1. **Stating the problem:** Solve the equation $$x^2 + \frac{1}{x^2} = \frac{17}{4}$$ for $x$. 2. **Use a substitution to simplify:** Let $$y = x + \frac{1}{x}$$. We know the identity: $$y^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$$ which implies $$x^2 + \frac{1}{x^2} = y^2 - 2$$ 3. **Rewrite the original equation using $y$:** $$y^2 - 2 = \frac{17}{4}$$ 4. **Solve for $y^2$:** $$y^2 = \frac{17}{4} + 2 = \frac{17}{4} + \frac{8}{4} = \frac{25}{4}$$ 5. **Find $y$:** $$y = \pm \frac{5}{2}$$ 6. **Recall $y = x + \frac{1}{x}$, so solve for $x$:** For $$y = \frac{5}{2}$$: $$x + \frac{1}{x} = \frac{5}{2}$$ Multiply both sides by $x$: $$x^2 + 1 = \frac{5}{2}x$$ Rearranged: $$x^2 - \frac{5}{2}x + 1 = 0$$ 7. **Solve quadratic equation:** Use quadratic formula: $$x = \frac{\frac{5}{2} \pm \sqrt{\left(\frac{5}{2}\right)^2 - 4 \cdot 1 \cdot 1}}{2} = \frac{\frac{5}{2} \pm \sqrt{\frac{25}{4} - 4}}{2} = \frac{\frac{5}{2} \pm \sqrt{\frac{25}{4} - \frac{16}{4}}}{2} = \frac{\frac{5}{2} \pm \sqrt{\frac{9}{4}}}{2}$$ Simplify the square root: $$\sqrt{\frac{9}{4}} = \frac{3}{2}$$ So: $$x = \frac{\frac{5}{2} \pm \frac{3}{2}}{2}$$ Calculate both solutions: - $$x = \frac{\frac{5}{2} + \frac{3}{2}}{2} = \frac{\frac{8}{2}}{2} = \frac{4}{2} = 2$$ - $$x = \frac{\frac{5}{2} - \frac{3}{2}}{2} = \frac{\frac{2}{2}}{2} = \frac{1}{2}$$ 8. **For $$y = -\frac{5}{2}$$:** $$x + \frac{1}{x} = -\frac{5}{2}$$ Multiply both sides by $x$: $$x^2 + 1 = -\frac{5}{2}x$$ Rearranged: $$x^2 + \frac{5}{2}x + 1 = 0$$ 9. **Solve quadratic:** $$x = \frac{-\frac{5}{2} \pm \sqrt{\left(\frac{5}{2}\right)^2 - 4}}{2} = \frac{-\frac{5}{2} \pm \sqrt{\frac{25}{4} - 4}}{2} = \frac{-\frac{5}{2} \pm \frac{3}{2}}{2}$$ Calculate both solutions: - $$x = \frac{-\frac{5}{2} + \frac{3}{2}}{2} = \frac{-\frac{2}{2}}{2} = \frac{-1}{2}$$ - $$x = \frac{-\frac{5}{2} - \frac{3}{2}}{2} = \frac{-\frac{8}{2}}{2} = \frac{-4}{2} = -2$$ 10. **Final solutions:** $$x = 2, \frac{1}{2}, -\frac{1}{2}, -2$$ These are the four values of $x$ satisfying the original equation.