Relation between roots and coefficients Questions in English

(B) Given the quadratic equation $3x^{2} + \lambda x - 1 = 0$ with roots $\alpha$ and $\beta$.
From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{\lambda}{3}$ and $\alpha\beta = -\frac{1}{3}$.
The sum of the squares of the reciprocals is given by $\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = 15$.
This simplifies to $\frac{\alpha^{2} + \beta^{2}}{(\alpha\beta)^{2}} = 15$,which is $\frac{(\alpha + \beta)^{2} - 2\alpha\beta}{(\alpha\beta)^{2}} = 15$.
Substituting the values: $\frac{(-\lambda/3)^{2} - 2(-1/3)}{(-1/3)^{2}} = 15$.
$\frac{\lambda^{2}/9 + 2/3}{1/9} = 15 \implies \lambda^{2} + 6 = 15 \implies \lambda^{2} = 9$.
Now,we need to calculate $6(\alpha^{3} + \beta^{3})^{2}$.
Recall that $\alpha^{3} + \beta^{3} = (\alpha + \beta)((\alpha + \beta)^{2} - 3\alpha\beta)$.
$\alpha^{3} + \beta^{3} = (-\frac{\lambda}{3})((-\frac{\lambda}{3})^{2} - 3(-\frac{1}{3})) = (-\frac{\lambda}{3})(\frac{\lambda^{2}}{9} + 1)$.
Since $\lambda^{2} = 9$,$\alpha^{3} + \beta^{3} = (-\frac{\lambda}{3})(\frac{9}{9} + 1) = (-\frac{\lambda}{3})(2) = -\frac{2\lambda}{3}$.
Then $6(\alpha^{3} + \beta^{3})^{2} = 6(-\frac{2\lambda}{3})^{2} = 6(\frac{4\lambda^{2}}{9}) = 6(\frac{4 \times 9}{9}) = 6 \times 4 = 24$.

(B) The given equation is $x^{4}-3x^{3}-2x^{2}+3x+1=10$,which simplifies to $x^{4}-3x^{3}-2x^{2}+3x-9=0$.
Dividing by $x^{2}$ (since $x=0$ is not a root),we get $x^{2}-3x-2+\frac{3}{x}-\frac{9}{x^{2}}=0$.
Rearranging terms: $(x^{2}-\frac{9}{x^{2}})-3(x-\frac{1}{x})-2=0$. This approach is complex,let's factorize directly.
$x^{4}-3x^{3}-2x^{2}+3x-9=0$
$x^{3}(x-3)-2(x^{2}-1.5x+4.5)$ is not direct. Let's group: $(x^{4}-9)-3x(x^{2}-1)-2x^{2}=0$.
Actually,$x^{4}-3x^{3}-2x^{2}+3x-9=0$ can be written as $(x^{2}-3)(x^{2}-3x+3)=0$ is incorrect. Let's test roots: $x=3$ gives $81-81-18+9-9 \neq 0$. $x=-1$ gives $1+3-2-3-9 \neq 0$. $x=3$ is a root of $x^{2}-3x-3=0$?
Let's use $x^{2}-3x-3=0$ and $x^{2}+3=0$. Roots are $x^{2}=3x+3$ and $x^{2}=-3$.
For $x^{2}=-3$,$x = \pm i\sqrt{3}$. Cubes are $-3i\sqrt{3}$ and $3i\sqrt{3}$,sum $= 0$.
For $x^{2}-3x-3=0$,roots $\alpha, \beta$ satisfy $\alpha+\beta=3, \alpha\beta=-3$.
Sum of cubes $\alpha^{3}+\beta^{3} = (\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta) = 3^{3}-3(-3)(3) = 27+27 = 54$.
Wait,the equation is $x^{4}-3x^{3}-2x^{2}+3x-9=0$. $(x^{2}-3)(x^{2}-3x+3)=0$ is $x^{4}-3x^{3}+3x^{2}-3x^{2}+9x-9 = x^{4}-3x^{3}+9x-9$. Not matching.
Correct factorization: $(x^{2}+3)(x^{2}-3x-3)=0$.
Roots of $x^{2}+3=0$ are $i\sqrt{3}, -i\sqrt{3}$. Cubes are $-3i\sqrt{3}, 3i\sqrt{3}$,sum $= 0$.
Roots of $x^{2}-3x-3=0$ are $\alpha, \beta$. $\alpha+\beta=3, \alpha\beta=-3$.
Sum of cubes $= (\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta) = 27-3(-3)(3) = 27+27 = 54$.

