1. **Problem:** Solve the equation $$6x^4 - 11x^2 + 3 = 0$$ 2. **Formula and approach:** This is a quartic equation but can be treated as a quadratic in terms of $x^2$. Let $y = x^2$, then the equation becomes: $$6y^2 - 11y + 3 = 0$$ 3. **Solve the quadratic in $y$:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=6$, $b=-11$, and $c=3$. 4. **Calculate the discriminant:** $$\Delta = (-11)^2 - 4 \times 6 \times 3 = 121 - 72 = 49$$ 5. **Find the roots for $y$:** $$y = \frac{11 \pm \sqrt{49}}{12} = \frac{11 \pm 7}{12}$$ 6. **Evaluate each root:** - For $y_1$: $$y_1 = \frac{11 + 7}{12} = \frac{18}{12} = \frac{3}{2}$$ - For $y_2$: $$y_2 = \frac{11 - 7}{12} = \frac{4}{12} = \frac{1}{3}$$ 7. **Recall $y = x^2$, so solve for $x$:** - For $y_1 = \frac{3}{2}$: $$x^2 = \frac{3}{2} \implies x = \pm \sqrt{\frac{3}{2}} = \pm \frac{\sqrt{6}}{2}$$ - For $y_2 = \frac{1}{3}$: $$x^2 = \frac{1}{3} \implies x = \pm \sqrt{\frac{1}{3}} = \pm \frac{\sqrt{3}}{3}$$ 8. **Final solution:** $$x = \pm \frac{\sqrt{6}}{2}, \quad x = \pm \frac{\sqrt{3}}{3}$$ This completes the solution for the first problem.