To determine which equation has exactly two real and two nonreal solutions, we can analyze each of the provided equations using the concept of the discriminant or by looking at the nature of their roots.
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Equation: \( x^4 - 36x^2 = 0 \)
- Factor to: \( x^2(x^2 - 36) = 0 \)
- Solutions: \( x^2 = 0 \) gives \( x = 0 \) (double root), and \( x^2 - 36 = 0 \) gives \( x = 6 \) and \( x = -6 \).
- Total real solutions: 3 (0, 6, -6) and no nonreal solutions.
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Equation: \( x^3 - x^2 + x - 4 = 0 \)
- This is a cubic equation. Cubic equations always have at least one real root.
- A numeric method (like a graph or synthetic division) may indicate the real roots are present, but without deeper analysis, we cannot directly determine if it has exactly two real roots and one nonreal.
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Equation: \( x^4 - 5x^2 - 36 = 0 \)
- Let \( y = x^2 \). The equation becomes \( y^2 - 5y - 36 = 0 \).
- The discriminant \( \Delta = (-5)^2 - 4(1)(-36) = 25 + 144 = 169 \).
- Since \( \Delta > 0\), there are two distinct \( y \) solutions.
- These yield: \( y = \frac{5 \pm 13}{2} \) → \( y = 9, y = -4 \).
- From \( y = 9 \), we get \( x = 3, -3 \) (2 real roots).
- From \( y = -4 \), we get no real roots (2 nonreal roots from \( x^2 = -4 \)).
- Therefore, it has exactly two real roots and two nonreal roots.
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Equation: \( x^3 - 2x^2 + x - 4 = 0 \)
- Similar to the second equation, this is also cubic and requires similar analysis.
- Generally, it too can have various configurations of real and nonreal roots depending on the discriminant.
Given this analysis:
The correct answer is: \( x^4 - 5x^2 - 36 = 0 \), as it has exactly two real and two nonreal solutions.