Find all real and imaginary solutions to the equation. Check your answer. \( x^{4}-16 x^{2}+63=0 \)

To find all real and imaginary solutions to the equation: \[ x^{4} - 16x^{2} + 63 = 0 \] we can solve it step-by-step as follows: 1. **Recognize the Equation as a Biquadratic:** The given equation is a biquadratic equation because it contains even powers of \( x \). We can simplify it by making a substitution. 2. **Substitute \( y = x^2 \):** Let \( y = x^2 \). The equation becomes: \[ y^2 - 16y + 63 = 0 \] 3. **Solve the Quadratic Equation for \( y \):** To solve \( y^2 - 16y + 63 = 0 \), we can use the quadratic formula: \[ y = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 63}}{2} = \frac{16 \pm \sqrt{256 - 252}}{2} = \frac{16 \pm \sqrt{4}}{2} = \frac{16 \pm 2}{2} \] This gives us two solutions for \( y \): \[ y_1 = \frac{16 + 2}{2} = 9 \quad \text{and} \quad y_2 = \frac{16 - 2}{2} = 7 \] 4. **Back-Substitute to Find \( x \):** Since \( y = x^2 \), we now have: \[ x^2 = 9 \quad \Rightarrow \quad x = \pm 3 \] \[ x^2 = 7 \quad \Rightarrow \quad x = \pm \sqrt{7} \] 5. **List All Solutions:** The equation has four real solutions: \[ x = 3, \quad x = -3, \quad x = \sqrt{7}, \quad x = -\sqrt{7} \] 6. **Verify the Solutions:** Substitute each solution back into the original equation to verify: - For \( x = 3 \): \[ 3^4 - 16 \cdot 3^2 + 63 = 81 - 144 + 63 = 0 \] - For \( x = -3 \): \[ (-3)^4 - 16 \cdot (-3)^2 + 63 = 81 - 144 + 63 = 0 \] - For \( x = \sqrt{7} \): \[ (\sqrt{7})^4 - 16 \cdot (\sqrt{7})^2 + 63 = 49 - 112 + 63 = 0 \] - For \( x = -\sqrt{7} \): \[ (-\sqrt{7})^4 - 16 \cdot (-\sqrt{7})^2 + 63 = 49 - 112 + 63 = 0 \] All solutions satisfy the original equation. **Final Answer:** All four roots are real numbers  3, –3, √7, and –√7. Thus, the solutions are  x = ± 3 and x = ± √7.