1. Problem 5: Find the values of constant $k$ for which $(2k-1)x^2 + 6x + k + 1 = 0$ has real roots. 2. For real roots, discriminant $\Delta \geq 0$. 3. Coefficients: $a = 2k-1$, $b = 6$, $c = k+1$. 4. Discriminant: $\Delta = b^2 - 4ac = 6^2 - 4(2k-1)(k+1) = 36 - 4(2k^2 + 2k - k -1) = 36 - 4(2k^2 + k -1) = 36 - 8k^2 -4k +4 = 40 - 8k^2 - 4k$. 5. Inequality: $40 - 8k^2 -4k \geq 0 \implies 8k^2 + 4k -40 \leq 0$. 6. Divide by 4: $2k^2 + k -10 \leq 0$. 7. Find roots of $2k^2 + k -10 = 0$: use quadratic formula $k = \frac{-1 \pm \sqrt{1 + 80}}{4} = \frac{-1 \pm 9}{4}$. 8. Roots: $k = 2$ and $k = -\frac{5}{2} = -2.5$. 9. Since coefficient of $k^2$ positive, inequality holds between roots: $-2.5 \leq k \leq 2$. 10. Problem 6(a): Write $5x^2 -14x +8$ in form $a(x+b)^2 + c$. 11. Complete the square: $5x^2 -14x +8 = 5(x^2 - \frac{14}{5}x) + 8$. 12. Half of $\frac{14}{5}$ is $\frac{7}{5}$, square is $\left(\frac{7}{5}\right)^2 = \frac{49}{25}$. 13. Add and subtract inside bracket: $$5\left(x^2 - \frac{14}{5}x + \frac{49}{25} - \frac{49}{25}\right) + 8 = 5\left(x - \frac{7}{5}\right)^2 - 5 \times \frac{49}{25} + 8 = 5\left(x - \frac{7}{5}\right)^2 - \frac{49}{5} + 8$$ 14. Simplify: $8 = \frac{40}{5}$, so $-\frac{49}{5} + \frac{40}{5} = -\frac{9}{5}$. 15. Final form: $5(x - \frac{7}{5})^2 - \frac{9}{5}$. 16. Problem 6(b): Stationary point coordinates are $(b, c)$ from vertex form. 17. Stationary point: $\left(\frac{7}{5}, -\frac{9}{5}\right)$. 18. Problem 7(a): Write $2x^2 + 5x + 3$ in form $2(x + a)^2 + b$. 19. Factor out 2 from the quadratic terms: $2(x^2 + \frac{5}{2}x) + 3$. 20. Half of $\frac{5}{2}$ is $\frac{5}{4}$, square is $\left(\frac{5}{4}\right)^2 = \frac{25}{16}$. 21. Complete the square: $$2\left(x^2 + \frac{5}{2}x + \frac{25}{16} - \frac{25}{16}\right) + 3 = 2\left(x + \frac{5}{4}\right)^2 - 2\times \frac{25}{16} + 3 = 2\left(x + \frac{5}{4}\right)^2 - \frac{25}{8} + 3$$ 22. Write 3 as $\frac{24}{8}$, sum: $-\frac{25}{8} + \frac{24}{8} = -\frac{1}{8}$. 23. So, final form: $2(x + \frac{5}{4})^2 - \frac{1}{8}$. 24. Problem 7(b): Stationary point is $\left(-\frac{5}{4}, -\frac{1}{8}\right)$. 25. Problem 7(c): Solve $2x^2 + 5x + 3 < \frac{15}{8}$. 26. Bring terms on one side: $$2x^2 + 5x + 3 - \frac{15}{8} < 0 \implies 2x^2 + 5x + \frac{9}{8} < 0$$ 27. Multiply entire inequality by 8 to clear denominator: $$16x^2 + 40x + 9 < 0$$ 28. Find roots of $16x^2 + 40x + 9 = 0$: $$x = \frac{-40 \pm \sqrt{40^2 - 4\times 16 \times 9}}{2 \times 16} = \frac{-40 \pm \sqrt{1600 - 576}}{32} = \frac{-40 \pm \sqrt{1024}}{32} = \frac{-40 \pm 32}{32}$$ 29. Roots: $$x_1 = \frac{-40 - 32}{32} = \frac{-72}{32} = -\frac{9}{4}, \quad x_2 = \frac{-40 + 32}{32} = \frac{-8}{32} = -\frac{1}{4}$$ 30. Since coefficient of $x^2$ positive, inequality is true between roots: $$-\frac{9}{4} < x < -\frac{1}{4}$$ 31. Problem 8: Find $a$ such that the line $y = (2a + 1)x -10$ is tangent to curve $y = ax^2 - 5x + 2$. 32. Equate line and curve: $$ax^2 - 5x + 2 = (2a + 1)x - 10$$ 33. Rearrange: $$ax^2 - 5x + 2 - (2a + 1)x + 10 = 0 \implies ax^2 - (5 + 2a + 1)x + 12 = 0$$ $$ax^2 - (2a + 6)x + 12 = 0$$ 34. For tangency, discriminant equals zero: $$\Delta = [-(2a+6)]^2 - 4(a)(12) = 0$$ 35. Compute: $$(2a+6)^2 - 48a = 0 \implies 4a^2 + 24a + 36 - 48a = 0 \implies 4a^2 - 24a + 36 = 0$$ 36. Divide whole equation by 4: $$a^2 - 6a + 9 = 0$$ 37. Factor: $$(a - 3)^2 = 0 \implies a = 3$$ 38. Problem 9: Show that $(2k + 1)x^2 - 4kx + 2k - 1 = 0$ has distinct real roots for $k \neq -\frac{1}{2}$. 39. Calculate discriminant: $$\Delta = (-4k)^2 - 4(2k+1)(2k-1) = 16k^2 - 4(4k^2 -1) = 16k^2 - 16k^2 + 4 = 4$$ 40. Since $\Delta = 4 > 0$, roots are distinct and real for all $k$ where $a = 2k + 1 \neq 0$, i.e., $k \neq -\frac{1}{2}$. 41. Problem 10: For equation $kx^2 + 4kx + 3k + 1 = 0$ to have two different real roots. 42. Since quadratic, $a= k$, roots real and distinct if discriminant $\Delta > 0$, and $a \neq 0$. 43. Compute discriminant: $$\Delta = (4k)^2 - 4(k)(3k+1) = 16k^2 - 4k(3k+1) = 16k^2 - 12k^2 - 4k = 4k^2 - 4k = 4k(k-1)$$ 44. For two different real roots: - $k \neq 0$ (else not quadratic), - $\Delta > 0 \implies 4k(k-1) > 0 \implies k(k-1) > 0$. 45. Inequality $k(k-1) > 0$ holds if $k < 0$ or $k > 1$. 46. Thus, $k < 0$ or $k > 1$.