyle -1<4x^2<1\).
BUT for all \(\displaystyle x\) \(\displaystyle 4x^2\ge 0\) so you have \(\displaystyle 0\le 4x^2<1\)
which implies \(\displaystyle -\frac{1}{2}<x<\frac{1}{2}\) which translates to \(\displaystyle |x|<\frac{1}{2}\).
If you were expected to know how to work this question the you should have known these properties.
If you have never seen these, you need to complain to school authorities.

 

[Dr.Peterson]

How to change [MATH]\mid 4x^2\mid<1[/MATH] to [MATH]\mid x\mid[/MATH] form?

I square root both side, [MATH]\mid \sqrt{4x^2} \mid<\pm \sqrt{1}[/MATH]then [MATH]\mid 2x \mid<\pm 1[/MATH][MATH]\mid x \mid<\pm \tfrac{1}{2}~[/MATH]I get plus and minus sign, then how to eliminate the minus sign?

There are several ways I can think of to do this.

One mistake in your method is that taking the square root of both sides of an inequality does not always retain the direction of the inequality; for example, [MATH](-2)^2 \lt (-3)^2[/MATH] but [MATH]-2 \nless -3[/MATH]. One result is that statements like [MATH]x <\pm a[/MATH] are ambiguous at best, and really wrong: Does it mean "less than both a and -a", or "either less than a or less than -a"? What it would have to mean to be correct is "between -a and a".

You can take the square root when you know that all numbers involved are positive (as the square function is monotonically increasing for positive x). In this case, if you observe that [MATH]4x^2 = |2x|^2[/MATH], where the base is positive, we can take the positive square roots and obtain |2x| < 1, from which the conclusion follows.

 

[mimie]

You can take the square root when you know that all numbers involved are positive (as the square function is monotonically increasing for positive x). In this case, if you observe that [MATH]4x^2 = |2x|^2[/MATH], where the base is positive, we can take the positive square roots and obtain |2x| < 1, from which the conclusion follows.

[MATH]|4x^2|<1[/MATH][MATH]−1<4x^2<1 [/MATH][MATH]-\frac{1}{4}<x^2<\frac{1}{4}[/MATH][MATH]\sqrt{-\frac{1}{4}}<\sqrt{x^2}<\sqrt{\frac{1}{4}}[/MATH][MATH]\sqrt{x^2}<\sqrt{\frac{1}{4}}[/MATH], reject [MATH]\sqrt{-\frac{1}{4}}[/MATH] because it is not real
[MATH]|x|<\sqrt{\frac{1}{4}}[/MATH][MATH]|x|<\frac{1}{2}[/MATH]
Is it correct if I write it like this, please check my working. Thanks.

 

[Dr.Peterson]

What did I say about taking the square root of an inequality? It is not valid! You can't do it without imposing conditions, and paying attention. The fact that you ended up claiming that an imaginary number is less than a real number (which of course is absolute nonsense) should have told you not only to ignore the imaginary number, but that your reasoning was wrong.

Look at the graph of [MATH]y = x^2[/MATH]. If one point on the graph is lower than another ([MATH]a^2 < b^2[/MATH]), it does not follow that [MATH]a < b[/MATH]. That's only true if you also know that both a and b are non-negative.

One way to rescue your work (besides doing what I said before) is to recognize that [MATH]x^2[/MATH] is never negative, so [MATH]-\frac{1}{4} < x^2 < \frac{1}{4}[/MATH] implies that [MATH]0 \le x^2 < \frac{1}{4}[/MATH]. Then continue from there, knowing that if a and b are non-negative, it is true that [MATH]a^2 < b^2[/MATH] implies [MATH]a < b[/MATH].

But to my mind it's cleaner just to see at the start that [MATH]|4x^2| = 4|x|^2[/MATH].

 

[mimie]

One way to rescue your work (besides doing what I said before) is to recognize that [MATH]x^2[/MATH] is never negative, so [MATH]-\frac{1}{4} < x^2 < \frac{1}{4}[/MATH] implies that [MATH]0 \le x^2 < \frac{1}{4}[/MATH]. Then continue from there, knowing that if a and b are non-negative, it is true that [MATH]a^2 < b^2[/MATH] implies [MATH]a < b[/MATH].

But to my mind it's cleaner just to see at the start that [MATH]|4x^2| = 4|x|^2[/MATH].

Sorry for bothering you, can you please check both of my working? Any mistake?
Sorry, I still need to ask some noob questions.
Why [MATH]0 \le x^2 < \frac{1}{4}[/MATH] become [MATH] x^2 < \frac{1}{4}[/MATH] ?
Is it because both [MATH] x^2 [/MATH] and [MATH] \frac{1}{4}[/MATH] are more then [MATH] 0[/MATH] so no need to restate?
Thanks.

 

[Dr.Peterson]

The first method is fine, though I would explicitly state why it is valid to take the square root.

In the second, which appears to be based on what pka said earlier, you're right that you can ignore the comparison to 0 because it "goes without saying". There's never anything wrong with ignoring something; in effect, you going from "A and B are true" to "B is true" because all you care about is B. Everything there is fine, because by explicitly using the radical, you are using only the positive root.

The method we usually recommend for solving something like x^2 < 4 is to factor: x^2 - 4 < 0, then (x + 2)(x - 2) < 0, so either both factors are positive (x > 2), or both are negative (x < -2). In this case, you could treat it as |x|^2 - 4 < 0, and ignore the negative case.

As I said at some point, there are lots of ways to carry this out.