(A) Let the quadratic polynomial be $f(x) = a(x - \alpha)(x - \beta)$. Since one root is $-1$,let $\alpha = -1$. Then $f(x) = a(x + 1)(x - \beta)$.
Given $f(-2) + f(3) = 0$.
$f(-2) = a(-2 + 1)(-2 - \beta) = a(-1)(-2 - \beta) = a(2 + \beta)$.
$f(3) = a(3 + 1)(3 - \beta) = a(4)(3 - \beta) = a(12 - 4\beta)$.
Summing these: $a(2 + \beta + 12 - 4\beta) = 0$.
Since $a \neq 0$,we have $14 - 3\beta = 0$,which gives $\beta = \frac{14}{3}$.
The roots are $-1$ and $\frac{14}{3}$.
The sum of the roots is $-1 + \frac{14}{3} = \frac{-3 + 14}{3} = \frac{11}{3}$.

(C) For the equation $x^{2}-8ax+2a=0$,the roots are $p$ and $r$. Thus,$p+r=8a$ and $pr=2a$.
Then $\frac{1}{p}+\frac{1}{r} = \frac{p+r}{pr} = \frac{8a}{2a} = 4$.
For the equation $x^{2}+12bx+6b=0$,the roots are $q$ and $s$. Thus,$q+s=-12b$ and $qs=6b$.
Then $\frac{1}{q}+\frac{1}{s} = \frac{q+s}{qs} = \frac{-12b}{6b} = -2$.
Let the $A$.$P$. be $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ with common difference $d$.
Then $\frac{1}{q} = \frac{1}{p}+d$,$\frac{1}{r} = \frac{1}{p}+2d$,and $\frac{1}{s} = \frac{1}{p}+3d$.
We have $\frac{1}{p}+\frac{1}{r} = \frac{2}{p}+2d = 4$,so $\frac{1}{p}+d = 2$. Thus $\frac{1}{q} = 2$.
We have $\frac{1}{q}+\frac{1}{s} = \frac{2}{q}+2d = -2$,so $\frac{1}{q}+d = -1$.
Since $\frac{1}{q}=2$,we have $2+d=-1$,so $d=-3$.
Then $\frac{1}{p} = \frac{1}{q}-d = 2-(-3) = 5$,so $p = \frac{1}{5}$.
Since $pr=2a$,$r = \frac{2a}{p} = 10a$.
Also $\frac{1}{r} = \frac{1}{p}+2d = 5+2(-3) = -1$,so $r = -1$.
Thus $10a = -1$,which gives $a = -\frac{1}{10}$,so $a^{-1} = -10$.
Also $\frac{1}{s} = \frac{1}{q}+2d = 2+2(-3) = -4$,so $s = -\frac{1}{4}$.
Since $qs=6b$,$q = \frac{1}{2}$,so $b = \frac{qs}{6} = \frac{(1/2)(-1/4)}{6} = -\frac{1}{48}$,so $b^{-1} = -48$.
Finally,$a^{-1}-b^{-1} = -10 - (-48) = 38$.

(C) Let the roots of the quadratic equation $x^{2}+(3-a)x+(1-2a)=0$ be $\alpha$ and $\beta$.
From the relation between roots and coefficients,we have $\alpha+\beta = -(3-a) = a-3$ and $\alpha\beta = 1-2a$.
The sum of the squares of the roots is given by $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2}-2\alpha\beta$.
Substituting the values,we get $f(a) = (a-3)^{2} - 2(1-2a)$.
Expanding this,$f(a) = a^{2} - 6a + 9 - 2 + 4a = a^{2} - 2a + 7$.
To find the minimum value,we complete the square: $f(a) = (a^{2} - 2a + 1) + 6 = (a-1)^{2} + 6$.
Since $(a-1)^{2} \geq 0$,the minimum value of $f(a)$ is $6$ when $a=1$.

