1. Problem: Given the roots of an equation are $\alpha$ and $\beta$, find the new equation whose roots are $\alpha - 2\beta$ and $\beta - 2\alpha$. Step 1: Let the original equation be $x^2 - (\alpha + \beta)x + \alpha\beta = 0$. Step 2: New roots are $r_1 = \alpha - 2\beta$ and $r_2 = \beta - 2\alpha$. Step 3: Sum of new roots: $$r_1 + r_2 = (\alpha - 2\beta) + (\beta - 2\alpha) = -\alpha - \beta = -(\alpha + \beta).$$ Step 4: Product of new roots: $$r_1 \cdot r_2 = (\alpha - 2\beta)(\beta - 2\alpha) = \alpha\beta - 2\alpha^2 - 2\beta^2 + 4\alpha\beta = 5\alpha\beta - 2(\alpha^2 + \beta^2).$$ Note that: $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta.$$ So: $$r_1 r_2 = 5\alpha\beta - 2[(\alpha + \beta)^2 - 2\alpha\beta] = 5\alpha\beta - 2(\alpha + \beta)^2 + 4\alpha\beta = 9\alpha\beta - 2(\alpha + \beta)^2.$$ Step 5: The new equation with roots $r_1$ and $r_2$ is: $$x^2 - (r_1 + r_2)x + r_1 r_2 = 0 $$ Substitute sums and products: $$x^2 + (\alpha + \beta)x + 2(\alpha + \beta)^2 - 9\alpha\beta = 0.$$ 2. Problem: If roots of $ax^2 + bx + c = 0$ are $\beta$ and $n\beta$, show that $ac(1+n)^2 = nb^2$. Step 1: Sum of roots: $$\beta + n\beta = \beta(1+n) = -\frac{b}{a}.$$ Step 2: Product of roots: $$\beta \cdot n\beta = n\beta^2 = \frac{c}{a}.$$ Step 3: From sum, solve for $\beta$: $$\beta = -\frac{b}{a(1+n)}.$$ Step 4: Substitute $\beta$ in product: $$n\left(-\frac{b}{a(1+n)}\right)^2 = \frac{c}{a}$$ Simplify: $$n \frac{b^2}{a^2(1+n)^2} = \frac{c}{a} \Rightarrow \frac{n b^2}{a^2 (1+n)^2} = \frac{c}{a}$$ Multiply both sides by $a^2(1+n)^2$: $$n b^2 = a c (1+n)^2$$ This proves the required relation. 3. Problem: Ways to arrange words. a. Number of ways to arrange PAINS: PAINS has 5 distinct letters, Number of arrangements = $5! = 120$. b. Number of ways to arrange BANANA: BANANA has letters B(1), A(3), N(2). Number of arrangements = $$\frac{6!}{3!2!1!} = \frac{720}{6\cdot 2} = 60.$$ 4. Problem: Committees of 5 from 4 men and 5 women. Total people = 9. (i) At least 3 women chosen. Possible numbers of women: 3, 4, or 5. - Women = 3, Men = 2: $\binom{5}{3} \times \binom{4}{2} = 10 \times 6 = 60$. - Women = 4, Men = 1: $\binom{5}{4} \times \binom{4}{1} = 5 \times 4 = 20$. - Women = 5, Men = 0: $\binom{5}{5} \times \binom{4}{0} = 1 \times 1 = 1$. Total = $60 + 20 + 1 = 81$. (ii) At most 2 men chosen. Men can be 0,1, or 2. - Men = 0, Women = 5: $\binom{4}{0} \times \binom{5}{5} = 1$. - Men = 1, Women = 4: $\binom{4}{1} \times \binom{5}{4} = 4 \times 5 = 20$. - Men = 2, Women = 3: $\binom{4}{2} \times \binom{5}{3} = 6 \times 10 = 60$. Total = $1 + 20 + 60 = 81$. 5. Problem: Solve equations. a. Solve $\frac{2}{3+x} \ge \frac{1}{4 - 3x}$. Step 1: Write inequality: $$\frac{2}{3+x} - \frac{1}{4 - 3x} \ge 0.$$ Step 2: Find common denominator: $$\frac{2(4 - 3x) - (3 + x)}{(3+x)(4 - 3x)} \ge 0.$$ Simplify numerator: $$8 - 6x - 3 - x = 5 - 7x.$$ Inequality becomes: $$\frac{5 - 7x}{(3+x)(4 - 3x)} \ge 0.$$ Step 3: Find critical points where numerator or denominator is zero: $$5 - 7x = 0 \Rightarrow x = \frac{5}{7},$$ $$3 + x = 0 \Rightarrow x = -3,$$ $$4 - 3x = 0 \Rightarrow x = \frac{4}{3}.$$ Step 4: Determine sign intervals around $x = -3, \frac{5}{7}, \frac{4}{3}$ and check where fraction is $\ge 0$. b. Solve $2(\ln x)^2 + \ln x = 3$. Step 1: Let $y = \ln x$, rewrite equation: $$2y^2 + y - 3 = 0.$$ Step 2: Solve quadratic: $$y = \frac{-1 \pm \sqrt{1 + 24}}{4} = \frac{-1 \pm 5}{4}.$$ Solutions: $$y = 1, \quad y = -\frac{3}{2}.$$ Step 3: Substitute back: $$\ln x = 1 \Rightarrow x = e,$$ $$\ln x = -\frac{3}{2} \Rightarrow x = e^{-3/2}.$$ Final answers: 1. New equation: $x^2 + (\alpha+\beta)x + 2(\alpha+\beta)^2 - 9\alpha\beta=0$ 2. $ac(1+n)^2 = nb^2$ 3a. 120 ways, 3b. 60 ways 4i. 81 committees, 4ii. 81 committees 5a. Solve inequality as in step 4, 5b. $x = e$ or $x = e^{-3/2}$.