Okay, so I need to solve the inequality -4 < x⁴ + 4x² < 21 for real values of x. Hmm, let me start by breaking this down. The inequality has two parts: the left part is -4 < x⁴ + 4x², and the right part is x⁴ + 4x² < 21. I'll need to solve both inequalities separately and then find the intersection of the solutions because both conditions have to be satisfied simultaneously.

First, let's look at the left inequality: -4 < x⁴ + 4x². Since x⁴ and x² are both non-negative for all real x, their sum x⁴ + 4x² is also non-negative. So, the left inequality simplifies to 0 < x⁴ + 4x². But wait, x⁴ + 4x² is always non-negative, right? Let me check when it's zero. If x⁴ + 4x² = 0, then x⁴ = 0 and 4x² = 0, which implies x = 0. So, at x = 0, the expression equals zero. Therefore, the inequality -4 < x⁴ + 4x² is always true except at x = 0 where it's equal to -4. So, the left inequality is satisfied for all real x except x = 0. Wait, no. Wait, the original inequality is -4 < x⁴ + 4x². Since x⁴ + 4x² is always non-negative, it's always greater than or equal to zero. So, -4 < x⁴ + 4x² is always true because the left side is -4, which is less than any non-negative number. Therefore, the left inequality doesn't impose any restrictions on x. So, the only restriction comes from the right inequality: x⁴ + 4x² < 21. Therefore, I just need to solve x⁴ + 4x² < 21.

Okay, so the problem simplifies to finding x such that x⁴ + 4x² < 21. Let me focus on that. Let me set y = x². Then the inequality becomes y² + 4y < 21. Let's rewrite this as y² + 4y - 21 < 0. Now, I need to solve the quadratic inequality y² + 4y - 21 < 0. To find the values of y where this inequality holds, first find the roots of the quadratic equation y² + 4y - 21 = 0. Using the quadratic formula: y = [-4 ± sqrt(16 + 84)] / 2 = [-4 ± sqrt(100)] / 2 = [-4 ± 10]/2. So, the roots are (6)/2 = 3 and (-14)/2 = -7. Therefore, the quadratic inequality y² + 4y - 21 < 0 holds when y is between -7 and 3. But since y = x², which is always non-negative, y cannot be less than 0. Therefore, the solution for y is 0 ≤ y < 3. Therefore, x² < 3. Taking square roots, this implies that |x| < sqrt(3). So, x is in the interval (-sqrt(3), sqrt(3)).

But wait, let me double-check. The quadratic inequality y² + 4y - 21 < 0. The parabola opens upwards (since the coefficient of y² is positive), so the inequality is satisfied between the roots. The roots are y = -7 and y = 3. So, y must be between -7 and 3. But since y is x², which is non-negative, the lower bound of -7 doesn't affect us. Therefore, the valid range for y is 0 ≤ y < 3. Therefore, x² < 3, which gives x in (-sqrt(3), sqrt(3)). So, that's the solution.

But hold on, let me verify with some test points. Let's take x = 0. Then x⁴ + 4x² = 0 + 0 = 0 < 21. That works. x = 1: 1 + 4 = 5 < 21. x = 2: 16 + 16 = 32, which is not less than 21. So, x = 2 is excluded. x = sqrt(3) ≈ 1.732. Then x² = 3, so x⁴ + 4x² = 9 + 12 = 21, which is not less than 21. So, x must be strictly less than sqrt(3). Similarly, x = -sqrt(3) is also excluded. So, the interval is indeed (-sqrt(3), sqrt(3)).

But wait, let me check another point. What about x = 1.5? Then x² = 2.25, x⁴ = (2.25)^2 = 5.0625. Then x⁴ + 4x² = 5.0625 + 9 = 14.0625 < 21. That's within the interval. So, that's good. What about x = sqrt(2) ≈ 1.414? x² = 2, x⁴ = 4, so 4 + 8 = 12 < 21. Still good. So, seems consistent.

Therefore, the solution is all real numbers x such that |x| < sqrt(3), which is the interval (-sqrt(3), sqrt(3)). Since the original inequality -4 < x⁴ + 4x² is always true except at x = 0 where it's equal to -4, but since the inequality is strict (-4 < ...), x = 0 is allowed as well because at x = 0, x⁴ + 4x² = 0, which is greater than -4. Wait, hold on. Wait, if x = 0, then x⁴ + 4x² = 0, which is indeed greater than -4. So, the inequality -4 < 0 is true. Therefore, x = 0 is included in the solution. But in our previous analysis, the solution was (-sqrt(3), sqrt(3)) which does include 0. So, that's okay. So, the interval (-sqrt(3), sqrt(3)) is correct.

But let me confirm once again. Let me test x = 0. The inequality becomes -4 < 0 < 21. Yes, that's true. So, x = 0 is allowed. So, the interval is (-sqrt(3), sqrt(3)). Since sqrt(3) is approximately 1.732, this is the interval where all x satisfy the original inequality.

Therefore, the final answer is (-sqrt(3), sqrt(3)).

But wait, just to make sure I didn't make any mistake in the substitution. Let me re-examine the substitution step. I set y = x², so the original inequality becomes y² + 4y < 21. Then solving y² + 4y - 21 < 0. The quadratic equation y² + 4y - 21 = 0 has roots at y = [-4 ± sqrt(16 + 84)]/2 = [-4 ± sqrt(100)]/2 = [-4 ± 10]/2. So, y = 6/2 = 3 or y = -14/2 = -7. So, the inequality is satisfied when y is between -7 and 3. Since y = x² ≥ 0, the valid interval is 0 ≤ y < 3. Therefore, x² < 3, so x ∈ (-sqrt(3), sqrt(3)). All steps check out. So, I think that's correct.

Therefore, the solution is all real numbers x between -sqrt(3) and sqrt(3), not including the endpoints. Expressed in interval notation, that's (-√3, √3).
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