(D) Given the equation: $x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0$.
Simplifying the coefficients:
$3^{\sqrt{\log _{3} 5}} = 3^{\sqrt{\log _{3} 5} \cdot \frac{\sqrt{\log _{3} 5}}{\sqrt{\log _{3} 5}}} = 3^{\log _{3} 5 \cdot \sqrt{\log _{5} 3}} = (3^{\log _{3} 5})^{\sqrt{\log _{5} 3}} = 5^{\sqrt{\log _{5} 3}}$.
Thus,the coefficient of $x$ is $-(5 + 5^{\sqrt{\log _{5} 3}} - 5^{\sqrt{\log _{5} 3}}) = -5$.
Similarly,$3^{\left(\log _{3} 5\right)^{\frac{1}{3}}} = 5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}$.
Thus,the constant term is $3(5^{\left(\log _{5} 3\right)^{\frac{2}{3}}} - 5^{\left(\log _{5} 3\right)^{\frac{2}{3}}} - 1) = -3$.
The equation simplifies to $x^{2}-5x-3=0$.
For this equation,$\alpha+\beta=5$ and $\alpha\beta=-3$.
The new roots are $\alpha+\frac{1}{\beta} = \frac{\alpha\beta+1}{\beta} = \frac{-3+1}{\beta} = \frac{-2}{\beta}$ and $\beta+\frac{1}{\alpha} = \frac{-2}{\alpha}$.
Let $t = \frac{-2}{\alpha}$,then $\alpha = \frac{-2}{t}$.
Substituting into $x^{2}-5x-3=0$:
$(\frac{-2}{t})^{2} - 5(\frac{-2}{t}) - 3 = 0 \Rightarrow \frac{4}{t^{2}} + \frac{10}{t} - 3 = 0$.
Multiplying by $t^{2}$: $4 + 10t - 3t^{2} = 0 \Rightarrow 3t^{2}-10t-4=0$.
Therefore,the required equation is $3x^{2}-10x-4=0$.

(C) Given the cubic equation $x^3+ax^2+bx+c=0$ with roots $a, b, c$.
By Vieta's formulas:
$a+b+c = -a \implies 2a+b+c = 0$ $(i)$
$ab+bc+ca = b$ (ii)
$abc = -c \implies ab = -1$ (since $c \neq 0$) (iii)
From (iii),$b = -\frac{1}{a}$.
Substitute $b$ into $(i)$: $2a - \frac{1}{a} + c = 0 \implies c = \frac{1}{a} - 2a$.
Substitute $b$ and $c$ into (ii):
$ab + c(a+b) = b$
$-1 + (\frac{1}{a} - 2a)(a - \frac{1}{a}) = -\frac{1}{a}$
$-1 + (1 - \frac{1}{a^2} - 2a^2 + 2) = -\frac{1}{a}$
$2 - \frac{1}{a^2} - 2a^2 = -\frac{1}{a}$
Multiply by $a^2$: $2a^2 - 1 - 2a^4 = -a$
$2a^4 - 2a^2 - a + 1 = 0$
$(a-1)(2a^3+2a^2-1) = 0$.
For $a=1$,$b=-1$,$c=-1$. Roots are $1, -1, -1$. Check: $x^3+x^2-x-1=0 \implies (x-1)(x+1)^2=0$. Roots are $1, -1, -1$. This works.
For $2a^3+2a^2-1=0$,there is one real root $a \approx 0.589$. This yields a second valid triple $(a, b, c)$.
Thus,there are exactly two such triples.

(B) Given $f(x) = x^6 - 2x^5 + x^3 + x^2 - x - 1$ and $g(x) = x^4 - x^3 - x^2 - 1$.
By performing polynomial division of $f(x)$ by $g(x)$,we get:
$f(x) = (x^2 - x)g(x) + (2x^2 - 2x - 1)$.
Since $a, b, c, d$ are roots of $g(x) = 0$,we have $g(a) = g(b) = g(c) = g(d) = 0$.
Thus,$f(a) = 2a^2 - 2a - 1$,$f(b) = 2b^2 - 2b - 1$,$f(c) = 2c^2 - 2c - 1$,and $f(d) = 2d^2 - 2d - 1$.
Summing these,we get $\sum f(a) = 2\sum a^2 - 2\sum a - 4$.
From $g(x) = x^4 - x^3 - x^2 - 1 = 0$,by Vieta's formulas,$\sum a = 1$ and $\sum ab = -1$.
We know $\sum a^2 = (\sum a)^2 - 2\sum ab = (1)^2 - 2(-1) = 1 + 2 = 3$.
Substituting these values,$\sum f(a) = 2(3) - 2(1) - 4 = 6 - 2 - 4 = 0$.

