1. **State the problem:** Solve the equation $$2(x^2 - 3) + \frac{7x^2}{x^2 + 3} - 6 = 0$$ for $x$. 2. **Rewrite and simplify:** Distribute the 2 in the first term: $$2x^2 - 6 + \frac{7x^2}{x^2 + 3} - 6 = 0$$ Combine the constants: $$2x^2 - 12 + \frac{7x^2}{x^2 + 3} = 0$$ 3. **Isolate the fraction:** Move the constant term to the right side: $$2x^2 + \frac{7x^2}{x^2 + 3} = 12$$ 4. **Multiply both sides by $x^2 + 3$ to clear the denominator:** $$\cancel{(x^2 + 3)}\left(2x^2 + \frac{7x^2}{\cancel{x^2 + 3}}\right) = 12(x^2 + 3)$$ This simplifies to: $$2x^2(x^2 + 3) + 7x^2 = 12x^2 + 36$$ 5. **Expand the left side:** $$2x^4 + 6x^2 + 7x^2 = 12x^2 + 36$$ Combine like terms: $$2x^4 + 13x^2 = 12x^2 + 36$$ 6. **Bring all terms to one side:** $$2x^4 + 13x^2 - 12x^2 - 36 = 0$$ Simplify: $$2x^4 + x^2 - 36 = 0$$ 7. **Substitute $y = x^2$ to solve the quartic as a quadratic in $y$:** $$2y^2 + y - 36 = 0$$ 8. **Use the quadratic formula:** $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-36)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 288}}{4} = \frac{-1 \pm \sqrt{289}}{4}$$ 9. **Calculate the roots for $y$:** $$y = \frac{-1 \pm 17}{4}$$ - For the plus sign: $$y = \frac{-1 + 17}{4} = \frac{16}{4} = 4$$ - For the minus sign: $$y = \frac{-1 - 17}{4} = \frac{-18}{4} = -\frac{9}{2}$$ 10. **Recall $y = x^2$, so solve for $x$:** - When $y = 4$, $$x^2 = 4 \implies x = \pm 2$$ - When $y = -\frac{9}{2}$, no real solution since $x^2$ cannot be negative. **Final answer:** $$x = \pm 2$$