1. Problem 1 part i: Statement: The roots of the equation $x^2+3x-7=0$ are $\alpha$ and $\beta$; find the new quadratic whose roots are $\alpha-2\beta$ and $\beta-2\alpha$. 2. Solution for 1.i: The sum of the original roots is $\alpha+\beta=-3$. 3. The product of the original roots is $\alpha\beta=-7$. 4. The sum of the new roots is $\alpha-2\beta+\beta-2\alpha=-\alpha-\beta$. 5. Using $\alpha+\beta=-3$ we get the sum of new roots equal to $3$. 6. The product of the new roots is $(\alpha-2\beta)(\beta-2\alpha)$. 7. Expand the product to get $\alpha\beta-2\alpha^2-2\beta^2+4\alpha\beta=5\alpha\beta-2(\alpha^2+\beta^2)$. 8. Use $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$ to rewrite the product as $5\alpha\beta-2((\alpha+\beta)^2-2\alpha\beta)$. 9. Simplify to get the product equal to $9\alpha\beta-2(\alpha+\beta)^2$. 10. Substitute $\alpha\beta=-7$ and $\alpha+\beta=-3$ to obtain the product $9(-7)-2(9)=-63-18=-81$. 11. Therefore the new quadratic with sum $3$ and product $-81$ is $x^2-3x-81=0$. 12. Problem 1 part ii: Statement: If the roots of $ax^2+bx+c=0$ are $\beta$ and $n\beta$ show that $ac(1+n)^2=nb^2$. 13. Solution for 1.ii: The sum of roots equals $\beta+n\beta=\beta(1+n)=-\dfrac{b}{a}$. 14. Hence $\beta=-\dfrac{b}{a(1+n)}$. 15. The product of roots equals $\beta\cdot n\beta=n\beta^2=\dfrac{c}{a}$. 16. Thus $\beta^2=\dfrac{c}{an}$. 17. Square the expression for $\beta$ from the sum to get $\beta^2=\dfrac{b^2}{a^2(1+n)^2}$. 18. Equate the two expressions for $\beta^2$ to obtain $\dfrac{b^2}{a^2(1+n)^2}=\dfrac{c}{an}$. 19. Rearranging gives $nb^2=ac(1+n)^2$, which is the required identity. 20. Problem 2 part a: Arrangements of the letters in the word PAINS. 21. Solution: PAINS has 5 distinct letters so the number of arrangements is $5!=120$. 22. Problem 2 part b: Arrangements of the letters in the word BANANA. 23. Solution: BANANA has 6 letters with A repeated 3 times and N repeated 2 times, so the number of distinct arrangements is $\dfrac{6!}{3!2!}=\dfrac{720}{12}=60$. 24. Problem 2 committee: Statement: A committee of 5 is chosen from 4 men and 5 women. 25. Part i: At least 3 women are chosen; we count by cases for number of women $w=3,4,5$. 26. For $w=3$ the number of committees is $\binom{5}{3}\binom{4}{2}=10\cdot 6=60$. 27. For $w=4$ the number is $\binom{5}{4}\binom{4}{1}=5\cdot 4=20$. 28. For $w=5$ the number is $\binom{5}{5}\binom{4}{0}=1\cdot 1=1$. 29. Summing these gives total $60+20+1=81$ committees when at least 3 women are chosen. 30. Part ii: At most 2 men are chosen; let the number of men be $m=0,1,2$. 31. For $m=0$ the number is $\binom{4}{0}\binom{5}{5}=1\cdot 1=1$. 32. For $m=1$ the number is $\binom{4}{1}\binom{5}{4}=4\cdot 5=20$. 33. For $m=2$ the number is $\binom{4}{2}\binom{5}{3}=6\cdot 10=60$. 34. Summing these gives total $1+20+60=81$ committees when at most 2 men are chosen. 35. Problem 3 part a: Solve the inequality $\dfrac{2}{3+x}\ge \dfrac{1}{4-3x}$. 36. Solution: Bring terms to one side to get $\dfrac{2}{3+x}-\dfrac{1}{4-3x}\ge 0$. 37. Combine into a single fraction to obtain $\dfrac{5-7x}{(3+x)(4-3x)}\ge 0$. 38. The critical points are zeros or poles at $x=\dfrac{5}{7}$, $x=-3$, and $x=\dfrac{4}{3}$. 39. Test intervals give the expression nonnegative on $(-3,\,5/7]$ and on $(4/3,\,\infty)$ with $x=-3$ and $x=4/3$ excluded because they make the denominator zero. 40. Therefore the solution set is $(-3,\,5/7]\cup(4/3,\,\infty)$. 41. Problem 3 part b: Solve $2(\ln x)^2+(\ln x)=3$ for $x$. 42. Solution: Let $t=\ln x$ which gives the quadratic $2t^2+t-3=0$. 43. Solve for $t$ using the quadratic formula to obtain $t=1$ or $t=-3/2$. 44. Back substitute to get $\ln x=1$ which gives $x=e$, and $\ln x=-3/2$ which gives $x=e^{-3/2}$. 45. Both solutions are valid since $x>0$ is required for $\ln x$, so the solutions are $x=e$ and $x=e^{-3/2}$.