(B) Let $f(x) = x^3 - 27x + k = 0$. Let the roots be $\alpha, \beta, \gamma$. Since the coefficient of $x^2$ is $0$,the sum of the roots $\alpha + \beta + \gamma = 0$. If two roots are distinct integers,the third root must also be an integer.
Let the roots be $x_1, x_2, x_3$. Then $x_1 + x_2 + x_3 = 0$ and $x_1 x_2 + x_2 x_3 + x_3 x_1 = -27$.
Substituting $x_3 = -(x_1 + x_2)$ into the second equation: $x_1 x_2 - (x_1 + x_2)^2 = -27$,which simplifies to $x_1^2 + x_1 x_2 + x_2^2 = 27$.
This is the equation of an ellipse. We look for integer solutions $(x_1, x_2)$.
If $x_1 = 0$,$x_2^2 = 27$ (no integer solution).
If $x_1 = 3$,$9 + 3x_2 + x_2^2 = 27$ $\Rightarrow x_2^2 + 3x_2 - 18 = 0$ $\Rightarrow (x_2+6)(x_2-3) = 0$. So $x_2 = 3$ or $x_2 = -6$.
If $x_2 = 3$,roots are $(3, 3, -6)$. Then $k = -(x_1 x_2 x_3) = -(3 \times 3 \times -6) = 54$.
If $x_2 = -6$,roots are $(3, -6, 3)$,same as above.
If $x_1 = -3$,$9 - 3x_2 + x_2^2 = 27$ $\Rightarrow x_2^2 - 3x_2 - 18 = 0$ $\Rightarrow (x_2-6)(x_2+3) = 0$. So $x_2 = 6$ or $x_2 = -3$.
If $x_2 = 6$,roots are $(-3, 6, -3)$. Then $k = -(-3 \times 6 \times -3) = -54$.
If $x_2 = -3$,roots are $(-3, -3, 6)$,same as above.
Thus,$k = 54$ or $k = -54$. There are $2$ such integers.

(C) Given $p(x)=ax^2+bx+c$ with $a, b, c > 0$ and $b-a=c-b$,which implies $2b=a+c$,so $a, b, c$ are in arithmetic progression.
Let $\alpha, \beta$ be the integer roots of $p(x)=0$. By Vieta's formulas,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Since $b = \frac{a+c}{2}$,we have $\alpha+\beta = -\frac{a+c}{2a} = -\frac{1}{2} - \frac{c}{2a}$.
Also,$\alpha\beta = \frac{c}{a}$. Substituting $\frac{c}{a} = \alpha\beta$ into the sum equation:
$\alpha+\beta = -\frac{1}{2} - \frac{\alpha\beta}{2}$ $\Rightarrow 2(\alpha+\beta) = -1 - \alpha\beta$ $\Rightarrow \alpha\beta + 2\alpha + 2\beta = -1$.
Adding $4$ to both sides to factorize:
$(\alpha+2)(\beta+2) = 3$.
Since $\alpha, \beta$ are integers,the possible pairs for $(\alpha+2, \beta+2)$ are $(1, 3), (3, 1), (-1, -3), (-3, -1)$.
Case $1$: $(\alpha+2, \beta+2) = (1, 3) \Rightarrow \alpha=-1, \beta=1$. Then $\alpha+\beta+\alpha\beta = -1+1+(-1)(1) = -1$ (Not in range).
Case $2$: $(\alpha+2, \beta+2) = (-1, -3) \Rightarrow \alpha=-3, \beta=-5$. Then $\alpha+\beta+\alpha\beta = -3-5+(-3)(-5) = -8+15 = 7$.
Since $7$ is in the range $0 \leq \alpha+\beta+\alpha\beta \leq 8$,the possible value is $7$.

