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For what real values of x is -4 < x^4 + 4x^2 < 21 satisfied? Express your answer in interval notation.

 Jul 1, 2024
edited by AHGSIGMA  Jul 1, 2024

Best Answer 

 #1
avatar+1267 
+1

Let's note something really important before we begin. 

To solve this problem, we must first do something quite important.

It's crucial that we split this inequality into two seperate ones before combining them back again. Thus, we have

\(x^4+4x^2>-4\\ x^4+4x^2<21\)

Now, note something about the first inequality. 

We have the terms x^4 and x^2 which is guarenteed to be positive. Since we are adding them, all terms of x work, so we don't worry about anything. 

 

Next, we have

\(x^4+4x^2<21\)

 

Subtracting both sides by 21 and rewriting in standard form, we have the inequality

\(x^4+4x^2-21<0\)

 

Factoring out +-\sqrt3, we have

\(\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\left(x^2+7\right)<0\)

 

Now, let's note the intervals. If x is greater than square root 3, then we have a positive number. If x is less than square root 3, we also have a positive number. 

If x is either, then we have 0. Thus, the interval is

\(-\sqrt{3} <\sqrt{3}\)

 

In interval notation, we have

\((-\sqrt{3},\sqrt{3})\)

 

So our answer is \((-\sqrt{3},\sqrt{3})\)

 

Thanks! :)

 Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
 #1
avatar+1267 
+1
Best Answer

Let's note something really important before we begin. 

To solve this problem, we must first do something quite important.

It's crucial that we split this inequality into two seperate ones before combining them back again. Thus, we have

\(x^4+4x^2>-4\\ x^4+4x^2<21\)

Now, note something about the first inequality. 

We have the terms x^4 and x^2 which is guarenteed to be positive. Since we are adding them, all terms of x work, so we don't worry about anything. 

 

Next, we have

\(x^4+4x^2<21\)

 

Subtracting both sides by 21 and rewriting in standard form, we have the inequality

\(x^4+4x^2-21<0\)

 

Factoring out +-\sqrt3, we have

\(\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\left(x^2+7\right)<0\)

 

Now, let's note the intervals. If x is greater than square root 3, then we have a positive number. If x is less than square root 3, we also have a positive number. 

If x is either, then we have 0. Thus, the interval is

\(-\sqrt{3} <\sqrt{3}\)

 

In interval notation, we have

\((-\sqrt{3},\sqrt{3})\)

 

So our answer is \((-\sqrt{3},\sqrt{3})\)

 

Thanks! :)

NotThatSmart Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024

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