Use Algebra To Solve Inequality

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harpazo

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Solve the inequality algebraically.

x^4 < 9x^2 < 0

x^2(x^2 - 9) < 0

x^2(x - 3) (x + 3) < 0

Stuck here....
 
Interpreting the problem exactly as written:

x^4 < 9x^2 < 0

We immediately know there's not going to be any real solutions to the problem because any (real number) squared must be positive, and 9 times a positive real number is still positive, so it cannot be the case that \(9x^2 < 0\) unless \(x\) is complex. Therefore, we can guarantee that \(x \ne 0\), such that dividing all sides of the inequality by \(x^2\) is a legal move. If we do that, we get \(x^2 < 9 < 0\) which is impossible.

On the other hand, your workings seem to indicate that the problem was actually:

\(\displaystyle x^4 \: {\color{red}{\mathbf{-}}} \: 9x^2 < 0\)

If that's the case, then everything you've done is good. Let's now temporarily suppose it was in equality and find the roots of:

\(\displaystyle x^2(x - 3) (x + 3) = 0\)

It should be clear that the roots are 0, 3, and -3. Next we can use this information to look at cases. What if \(x < -3\)? We'd have:

\(\displaystyle x^2 > 0 \\ x - 3 < 0 \\ x + 3 < 0 \\~\\
\text{{Positive}} \cdot \text{{Negative}} \cdot \text{{Negative}} = \text{{Positive}}\)
Thus, we can conclude that \(x^2(x-3)(x+3) > 0\) in this interval. Now let's consider the next interval. What if \(x\) is between the roots? That is, what if \(-3 < x < 0\)? I'll leave this and the rest of the intervals to you to work out, but see what you find when all is said and done.
 
In x^2(x - 3) (x + 3) → x^2 is always positive.

So you must have (x - 3) (x + 3) must be negative (<0) for x^2(x - 3) (x + 3) < 0

Now think.....
 
x^4 < 9x^2 < 0

x^2(x^2 - 9) < 0

x^2(x - 3) (x + 3) < 0

x^2 < 0

sqrt{x^2} < sqrt{0}

x < 0

x - 3 < 0

x < 3

x + 3 < 0

x < -3

Replacing x with 0, -3 and 3 leads to 0 < 0, which is a false statement. I say no solutions exist.
 
x^4 < 9x^2 < 0

x^2(x^2 - 9) < 0

x^2(x - 3) (x + 3) < 0

x^2 < 0

sqrt{x^2} < sqrt{0}

x < 0

x - 3 < 0

x < 3

x + 3 < 0

x < -3

Replacing x with 0, -3 and 3 leads to 0 < 0, which is a false statement. I say no solutions exist.
That is incorrect.
 
Solve the inequality algebraically.

x^4 < 9x^2 < 0

x^2(x^2 - 9) < 0
Are you sure there isn't a typo here? The second line does not follow from the first. Could the original be
\(\displaystyle x^4 - 9x^2 < 0\)

The problem looks more realistic this way.

-Dan
 
Are you sure there isn't a typo here? The second line does not follow from the first. Could the original be
\(\displaystyle x^4 - 9x^2 < 0\)

The problem looks more realistic this way.

-Dan

It may be a typo. I will look in the textbook again.
 
Are you sure there isn't a typo here? The second line does not follow from the first. Could the original be
\(\displaystyle x^4 - 9x^2 < 0\)

The problem looks more realistic this way.

-Dan

Yes, you got it. It is x^4 - 9x^2 < 0....
 
Thank you everyone. The work was completed on paper and double checked online.
 
It's unfortunate that you skipped post #2.

It's disappointing that you're still not proofreading your exercise statements. (You said you would.) It's easy to do. It builds character.

\(\;\)

I gotta pay more attention to replies.
 
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