1. **State the problem:** Solve the equation $x^5 + x - 1 = 0$ for $x$. 2. **Formula and rules:** This is a polynomial equation of degree 5. There is no general algebraic formula for roots of quintic equations, so we use numerical methods or approximations. 3. **Check for obvious roots:** Try $x=0$: $0^5 + 0 - 1 = -1 \neq 0$. Try $x=1$: $1^5 + 1 - 1 = 1 \neq 0$. 4. **Use Intermediate Value Theorem:** Evaluate at $x=0$ and $x=1$: $$f(0) = -1 < 0, \quad f(1) = 1 > 0$$ Since $f(x)$ is continuous, there is at least one root between 0 and 1. 5. **Approximate root using Newton's method:** Define $f(x) = x^5 + x - 1$ and $f'(x) = 5x^4 + 1$. Start with $x_0 = 0.5$. Iteration 1: $$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0.5 - \frac{0.5^5 + 0.5 - 1}{5 \times 0.5^4 + 1} = 0.5 - \frac{0.03125 + 0.5 - 1}{5 \times 0.0625 + 1} = 0.5 - \frac{-0.46875}{1.3125} \approx 0.8571$$ Iteration 2: $$x_2 = 0.8571 - \frac{0.8571^5 + 0.8571 - 1}{5 \times 0.8571^4 + 1} \approx 0.8571 - \frac{0.464 + 0.8571 - 1}{5 \times 0.539 + 1} = 0.8571 - \frac{0.3211}{3.695} \approx 0.7701$$ Iteration 3: $$x_3 = 0.7701 - \frac{0.7701^5 + 0.7701 - 1}{5 \times 0.7701^4 + 1} \approx 0.7701 - \frac{0.254 + 0.7701 - 1}{5 \times 0.351 + 1} = 0.7701 - \frac{0.0241}{2.755} \approx 0.7613$$ Iteration 4: $$x_4 = 0.7613 - \frac{0.7613^5 + 0.7613 - 1}{5 \times 0.7613^4 + 1} \approx 0.7613 - \frac{0.243 + 0.7613 - 1}{5 \times 0.337 + 1} = 0.7613 - \frac{0.0043}{2.685} \approx 0.7597$$ 6. **Conclusion:** The root is approximately $x \approx 0.76$. This is the real root of the equation $x^5 + x - 1 = 0$.