<Question>
For what real values of $x$ is $-4<x^{4}+4x^{2}<21$ satisfied? Express your answer in interval notation.
</Question>

<think>
<Plan_1>
Okay, so I need to find all real numbers x such that -4 < x⁴ + 4x² < 21. Hmm, let's break this down step by step. First, the inequality is a compound inequality, meaning both parts have to be true at the same time. So, x has to satisfy both -4 < x⁴ + 4x² and x⁴ + 4x² < 21. Maybe I can tackle each part separately and then find the intersection of the solutions.

Let me start with the left part: -4 < x⁴ + 4x². Since x⁴ and 4x² are both non-negative for all real x (because any real number raised to an even power is non-negative), their sum is also non-negative.
So, x⁴ + 4x² ≥ 0. Therefore, -4 < x⁴ + 4x² is always true because 0 is greater than -4, and the left side of the inequality is just saying something that's always true.

Wait, but hold on. If x⁴ + 4x² is always non-negative, then it's always greater than -4. So, the first inequality doesn't actually restrict any values of x. That means the only part that matters is the second inequality: x⁴ + 4x² < 21.

Is that right? Let me confirm.

<prompt_1.1> Confirm whether the inequality -4 < x⁴ + 4x² is always true. </prompt_1.1>

So, the problem reduces to solving x⁴ + 4x² < 21.

To solve this, maybe substitution will help. Let me set y = x². Then, since x⁴ is (x²)², that becomes y². So, substituting, the inequality becomes y² + 4y < 21. Let's rewrite that as y² + 4y - 21 < 0.

<prompt_1.2> Use substitution to solve the inequality x⁴ + 4x² < 21. </prompt_1.2>
-----

But let me check another value between -3 and 3, say x=0. Then x⁴ +4x²=0+0=0, which is less than 21. Okay, that's fine. How about x=1?

<prompt_1.3> Check specific values of x to verify the solution interval. </prompt_1.3>

But wait, let me check the endpoints. When x=3, x⁴ +4x²=81 + 36=117, which is greater than 21, so x=3 is excluded, which matches the interval. Similarly, x=-3 gives the same result, so it's excluded.

Is there any other value I need to check? Maybe a value larger than 3, like x=4.

<prompt_1.4> Check additional values to ensure the solution interval is correct. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
Yes, because x⁴ is always ≥ 0 and 4x² is also ≥ 0, so their sum is ≥ 0. Since -4 is less than 0, the inequality -4 < non-negative number is always true. So, indeed, the first part of the compound inequality doesn't impose any restrictions. Therefore, the entire solution set is determined by x⁴ + 4x² < 21.
</execution_1.1>

<execution_1.2>
Now, we need to solve the quadratic inequality y² + 4y - 21 < 0. To do this, first find the roots of the quadratic equation y² + 4y - 21 = 0. Using the quadratic formula: y = [-4 ± √(16 + 84)] / 2 = [-4 ± √100] / 2 = [-4 ± 10]/2. So, the roots are (-4 + 10)/2 = 6/2 = 3, and (-4 - 10)/2 = -14/2 = -7. Therefore, the quadratic can be factored as (y - 3)(y + 7) < 0.

Now, to solve (y - 3)(y + 7) < 0, we consider the critical points y = -7 and y = 3. These divide the number line into three intervals: (-∞, -7), (-7, 3), and (3, ∞). We test each interval:

1. For y < -7, say y = -8: (-8 - 3)(-8 + 7) = (-11)(-1) = 11 > 0.
2. For -7 < y < 3, say y = 0: (0 - 3)(0 + 7) = (-3)(7) = -21 < 0.
3. For y > 3, say y = 4: (4 - 3)(4 + 7) = (1)(11) = 11 > 0.

Since we want where the product is less than 0, the solution is -7 < y < 3.
</execution_1.2>

<execution_1.3>
x⁴ +4x²=1 +4=5 <21, which is true. How about x=2? 16 +16=32, which is greater than 21, so x=2 should not be included, which matches the interval. Similarly, x=-2 would also be excluded.
</execution_1.3>

<execution_1.4>
x⁴ +4x² would be 256 + 16=272, which is way more than 21. What about a value between 0 and 3, like x=2? 16 + 16=32, still more than 21. Wait, but 2 is within the interval (-3,3). Wait, no, 2 is between -3 and 3. Wait, 2 is 4 + 16=20. Wait, x=2: x⁴ is 16, 4x² is 16, so 16 +16=32. Oh, right, 32 is greater than 21. So x=2 is excluded, which is within (-3,3). So that's correct.

So, the solution set is all real numbers between -3 and 3, not including -3 and 3.
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.2:
Remember, y was substituted for x², so y = x². Therefore, we have -7 < x² < 3.

<prompt_2.1> Convert the solution for y back to x and determine the valid intervals. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Now, let's analyze this. Since x² is always non-negative, the left inequality -7 < x² is always true because x² ≥ 0 > -7. Therefore, the inequality x² < 3 is the one that actually constrains x. So, x² < 3.

Solving x² < 3, we take square roots: |x| < √3. Which means -√3 < x < √3. Therefore, the solution in terms of x is the interval (-√3, √3).
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But hold on, let me verify this. Let's test a value inside the interval, say x=1.

<prompt_3.1> Verify the solution by testing values inside and outside the interval. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Then x⁴ +4x² =1 +4=5, which is less than 21. What about x=√3, which is approximately 1.732. Let's compute x⁴ +4x² when x=√3. x⁴ is (√3)^4 = (3)^2=9, and 4x²=4*(3)=12. So 9 +12=21. But the inequality is strict: less than 21. Therefore, x=√3 and x=-√3 are excluded, which matches our interval notation.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But just to be thorough, let's check another value, say x=1.5, which is less than √3 (since √3 ≈1.732). x=1.5: x²=2.25, x⁴=(2.25)^2=5.0625.
So x⁴ +4x²=5.0625 +9=14.0625, which is less than 21. That's good. How about x=2, which is outside the interval: x⁴ +4x²=16 +16=32>21. Correct. How about x=0: 0 +0=0 <21. Correct.

So, putting it all together, the solution is all real numbers x where -√3 <x<√3. In interval notation, that's (-√3, √3).

But let me just recap to make sure I didn't skip any steps. Original inequality: -4 <x⁴ +4x² <21. As x⁴ +4x² is always ≥0, the left inequality is always true. Therefore, we only need to solve x⁴ +4x² <21. Substituting y=x², we get y² +4y -21 <0. Solving the quadratic inequality gives y between -7 and 3. Since y=x² can't be negative, we discard -7 and focus on y<3, leading to x² <3, hence |x| <√3. Therefore, x is in (-√3, √3).

Is there any possibility that I missed something? For example, could there be another interval where x⁴ +4x² <21? Let's see. Since x⁴ and 4x² are both positive, the expression x⁴ +4x² is symmetric for x and -x. So, the behavior for positive and negative x is the same. Therefore, it's sufficient to consider x ≥0 and then mirror it to x ≤0. However, since we already considered x² <3, which covers all x between -√3 and √3, both positive and negative, that's already the complete interval.

Another check: Let's pick x approaching √3 from the left, say x=√3 - ε for very small ε>0. Then x⁴ +4x² approaches 21 from below, which satisfies the inequality. Similarly, approaching from the right would exceed 21. So the endpoints are excluded.

Therefore, the final answer is the open interval from -√3 to √3.
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
$$x \in \boxed{(-\sqrt{3}, \sqrt{3})}.
$$
</Final_answer>
</think>