(C) Let the roots of the equation $x^2+bx+a=0$ be $\alpha$ and $\beta$.
From the properties of roots,we have $\alpha+\beta = -b$ and $\alpha\beta = a$.
Given that the roots of the equation $x^2+ax+b=0$ are each $1$ greater than the roots of the first equation,the roots are $(\alpha+1)$ and $(\beta+1)$.
From the properties of roots for the second equation,we have:
$(\alpha+1)+(\beta+1) = -a \implies \alpha+\beta+2 = -a$.
Substituting $\alpha+\beta = -b$,we get $-b+2 = -a$,which simplifies to $a-b = -2$ (Equation $1$).
Also,$(\alpha+1)(\beta+1) = b \implies \alpha\beta + \alpha+\beta + 1 = b$.
Substituting $\alpha\beta = a$ and $\alpha+\beta = -b$,we get $a-b+1 = b$,which simplifies to $a-2b = -1$ (Equation $2$).
Subtracting Equation $2$ from Equation $1$: $(a-b) - (a-2b) = -2 - (-1) \implies b = -1$.
Substituting $b = -1$ into Equation $1$: $a - (-1) = -2 \implies a+1 = -2 \implies a = -3$.
Therefore,$a+b = -3 + (-1) = -4$.

(C) For the equation $14 x^2-31 x+3 \lambda=0$,we have $\alpha+\beta=\frac{31}{14}$ and $\alpha \beta=\frac{3 \lambda}{14}$.
For the equation $35 x^2-53 x+4 \lambda=0$,we have $\alpha+\gamma=\frac{53}{35}$ and $\alpha \gamma=\frac{4 \lambda}{35}$.
Subtracting the two equations for $\alpha$: $(\alpha+\beta)-(\alpha+\gamma) = \frac{31}{14}-\frac{53}{35}$ $\Rightarrow \beta-\gamma = \frac{155-106}{70} = \frac{49}{70} = \frac{7}{10}$.
From $\alpha \beta = \frac{3 \lambda}{14}$ and $\alpha \gamma = \frac{4 \lambda}{35}$,we get $\frac{\beta}{\gamma} = \frac{3 \lambda}{14} \times \frac{35}{4 \lambda} = \frac{15}{8}$,so $\beta = \frac{15}{8} \gamma$.
Substituting $\beta$ in $\beta-\gamma = \frac{7}{10}$: $\frac{15}{8} \gamma - \gamma = \frac{7}{10}$ $\Rightarrow \frac{7}{8} \gamma = \frac{7}{10}$ $\Rightarrow \gamma = \frac{4}{5}$.
Then $\beta = \frac{15}{8} \times \frac{4}{5} = \frac{3}{2}$ and $\alpha = \frac{31}{14} - \frac{3}{2} = \frac{31-21}{14} = \frac{10}{14} = \frac{5}{7}$.
Now,$\lambda = \frac{14 \alpha \beta}{3} = \frac{14}{3} \times \frac{5}{7} \times \frac{3}{2} = 5$.
The roots of the required equation are $x_1 = \frac{3 \alpha}{\beta} = \frac{3(5/7)}{3/2} = \frac{15/7}{3/2} = \frac{10}{7}$ and $x_2 = \frac{4 \alpha}{\gamma} = \frac{4(5/7)}{4/5} = \frac{20/7}{4/5} = \frac{25}{7}$.
Sum of roots: $x_1+x_2 = \frac{10}{7} + \frac{25}{7} = \frac{35}{7} = 5$.
Product of roots: $x_1 x_2 = \frac{10}{7} \times \frac{25}{7} = \frac{250}{49}$.
The required equation is $x^2 - (x_1+x_2)x + x_1 x_2 = 0$ $\Rightarrow x^2 - 5x + \frac{250}{49} = 0$ $\Rightarrow 49 x^2 - 245 x + 250 = 0$.

