1. **Problem (a):** Find the range of values of $x$ such that $$x^2 - 4x - 5 > 0$$ using the table method. 2. **Step 1:** Factorize the quadratic expression: $$x^2 - 4x - 5 = (x - 5)(x + 1)$$ 3. **Step 2:** Identify the roots where the expression equals zero: $$x - 5 = 0 \Rightarrow x = 5$$ $$x + 1 = 0 \Rightarrow x = -1$$ 4. **Step 3:** Use the roots to divide the number line into intervals: $$(-\infty, -1), (-1, 5), (5, \infty)$$ 5. **Step 4:** Test a value from each interval in the inequality $x^2 - 4x - 5 > 0$: - For $x = -2$ (in $(-\infty, -1)$): $$(-2)^2 - 4(-2) - 5 = 4 + 8 - 5 = 7 > 0$$ (True) - For $x = 0$ (in $(-1, 5)$): $$0^2 - 4(0) - 5 = -5 > 0$$ (False) - For $x = 6$ (in $(5, \infty)$): $$6^2 - 4(6) - 5 = 36 - 24 - 5 = 7 > 0$$ (True) 6. **Step 5:** Conclusion for (a): The inequality holds for $$x < -1$$ or $$x > 5$$. --- 7. **Problem (b):** Given that $\alpha$ and $\beta$ are roots of the quadratic equation $$2x^2 + kx - 2 = 0,$$ form a quadratic equation with roots $$\frac{1}{\alpha + 1}$$ and $$\frac{1}{\beta + 1}$$ in terms of $k$. 8. **Step 1:** Recall sum and product of roots for the original equation: $$\alpha + \beta = -\frac{k}{2}$$ $$\alpha \beta = -\frac{2}{2} = -1$$ 9. **Step 2:** Let the new roots be $$r_1 = \frac{1}{\alpha + 1}, \quad r_2 = \frac{1}{\beta + 1}$$ 10. **Step 3:** Find the sum of new roots: $$r_1 + r_2 = \frac{1}{\alpha + 1} + \frac{1}{\beta + 1} = \frac{(\beta + 1) + (\alpha + 1)}{(\alpha + 1)(\beta + 1)} = \frac{\alpha + \beta + 2}{(\alpha + 1)(\beta + 1)}$$ 11. **Step 4:** Calculate the denominator: $$(\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1 = -1 + ( -\frac{k}{2} ) + 1 = -\frac{k}{2}$$ 12. **Step 5:** Substitute values: $$r_1 + r_2 = \frac{-\frac{k}{2} + 2}{-\frac{k}{2}} = \frac{2 - \frac{k}{2}}{-\frac{k}{2}} = \frac{4 - k}{-k} = \frac{k - 4}{k}$$ 13. **Step 6:** Find the product of new roots: $$r_1 r_2 = \frac{1}{(\alpha + 1)(\beta + 1)} = \frac{1}{-\frac{k}{2}} = -\frac{2}{k}$$ 14. **Step 7:** Form the quadratic equation with roots $r_1$ and $r_2$: $$x^2 - (r_1 + r_2)x + r_1 r_2 = 0$$ Substitute the sum and product: $$x^2 - \frac{k - 4}{k} x - \frac{2}{k} = 0$$ 15. **Step 8:** Multiply through by $k$ to clear denominators: $$k x^2 - (k - 4) x - 2 = 0$$ **Final answers:** (a) $$x < -1 \quad \text{or} \quad x > 5$$ (b) Quadratic equation with roots $$\frac{1}{\alpha + 1}$$ and $$\frac{1}{\beta + 1}$$ is $$k x^2 - (k - 4) x - 2 = 0$$