1. **Problem:** Find the approximate root at the 3rd iteration for the equation $f(x) = 2x^3 + 5x + 20 = 0$ using the secant method, then determine the relative error. 2. **Formula and rules:** The secant method formula is: $$x_{n+1} = x_n - f(x_n) \times \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})}$$ This method uses two initial guesses and iteratively improves the root approximation. 3. **Choose initial guesses:** Let $x_0 = -2$ and $x_1 = -1$ (chosen to bracket the root roughly). 4. **Calculate $f(x_0)$ and $f(x_1)$:** $$f(-2) = 2(-2)^3 + 5(-2) + 20 = 2(-8) - 10 + 20 = -16 - 10 + 20 = -6$$ $$f(-1) = 2(-1)^3 + 5(-1) + 20 = 2(-1) - 5 + 20 = -2 - 5 + 20 = 13$$ 5. **Iteration 2:** $$x_2 = x_1 - f(x_1) \times \frac{x_1 - x_0}{f(x_1) - f(x_0)} = -1 - 13 \times \frac{-1 - (-2)}{13 - (-6)} = -1 - 13 \times \frac{1}{19} = -1 - \frac{13}{19} = -1 - 0.6842 = -1.6842$$ 6. **Calculate $f(x_2)$:** $$f(-1.6842) = 2(-1.6842)^3 + 5(-1.6842) + 20$$ Calculate $(-1.6842)^3 = -4.778$ (approx), so: $$f(-1.6842) = 2(-4.778) + (-8.421) + 20 = -9.556 - 8.421 + 20 = 2.023$$ 7. **Iteration 3:** $$x_3 = x_2 - f(x_2) \times \frac{x_2 - x_1}{f(x_2) - f(x_1)} = -1.6842 - 2.023 \times \frac{-1.6842 - (-1)}{2.023 - 13} = -1.6842 - 2.023 \times \frac{-0.6842}{-10.977}$$ Simplify fraction: $$\frac{-0.6842}{-10.977} = 0.0623$$ So: $$x_3 = -1.6842 - 2.023 \times 0.0623 = -1.6842 - 0.126 = -1.8102$$ 8. **Calculate relative error:** Relative error $= \left| \frac{x_3 - x_2}{x_3} \right| = \left| \frac{-1.8102 - (-1.6842)}{-1.8102} \right| = \left| \frac{-0.126}{-1.8102} \right| = 0.0696$ or 6.96%. **Final answer:** The approximate root at iteration 3 is $x_3 \approx -1.8102$ with a relative error of about 6.96%.