(B) Given the equation $x^2-x-1=0$,the roots $\alpha$ and $\beta$ satisfy $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$.
Given $S_n = 2023 \alpha^n + 2024 \beta^n$.
Consider $S_{n-1} + S_{n-2} = (2023 \alpha^{n-1} + 2024 \beta^{n-1}) + (2023 \alpha^{n-2} + 2024 \beta^{n-2})$.
$= 2023 \alpha^{n-2}(\alpha + 1) + 2024 \beta^{n-2}(\beta + 1)$.
Since $\alpha + 1 = \alpha^2$ and $\beta + 1 = \beta^2$,we have:
$= 2023 \alpha^{n-2}(\alpha^2) + 2024 \beta^{n-2}(\beta^2) = 2023 \alpha^n + 2024 \beta^n = S_n$.
Thus,$S_n = S_{n-1} + S_{n-2}$.
For $n=12$,we get $S_{12} = S_{11} + S_{10}$.

(A) Given the quadratic equation $px^2+qx-r=0$. Since $p, q, r$ are consecutive terms of a non-constant $G$.$P$.,we can write $q=pr$ and $r=pr^2$ (or using common ratio $k$,$q=pk, r=pk^2$).
Substituting these into the equation: $px^2+pkx-pk^2=0$.
Dividing by $p$ $(p \neq 0)$: $x^2+kx-k^2=0$.
From the roots $\alpha, \beta$,we have $\alpha+\beta = -k$ and $\alpha\beta = -k^2$.
Given $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$,we have $\frac{\alpha+\beta}{\alpha\beta} = \frac{-k}{-k^2} = \frac{1}{k} = \frac{3}{4}$,so $k = \frac{4}{3}$.
We need to find $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$.
$(\alpha-\beta)^2 = (-k)^2 - 4(-k^2) = k^2 + 4k^2 = 5k^2$.
Substituting $k = \frac{4}{3}$: $(\alpha-\beta)^2 = 5 \times (\frac{4}{3})^2 = 5 \times \frac{16}{9} = \frac{80}{9}$.

(D) Given the roots of $ax^2 + bx + 1 = 0$ are $2$ and $6$.
Sum of roots: $2 + 6 = 8 = -\frac{b}{a} \implies b = -8a$.
Product of roots: $2 \times 6 = 12 = \frac{1}{a} \implies a = \frac{1}{12}$.
Substituting $a$ into $b = -8a$: $b = -8 \times \frac{1}{12} = -\frac{2}{3}$.
Now,calculate the new roots:
Root $1 = \frac{1}{2a + b} = \frac{1}{2(\frac{1}{12}) - \frac{2}{3}} = \frac{1}{\frac{1}{6} - \frac{4}{6}} = \frac{1}{-\frac{3}{6}} = -2$.
Root $2 = \frac{1}{6a + b} = \frac{1}{6(\frac{1}{12}) - \frac{2}{3}} = \frac{1}{\frac{1}{2} - \frac{2}{3}} = \frac{1}{-\frac{1}{6}} = -6$.
The quadratic equation with roots $-2$ and $-6$ is $(x + 2)(x + 6) = 0$.
$x^2 + 8x + 12 = 0$.

(D) Given that $\alpha$ and $\beta$ are roots of $x^2 + \sqrt{2}x - 8 = 0$,they satisfy the equation:
$\alpha^2 + \sqrt{2}\alpha - 8 = 0 \implies \alpha^2 + \sqrt{2}\alpha = 8$
$\beta^2 + \sqrt{2}\beta - 8 = 0 \implies \beta^2 + \sqrt{2}\beta = 8$
We need to evaluate the expression:
$E = \frac{U_{10} + \sqrt{2}U_9}{2U_8} = \frac{(\alpha^{10} + \beta^{10}) + \sqrt{2}(\alpha^9 + \beta^9)}{2(\alpha^8 + \beta^8)}$
Rearranging the numerator:
$E = \frac{\alpha^8(\alpha^2 + \sqrt{2}\alpha) + \beta^8(\beta^2 + \sqrt{2}\beta)}{2(\alpha^8 + \beta^8)}$
Substituting the values $\alpha^2 + \sqrt{2}\alpha = 8$ and $\beta^2 + \sqrt{2}\beta = 8$:
$E = \frac{\alpha^8(8) + \beta^8(8)}{2(\alpha^8 + \beta^8)}$
$E = \frac{8(\alpha^8 + \beta^8)}{2(\alpha^8 + \beta^8)} = \frac{8}{2} = 4$

