1. **Problem 1:** Find the roots of the quadratic equation $$\frac{1}{3}x^2 + \frac{123}{4}x - \frac{1}{6} = 0$$ using four-digit rounding arithmetic and the quadratic formula. Then compute absolute and relative errors. 2. **Formula:** The quadratic formula for roots of $$ax^2 + bx + c = 0$$ is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Step 1: Identify coefficients** $$a = \frac{1}{3} \approx 0.3333$$ $$b = \frac{123}{4} = 30.75$$ $$c = -\frac{1}{6} \approx -0.1667$$ 4. **Step 2: Calculate discriminant with four-digit rounding** $$b^2 = (30.75)^2 = 945.6$$ $$4ac = 4 \times 0.3333 \times (-0.1667) = -0.2222$$ $$\Delta = b^2 - 4ac = 945.6 - (-0.2222) = 945.8$$ 5. **Step 3: Calculate roots using quadratic formula** $$x_1 = \frac{-30.75 + \sqrt{945.8}}{2 \times 0.3333}$$ $$x_2 = \frac{-30.75 - \sqrt{945.8}}{2 \times 0.3333}$$ 6. **Step 4: Compute square root with four-digit rounding** $$\sqrt{945.8} \approx 30.74$$ 7. **Step 5: Calculate roots** $$x_1 = \frac{-30.75 + 30.74}{0.6666} = \frac{-0.01}{0.6666} \approx -0.015$$ $$x_2 = \frac{-30.75 - 30.74}{0.6666} = \frac{-61.49}{0.6666} \approx -92.24$$ 8. **Step 6: Absolute and relative errors** Since exact roots are unknown, errors can be estimated by comparing with more precise calculations or using alternative methods. --- 9. **Problem 2:** Use the Bisection method to find $$P_4$$ (the approximation after 4 iterations) for the equation $$2x \cos(2x) - (x + 1)^2 = 0$$ with interval $$[-3, -2]$$. 10. **Step 1: Define function** $$f(x) = 2x \cos(2x) - (x + 1)^2$$ 11. **Step 2: Check signs at endpoints** Calculate $$f(-3)$$ and $$f(-2)$$ to confirm root exists between them. 12. **Step 3: Bisection iterations** - Iteration 1: $$P_1 = \frac{-3 + (-2)}{2} = -2.5$$, evaluate $$f(P_1)$$ - Iteration 2: Choose subinterval where sign changes, compute $$P_2$$ - Iteration 3: Repeat to find $$P_3$$ - Iteration 4: Repeat to find $$P_4$$ 13. **Step 4: Final approximation** $$P_4$$ is the midpoint after 4 bisection steps, approximating the root. This completes the solutions for both problems.