1. Problem statement: We are given the quadratic $x^2+3x-7=0$ with roots $\alpha$ and $\beta$, and we must find a new quadratic whose roots are $\alpha-2\beta$ and $\beta-2\alpha$, then show a relation when the roots of $ax^2+bx+c=0$ are $\beta$ and $n\beta$, count permutations of PAINS and BANANA, compute committee counts from 4 men and 5 women under specified constraints, and solve a rational inequality and a logarithmic equation. 2. Q1(i) Find the quadratic with roots $\alpha-2\beta$ and $\beta-2\alpha$. Let the roots of $x^2+3x-7=0$ be $\alpha$ and $\beta$. By Vieta's formulas we have $\alpha+\beta=-3$ and $\alpha\beta=-7$. Set the new roots $r_1=\alpha-2\beta$ and $r_2=\beta-2\alpha$. Compute the sum $S=r_1+r_2=\alpha-2\beta+\beta-2\alpha=-(\alpha+\beta)$. Using $\alpha+\beta=-3$ gives $S=3$. Compute the product $P=r_1r_2=(\alpha-2\beta)(\beta-2\alpha)=\alpha\beta-2\alpha^2-2\beta^2+4\alpha\beta=5\alpha\beta-2(\alpha^2+\beta^2)$. Compute $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=9+14=23$. Substitute $\alpha\beta=-7$ and $\alpha^2+\beta^2=23$ to get $P=5(-7)-2(23)=-35-46=-81$. Thus the required monic quadratic is $x^2-Sx+P=x^2-3x-81=0$. 3. Q1(ii) If the roots of $ax^2+bx+c=0$ are $\beta$ and $n\beta$, prove $ac(1+n)^2=nb^2$. By Vieta $\beta+n\beta=\beta(1+n)=-\frac{b}{a}$ and $n\beta^2=\frac{c}{a}$. From the sum equation $\beta=-\frac{b}{a(1+n)}$. Squaring gives $\beta^2=\frac{b^2}{a^2(1+n)^2}$. Substitute into $n\beta^2=\frac{c}{a}$ to obtain $n\frac{b^2}{a^2(1+n)^2}=\frac{c}{a}$. Multiply both sides by $a^2(1+n)^2$ to get $nb^2=ac(1+n)^2$. This is the desired identity $ac(1+n)^2=nb^2$. 4. Q2(a) Permutations of the word PAINS. PAINS has 5 distinct letters, so the number of arrangements is $5!=120$. 5. Q2(b) Permutations of the word BANANA. BANANA has 6 letters with multiplicities $A:3$, $N:2$, $B:1$. The number of distinct permutations is $\frac{6!}{3!2!1!}=\frac{720}{12}=60$. 6. Q2 committee selections from 4 men and 5 women to form a committee of 5. (i) At least 3 women are chosen, so number of women $k=3,4,5$ and men $5-k$. Total ways $=\sum_{k=3}^{5}\binom{5}{k}\binom{4}{5-k}$. Compute terms: $k=3:\binom{5}{3}\binom{4}{2}=10\times6=60$. $k=4:\binom{5}{4}\binom{4}{1}=5\times4=20$. $k=5:\binom{5}{5}\binom{4}{0}=1\times1=1$. So total $=60+20+1=81$ committees. (ii) At most 2 men are chosen, so men $m=0,1,2$ and women $5-m$. Total ways $=\sum_{m=0}^{2}\binom{4}{m}\binom{5}{5-m}$. Compute terms: $m=0:\binom{4}{0}\binom{5}{5}=1$. $m=1:\binom{4}{1}\binom{5}{4}=4\times5=20$. $m=2:\binom{4}{2}\binom{5}{3}=6\times10=60$. So total $=1+20+60=81$ committees, the same as part (i). 7. Q3(a) Solve the inequality $\dfrac{2}{8+x}\ge\dfrac{1}{4-3x}$. Domain restrictions are $x\neq -8$ and $x\neq \frac{4}{3}$. Bring terms to one side and combine over a common denominator to obtain $\dfrac{2}{8+x}-\dfrac{1}{4-3x}=\dfrac{-7x}{(8+x)(4-3x)}$. Thus the inequality is $\dfrac{-7x}{(8+x)(4-3x)}\ge 0$. Multiply both sides by $-1$ and reverse the inequality to get $\dfrac{7x}{(8+x)(4-3x)}\le 0$. Since $7>0$ this is equivalent to $\dfrac{x}{(8+x)(4-3x)}\le 0$. Critical points are $x=-8$, $x=0$, and $x=\frac{4}{3}$ which partition the real line. Testing signs on the intervals yields the solution set $(-8,0]\cup\left(\frac{4}{3},\infty\right)$ with $x\neq -8$ and $x\neq \frac{4}{3}$. 8. Q3(b) Solve $2(\ln x)^2+\ln(2x)=3$ for $x>0$. Let $y=\ln x$ so that $\ln(2x)=\ln 2+\ln x=\ln 2+y$. Substitute to obtain the quadratic $2y^2+y+\ln 2-3=0$. Apply the quadratic formula to get $y=\frac{-1\pm\sqrt{1-8(\ln 2-3)}}{4}=\frac{-1\pm\sqrt{25-8\ln 2}}{4}$. Therefore $x=\exp\left(\frac{-1\pm\sqrt{25-8\ln 2}}{4}\right)$ and both roots are positive and valid. Numerically $\ln 2\approx0.693147$, $\sqrt{25-8\ln 2}\approx4.411$, giving $y\approx0.8528$ or $y\approx-1.3528$ and hence $x\approx2.346$ or $x\approx0.2586$. 9. Final answers summary: The new quadratic in Q1(i) is $x^2-3x-81=0$. The identity in Q1(ii) is $ac(1+n)^2=nb^2$. Arrangements: PAINS gives 120 and BANANA gives 60. Committee counts in Q2 are 81 for each constrained case. The solution to Q3(a) is $(-8,0]\cup\left(\frac{4}{3},\infty\right)$. The solutions to Q3(b) are $x=\exp\left(\dfrac{-1\pm\sqrt{25-8\ln 2}}{4}\right)\approx2.346\text{ or }0.2586$.