(C) Given equation: $x^2+2 \sqrt{2} x-1=0$
Sum of roots: $\alpha+\beta = -2 \sqrt{2}$
Product of roots: $\alpha \beta = -1$
Calculate $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha \beta = (-2 \sqrt{2})^2 - 2(-1) = 8+2 = 10$
Calculate $\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha \beta)^2 = (10)^2 - 2(-1)^2 = 100-2 = 98$
Calculate $\alpha^6+\beta^6 = (\alpha^2+\beta^2)(\alpha^4 - \alpha^2 \beta^2 + \beta^4) = (10)(98 - (-1)^2) = 10(97) = 970$
Roots of the new equation are $98$ and $\frac{1}{10}(970) = 97$
Sum of new roots: $98+97 = 195$
Product of new roots: $98 \times 97 = 9506$
The required quadratic equation is $x^2 - (\text{sum})x + (\text{product}) = 0$,which is $x^2 - 195x + 9506 = 0$.

(B) Given the equation $x^2 - \sqrt{2}x - \sqrt{3} = 0$,the roots $\alpha$ and $\beta$ satisfy:
$\alpha^2 - \sqrt{2}\alpha - \sqrt{3} = 0 \Rightarrow \alpha^{n+2} - \sqrt{2}\alpha^{n+1} - \sqrt{3}\alpha^n = 0$
$\beta^2 - \sqrt{2}\beta - \sqrt{3} = 0 \Rightarrow \beta^{n+2} - \sqrt{2}\beta^{n+1} - \sqrt{3}\beta^n = 0$
Subtracting these equations,we get:
$P_{n+2} - \sqrt{2}P_{n+1} - \sqrt{3}P_n = 0 \Rightarrow P_{n+2} = \sqrt{2}P_{n+1} + \sqrt{3}P_n$
For $n=10$,$P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}$
For $n=9$,$P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9 \Rightarrow \sqrt{3}P_9 = P_{11} - \sqrt{2}P_{10}$
Now,consider the expression $E = (11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}$
Substitute $P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}$:
$E = 11\sqrt{3}P_{10} - 10\sqrt{2}P_{10} + 11\sqrt{2}P_{11} + 10P_{11} - 11(\sqrt{2}P_{11} + \sqrt{3}P_{10})$
$E = 11\sqrt{3}P_{10} - 10\sqrt{2}P_{10} + 11\sqrt{2}P_{11} + 10P_{11} - 11\sqrt{2}P_{11} - 11\sqrt{3}P_{10}$
$E = -10\sqrt{2}P_{10} + 10P_{11} = 10(P_{11} - \sqrt{2}P_{10})$
Since $P_{11} - \sqrt{2}P_{10} = \sqrt{3}P_9$,we have $E = 10\sqrt{3}P_9$.

(B, D) $(1)$ Since $\alpha$ and $\beta$ are roots of $x^2-x-1=0$,we have $\alpha^2 = \alpha+1$ and $\beta^2 = \beta+1$.
Multiplying by $\alpha^{n-2}$ and $\beta^{n-2}$ respectively: $\alpha^n = \alpha^{n-1} + \alpha^{n-2}$ and $\beta^n = \beta^{n-1} + \beta^{n-2}$.
Multiplying by $p$ and $q$ and adding: $p\alpha^n + q\beta^n = (p\alpha^{n-1} + q\beta^{n-1}) + (p\alpha^{n-2} + q\beta^{n-2})$.
Thus,$a_n = a_{n-1} + a_{n-2}$. For $n=12$,$a_{12} = a_{11} + a_{10}$. Option $B$ is correct.
$(2)$ We have $a_0 = p+q$,$a_1 = p\alpha + q\beta$,$a_2 = p\alpha^2 + q\beta^2 = p(\alpha+1) + q(\beta+1) = a_1 + a_0$,$a_3 = a_2 + a_1 = 2a_1 + a_0$,$a_4 = a_3 + a_2 = 3a_1 + 2a_0$.
Given $a_4 = 3(p\alpha + q\beta) + 2(p+q) = 28$.
Since $\alpha = \frac{1+\sqrt{5}}{2}$ and $\beta = \frac{1-\sqrt{5}}{2}$,$3p(\frac{1+\sqrt{5}}{2}) + 3q(\frac{1-\sqrt{5}}{2}) + 2p + 2q = 28$.
$\frac{3p+3p\sqrt{5}+3q-3q\sqrt{5}}{2} + 2p + 2q = 28$.
$(3.5p + 3.5q) + \frac{3\sqrt{5}}{2}(p-q) = 28$.
Since $28$ is rational,$p-q=0 \Rightarrow p=q$.
$7p = 28 \Rightarrow p=4, q=4$.
Then $p+2q = 4 + 2(4) = 12$. Option $D$ is correct.

