1. **Problem Statement:** Find the root of the equation $$x^3 - 2x - 5 = 0$$ using the bisection method correct up to 3 decimal places. 2. **Formula and Method:** The bisection method is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie. It requires two initial points $$a$$ and $$b$$ such that $$f(a)\cdot f(b) < 0$$. 3. **Steps:** - Define the function: $$f(x) = x^3 - 2x - 5$$. - Choose initial interval: test values to find $$a$$ and $$b$$ such that $$f(a)\cdot f(b) < 0$$. 4. **Finding initial interval:** - Evaluate $$f(2) = 2^3 - 2\times2 - 5 = 8 - 4 - 5 = -1$$. - Evaluate $$f(3) = 3^3 - 2\times3 - 5 = 27 - 6 - 5 = 16$$. Since $$f(2) < 0$$ and $$f(3) > 0$$, the root lies between 2 and 3. 5. **Bisection iterations:** - Compute midpoint $$c = \frac{2 + 3}{2} = 2.5$$. - Evaluate $$f(2.5) = 2.5^3 - 2\times2.5 - 5 = 15.625 - 5 - 5 = 5.625 > 0$$. Since $$f(2) < 0$$ and $$f(2.5) > 0$$, root lies between 2 and 2.5. - Next midpoint $$c = \frac{2 + 2.5}{2} = 2.25$$. - Evaluate $$f(2.25) = 2.25^3 - 2\times2.25 - 5 = 11.39 - 4.5 - 5 = 1.89 > 0$$. Root lies between 2 and 2.25. - Next midpoint $$c = \frac{2 + 2.25}{2} = 2.125$$. - Evaluate $$f(2.125) = 2.125^3 - 2\times2.125 - 5 = 9.59 - 4.25 - 5 = 0.34 > 0$$. Root lies between 2 and 2.125. - Next midpoint $$c = \frac{2 + 2.125}{2} = 2.0625$$. - Evaluate $$f(2.0625) = 2.0625^3 - 2\times2.0625 - 5 = 8.78 - 4.125 - 5 = -0.34 < 0$$. Root lies between 2.0625 and 2.125. - Next midpoint $$c = \frac{2.0625 + 2.125}{2} = 2.09375$$. - Evaluate $$f(2.09375) = 2.09375^3 - 2\times2.09375 - 5 = 9.18 - 4.19 - 5 = 0.01 > 0$$. Root lies between 2.0625 and 2.09375. - Next midpoint $$c = \frac{2.0625 + 2.09375}{2} = 2.078125$$. - Evaluate $$f(2.078125) = 2.078125^3 - 2\times2.078125 - 5 = 8.98 - 4.16 - 5 = -0.16 < 0$$. Root lies between 2.078125 and 2.09375. - Next midpoint $$c = \frac{2.078125 + 2.09375}{2} = 2.0859375$$. - Evaluate $$f(2.0859375) = 2.0859375^3 - 2\times2.0859375 - 5 = 9.08 - 4.17 - 5 = -0.07 < 0$$. Root lies between 2.0859375 and 2.09375. - Next midpoint $$c = \frac{2.0859375 + 2.09375}{2} = 2.08984375$$. - Evaluate $$f(2.08984375) = 2.08984375^3 - 2\times2.08984375 - 5 = 9.13 - 4.18 - 5 = -0.03 < 0$$. Root lies between 2.08984375 and 2.09375. - Next midpoint $$c = \frac{2.08984375 + 2.09375}{2} = 2.091796875$$. - Evaluate $$f(2.091796875) = 2.091796875^3 - 2\times2.091796875 - 5 = 9.15 - 4.18 - 5 = -0.01 < 0$$. Root lies between 2.091796875 and 2.09375. - Next midpoint $$c = \frac{2.091796875 + 2.09375}{2} = 2.0927734375$$. - Evaluate $$f(2.0927734375) = 2.0927734375^3 - 2\times2.0927734375 - 5 = 9.16 - 4.18 - 5 = 0.00 > 0$$. Root lies between 2.091796875 and 2.0927734375. 6. **Stopping criteria:** The interval length is less than 0.001, so the root is approximately $$2.092$$. **Final answer:** The root of the equation $$x^3 - 2x - 5 = 0$$ correct up to 3 decimal places is $$\boxed{2.092}$$.