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 rules: For a quadratic $ax^2+bx+c=0$ with real coefficients the roots satisfy $w_1+w_2=-\frac{b}{a}$ and complex roots occur in conjugate pairs. 3. Intermediate work: Here $a=1$ and $b=2$ so $w_1+w_2=-\frac{2}{1}=-2$. 4. Compute $w_2$: $w_2=-2-w_1=-2-(-1+2i)=-1-2i$. 5. Relationship: $w_2$ is the complex conjugate of $w_1$ since $w_2=\overline{w_1}=-1-2i$. 6. Problem (iii): Determine the modulus of $w_1=-1+2i$. 7. Formula and rules: The modulus of $w=a+bi$ is $|w|=\sqrt{a^2+b^2}$. 8. Intermediate work: For $w_1$ we have $a=-1$ and $b=2$ so $|w_1|=\sqrt{(-1)^2+2^2}=\sqrt{1+4}=\sqrt{5}$. 9. Final: $|w_1|=\sqrt{5}$. 10. Problem: Solve the inequality $|3x+2|\le 5$. 11. Formula and rules: $|u|\le k$ with $k\ge0$ is equivalent to $-k\le u\le k$. 12. Intermediate work: Apply with $u=3x+2$ and $k=5$ to get $-5\le 3x+2\le 5$. 13. Solve: Subtract 2: $-7\le 3x\le 3$. 14. Divide by 3: $-\frac{7}{3}\le x\le 1$. 15. Final: The solution set is $-\frac{7}{3}\le x\le 1$. 16. Problem: Find the coefficient of the $x$ term in the expansion of $(5-\frac{x}{3})^3$. 17. Formula and rules: Use the binomial expansion $(a+b)^3=a^3+3a^2b+3ab^2+b^3$. 18. Intermediate work: Take $a=5$ and $b=-\frac{x}{3}$. 19. Compute terms: $a^3=125$ which is constant, $3a^2b=3\cdot25\cdot(-\frac{x}{3})=25\cdot(-x)=-25x$, $3ab^2=3\cdot5\cdot\left(-\frac{x}{3}\right)^2=15\cdot\frac{x^2}{9}=\frac{5}{3}x^2$, $b^3= -\frac{x^3}{27}$. 20. Extract coefficient of $x$: The only linear term is $-25x$ so the coefficient is $-25$. 21. Problem (d)(i): John bought stationery at fixed prices: first purchase $4x+3y+3z=74$. Use the information to state two linear equations for the other purchases. 22. Statement: Second purchase (3 pencils, 1 pen, 1 eraser) gives $3x+y+z=33$. 23. Statement: Third purchase (1 pencil and 4 pens) gives $x+4y=45$. 24. Problem (d)(ii): Construct the augmented matrix $[A|b]$ for the system above. 25. Definition: $A$ is the $3\times3$ coefficient matrix and $b$ the $3\times1$ constants column. 26. Augmented matrix (displayed): $$[A|b]=\begin{bmatrix}4 & 3 & 3 & | & 74\\3 & 1 & 1 & | & 33\\1 & 4 & 0 & | & 45\end{bmatrix}$$ 27. Problem (e): The midpoint of segment $PQ$ is $(-7,-2)$ and $P=(6,8)$. Find $Q$. 28. Formula and rules: Midpoint $M=\left(\frac{x_P+x_Q}{2},\frac{y_P+y_Q}{2}\right)$ so $x_Q=2x_M-x_P$ and $y_Q=2y_M-y_P$. 29. Intermediate work: Compute $x_Q=2(-7)-6=-14-6=-20$. 30. Intermediate work: Compute $y_Q=2(-2)-8=-4-8=-12$. 31. Final: $Q=(-20,-12)$. Final answers summary: - Other root: $w_2=-1-2i$ and $w_2=\overline{w_1}$. - Modulus: $|w_1|=\sqrt{5}$. - Inequality solution: $-\frac{7}{3}\le x\le 1$. - Coefficient of $x$ in $(5-\frac{x}{3})^3$: $-25$. - Other purchase equations: $3x+y+z=33$ and $x+4y=45$. - Augmented matrix: displayed above. - Midpoint result: $Q=(-20,-12)$.