1. **State the problem:** We want to find the approximated root after the second iteration of the Newton-Raphson method for the equation $$x^3 + 4x^2 - 10 = 0$$ starting with the initial guess $$x_0 = 1.5$$. 2. **Recall the Newton-Raphson formula:** $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ where $$f(x) = x^3 + 4x^2 - 10$$ and $$f'(x)$$ is its derivative. 3. **Find the derivative:** $$f'(x) = 3x^2 + 8x$$ 4. **Calculate the first iteration $$x_1$$:** - Evaluate $$f(1.5) = (1.5)^3 + 4(1.5)^2 - 10 = 3.375 + 9 - 10 = 2.375$$ - Evaluate $$f'(1.5) = 3(1.5)^2 + 8(1.5) = 3(2.25) + 12 = 6.75 + 12 = 18.75$$ - Apply formula: $$x_1 = 1.5 - \frac{2.375}{18.75} = 1.5 - 0.1267 = 1.3733$$ 5. **Calculate the second iteration $$x_2$$:** - Evaluate $$f(1.3733) = (1.3733)^3 + 4(1.3733)^2 - 10 \approx 2.588 + 7.544 - 10 = 0.132$$ - Evaluate $$f'(1.3733) = 3(1.3733)^2 + 8(1.3733) \approx 3(1.886) + 10.986 = 5.658 + 10.986 = 16.644$$ - Apply formula: $$x_2 = 1.3733 - \frac{0.132}{16.644} = 1.3733 - 0.0079 = 1.3654$$ 6. **Conclusion:** The approximated root after the second iteration is approximately $$1.3653$$, which corresponds to option iv. **Final answer:** iv. 1.3653