1. **Problem (a): Express $16x^2 - 24x + 10$ in the form $(4x + a)^2 + b$.** 2. Start with the expression: $$16x^2 - 24x + 10$$ 3. Factor out the coefficient of $x^2$ from the first two terms: $$16x^2 - 24x = 16\left(x^2 - \frac{24}{16}x\right) = 16\left(x^2 - \frac{3}{2}x\right)$$ 4. Complete the square inside the parentheses. Recall the formula: $$x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2$$ Here, $b = -\frac{3}{2}$, so: $$\left(x - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 = x^2 - \frac{3}{2}x$$ 5. Substitute back: $$16\left(\left(x - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2\right) + 10 = 16\left(x - \frac{3}{4}\right)^2 - 16\times \frac{9}{16} + 10$$ 6. Simplify the constants: $$16\times \frac{9}{16} = 9$$ So the expression becomes: $$16\left(x - \frac{3}{4}\right)^2 - 9 + 10 = 16\left(x - \frac{3}{4}\right)^2 + 1$$ 7. Rewrite in the form $(4x + a)^2 + b$: Note that: $$16\left(x - \frac{3}{4}\right)^2 = (4x - 3)^2$$ Therefore: $$(4x - 3)^2 + 1$$ So, $a = -3$ and $b = 1$. --- 8. **Problem (b): Given $16x^2 - 24x + 10 = k$ has exactly one root, find the root.** 9. For a quadratic equation $ax^2 + bx + c = k$ to have exactly one root, the discriminant must be zero: $$\Delta = b^2 - 4a(c - k) = 0$$ Here, $a=16$, $b=-24$, and $c=10$. 10. Substitute values: $$(-24)^2 - 4 \times 16 \times (10 - k) = 0$$ $$576 - 64(10 - k) = 0$$ 11. Expand and solve for $k$: $$576 - 640 + 64k = 0$$ $$64k - 64 = 0$$ $$64k = 64$$ $$k = 1$$ 12. Now solve for the root when $16x^2 - 24x + 10 = 1$: $$16x^2 - 24x + 9 = 0$$ 13. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=16$, $b=-24$, $c=9$. 14. Calculate discriminant: $$(-24)^2 - 4 \times 16 \times 9 = 576 - 576 = 0$$ 15. Since discriminant is zero, one root: $$x = \frac{24}{2 \times 16} = \frac{24}{32} = \frac{3}{4}$$ --- 16. **Problem (14): Find $k$ and the tangent point where the line $y = kx - k$ is tangent to the curve $y = -\frac{1}{2x}$.** 17. At the tangent point, the line and curve share the same $x$, $y$, and slope. 18. Set the line and curve equal: $$kx - k = -\frac{1}{2x}$$ Multiply both sides by $2x$ to clear denominator: $$2x(kx - k) = -1$$ $$2kx^2 - 2kx = -1$$ 19. Rearrange: $$2kx^2 - 2kx + 1 = 0$$ 20. The line is tangent to the curve, so this quadratic in $x$ has exactly one solution. Discriminant zero: $$\Delta = (-2k)^2 - 4 \times 2k \times 1 = 0$$ $$4k^2 - 8k = 0$$ 21. Factor: $$4k(k - 2) = 0$$ Since $k$ is positive, $k = 2$. 22. Substitute $k=2$ back into the quadratic: $$2 \times 2 x^2 - 2 \times 2 x + 1 = 0$$ $$4x^2 - 4x + 1 = 0$$ 23. Solve for $x$: Discriminant: $$(-4)^2 - 4 \times 4 \times 1 = 16 - 16 = 0$$ One root: $$x = \frac{4}{2 \times 4} = \frac{4}{8} = \frac{1}{2}$$ 24. Find $y$ coordinate on the curve: $$y = -\frac{1}{2x} = -\frac{1}{2 \times \frac{1}{2}} = -1$$ 25. Check $y$ on the line: $$y = kx - k = 2 \times \frac{1}{2} - 2 = 1 - 2 = -1$$ Matches, confirming the tangent point. --- **Final answers:** (a) $16x^2 - 24x + 10 = (4x - 3)^2 + 1$ (b) The value of $k$ for one root is $1$, and the root is $x = \frac{3}{4}$. (14) $k = 2$, tangent point is $\left(\frac{1}{2}, -1\right)$.