1. **State the problem:** We want to find the 5th root of a number $n$ using the Newton-Raphson method, starting with an initial guess $x_0=1.5$. Specifically, estimate the 5th root of 20 correct to 3 significant figures. 2. **Formula and method:** The 5th root of $n$ is the solution to the equation $$f(x) = x^5 - n = 0.$$ Newton-Raphson iterative formula is given by: $$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}$$ where $f'(x) = 5x^4$. 3. **Derive the iterative formula:** $$x_{k+1} = x_k - \frac{x_k^5 - n}{5x_k^4} = x_k - \frac{x_k^5}{5x_k^4} + \frac{n}{5x_k^4} = x_k - \frac{x_k}{5} + \frac{n}{5x_k^4} = \frac{4}{5}x_k + \frac{n}{5x_k^4}.$$ So the iterative formula is: $$x_{k+1} = \frac{4}{5}x_k + \frac{n}{5x_k^4}.$$ 4. **Apply the formula to estimate 5th root of 20:** Start with $x_0 = 1.5$ and $n=20$. - Iteration 1: $$x_1 = \frac{4}{5} \times 1.5 + \frac{20}{5 \times (1.5)^4} = 1.2 + \frac{20}{5 \times 5.0625} = 1.2 + \frac{20}{25.3125} = 1.2 + 0.7901 = 1.9901.$$ - Iteration 2: $$x_2 = \frac{4}{5} \times 1.9901 + \frac{20}{5 \times (1.9901)^4} = 1.5921 + \frac{20}{5 \times 15.681} = 1.5921 + \frac{20}{78.405} = 1.5921 + 0.2552 = 1.8473.$$ - Iteration 3: $$x_3 = \frac{4}{5} \times 1.8473 + \frac{20}{5 \times (1.8473)^4} = 1.4778 + \frac{20}{5 \times 11.644} = 1.4778 + \frac{20}{58.22} = 1.4778 + 0.3435 = 1.8213.$$ - Iteration 4: $$x_4 = \frac{4}{5} \times 1.8213 + \frac{20}{5 \times (1.8213)^4} = 1.4570 + \frac{20}{5 \times 11.000} = 1.4570 + \frac{20}{55.0} = 1.4570 + 0.3636 = 1.8206.$$ 5. **Conclusion:** The value stabilizes around $1.82$ to 3 significant figures. **Final answer:** The 5th root of 20 is approximately **1.82**.