1. **State the problem:** Solve the equation $$2x^4 - 3x^2 + 1 = 0$$ for $x$. 2. **Identify the type of equation:** This is a quartic equation but can be treated as a quadratic in terms of $x^2$. 3. **Substitute:** Let $y = x^2$. Then the equation becomes: $$2y^2 - 3y + 1 = 0$$ 4. **Use the quadratic formula:** For $ay^2 + by + c = 0$, the solutions are: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=2$, $b=-3$, $c=1$. 5. **Calculate the discriminant:** $$\Delta = (-3)^2 - 4 \times 2 \times 1 = 9 - 8 = 1$$ 6. **Find $y$ values:** $$y = \frac{3 \pm \sqrt{1}}{4}$$ So, $$y_1 = \frac{3 + 1}{4} = 1$$ $$y_2 = \frac{3 - 1}{4} = \frac{1}{2}$$ 7. **Recall $y = x^2$, solve for $x$:** - For $y_1 = 1$: $$x^2 = 1 \implies x = \pm 1$$ - For $y_2 = \frac{1}{2}$: $$x^2 = \frac{1}{2} \implies x = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2}$$ 8. **Final solution set:** $$\left\{ -1, 1, -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\}$$ 9. **Match with options:** This corresponds to option D: $\left\{ -\frac{\sqrt{2}}{2}, -1, \frac{\sqrt{2}}{2}, 1 \right\}$. **Answer:** D