1. Problem (ii): One complex solution to $f(x)=x^2+2x+5$ is $w_1=-1+2i$. Determine the other solution $w_2$ and state the relationship between $w_1$ and $w_2$. 2. Formula and rule: For a quadratic $ax^2+bx+c=0$ the sum of the roots is $\alpha+\beta=-\frac{b}{a}$ and the product is $\alpha\beta=\frac{c}{a}$. Complex roots of polynomials with real coefficients occur in conjugate pairs. 3. Apply to $x^2+2x+5$: here $a=1$, $b=2$, $c=5$. Thus the sum of roots is $-2$ and the product is $5$. 4. Compute $w_2$ using $w_1+w_2=-2$. So $w_2=-2-w_1$. Substitute $w_1=-1+2i$ to get $w_2=-2-(-1+2i)=-1-2i$. 5. Final answer for (ii): $w_2=-1-2i$. Relationship: $w_2=\overline{w_1}$, they are complex conjugates, and $w_1+w_2=-2$, $w_1w_2=5$. 6. Problem (iii): Determine the modulus of $w_1=-1+2i$. 7. Formula: The modulus of $a+bi$ is $|a+bi|=\sqrt{a^2+b^2}$. 8. Apply: $|w_1|=\sqrt{(-1)^2+2^2}=\sqrt{1+4}=\sqrt{5}$. 9. Final answer for (iii): $|w_1|=\sqrt{5}$. 10. Problem (b)(i): Solve $|3x+2|\le 5$ for $x$. 11. Rule: $|u|\le k$ with $k\ge 0$ is equivalent to $-k\le u\le k$. 12. Apply: $-5\le 3x+2\le 5$. 13. Subtract 2: $-7\le 3x\le 3$. 14. Divide by 3: $-\frac{7}{3}\le x\le 1$. 15. Final answer for (b)(i): $x\in[-\frac{7}{3},1]$. 16. Problem: Find the coefficient of $x$ in the expansion of $\left(5-\frac{x}{3}\right)^3$. 17. Formula: Use the binomial theorem $ (a+b)^3=a^3+3a^2b+3ab^2+b^3$. 18. Set $a=5$ and $b=-\frac{x}{3}$ and compute each term. 19. Compute $a^3=125$. 20. Compute $3a^2b=3\cdot25\cdot\left(-\frac{x}{3}\right)=75\left(-\frac{x}{3}\right)=-25x$. 21. Compute $3ab^2=3\cdot5\cdot\left(\frac{x^2}{9}\right)=15\cdot\frac{x^2}{9}=\frac{5}{3}x^2$. 22. Compute $b^3=\left(-\frac{x}{3}\right)^3=-\frac{x^3}{27}$. 23. So the expansion is $125-25x+\frac{5}{3}x^2-\frac{x^3}{27}$. 24. Final answer: the coefficient of $x$ is $-25$. 25. Problem (d)(i): Given purchases, write equations with $x,y,z$ the prices of pencil, pen, eraser respectively. 26. First purchase gives $4x+3y+3z=74$ which is given. 27. Second purchase: 3 pencils, 1 pen, 1 eraser for 33 gives $3x+y+z=33$. 28. Third purchase: 1 pencil and 4 pens for 45 gives $x+4y=45$ which can be written $x+4y+0z=45$. 29. Problem (d)(ii): Construct the augmented matrix $[A|b]$ for the system above. 30. The coefficient matrix $A$ and column $b$ yield the augmented matrix. 31. In display form: $$[A|b]=\begin{bmatrix}4 & 3 & 3 & 74\\3 & 1 & 1 & 33\\1 & 4 & 0 & 45\end{bmatrix}$$. 32. Final answer for (d)(ii): the augmented matrix is as shown above. 33. Problem (a)(i): Simplify $\frac{6-7i}{1-2i}$ into standard form $a+bi$. 34. Method: Multiply numerator and denominator by the complex conjugate of the denominator. 35. Compute $(6-7i)(1+2i)=6+12i-7i-14i^2=6+5i+14=20+5i$. 36. Denominator $(1-2i)(1+2i)=1+4=5$. 37. Divide to get $\frac{20+5i}{5}=4+i$. 38. Final answer for (a)(i): $4+i$.