(B) Given $\alpha+\beta = -p$ and $\alpha^3+\beta^3 = q$.
Using the identity $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$,we have $q = (-p)^3 - 3\alpha\beta(-p) = -p^3 + 3p\alpha\beta$.
Thus,$3p\alpha\beta = p^3+q$,which implies $\alpha\beta = \frac{p^3+q}{3p}$.
The quadratic equation with roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is $x^2 - (\frac{\alpha}{\beta} + \frac{\beta}{\alpha})x + 1 = 0$.
This simplifies to $x^2 - (\frac{\alpha^2+\beta^2}{\alpha\beta})x + 1 = 0$.
Since $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (-p)^2 - 2(\frac{p^3+q}{3p}) = p^2 - \frac{2(p^3+q)}{3p} = \frac{3p^3 - 2p^3 - 2q}{3p} = \frac{p^3-2q}{3p}$.
Substituting these into the equation: $x^2 - (\frac{(p^3-2q)/3p}{(p^3+q)/3p})x + 1 = 0$.
$x^2 - (\frac{p^3-2q}{p^3+q})x + 1 = 0$.
Multiplying by $(p^3+q)$,we get $(p^3+q)x^2 - (p^3-2q)x + (p^3+q) = 0$.

(C) Since $\alpha$ and $\beta$ are the roots of the equation $x^2-6x-2=0$,they satisfy the equation:
$\alpha^2-6\alpha-2=0 \implies \alpha^2 = 6\alpha + 2$
$\beta^2-6\beta-2=0 \implies \beta^2 = 6\beta + 2$
Given $a_n = \alpha^n - \beta^n$,we have $a_{10} = \alpha^{10} - \beta^{10}$ and $a_8 = \alpha^8 - \beta^8$.
Multiplying the root equations by $\alpha^8$ and $\beta^8$ respectively:
$\alpha^{10} = 6\alpha^9 + 2\alpha^8$
$\beta^{10} = 6\beta^9 + 2\beta^8$
Subtracting these equations:
$a_{10} = \alpha^{10} - \beta^{10} = 6(\alpha^9 - \beta^9) + 2(\alpha^8 - \beta^8)$
$a_{10} = 6a_9 + 2a_8$
Rearranging the terms:
$a_{10} - 2a_8 = 6a_9$
Dividing by $2a_9$:
$\frac{a_{10}-2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3$

(A) Given $x^2-x-1=0$,roots are $\alpha, \beta$. Thus $\alpha+\beta=1$ and $\alpha\beta=-1$.
$a_n = \frac{\alpha^n-\beta^n}{\alpha-\beta}$.
For $(1)$: $a_{n+2}-a_{n+1} = \frac{\alpha^{n+2}-\beta^{n+2}-\alpha^{n+1}+\beta^{n+1}}{\alpha-\beta} = \frac{\alpha^{n+1}(\alpha-1)-\beta^{n+1}(\beta-1)}{\alpha-\beta}$. Since $\alpha^2-\alpha-1=0 \implies \alpha^2-1=\alpha$,we have $a_{n+2}-a_{n+1} = \frac{\alpha^{n+1}(\alpha^2-1)-\beta^{n+1}(\beta^2-1)}{\alpha-\beta}$ is incorrect logic; rather $a_{n+2} = a_{n+1}+a_n$. Thus $\sum_{i=1}^n a_i = a_{n+2}-a_2 = a_{n+2}-1$. Statement $(1)$ is correct.
For $(2)$: $\sum_{n=1}^{\infty} \frac{a_n}{10^n} = \frac{1}{\alpha-\beta} \left( \sum (\frac{\alpha}{10})^n - \sum (\frac{\beta}{10})^n \right) = \frac{1}{\alpha-\beta} \left( \frac{\alpha/10}{1-\alpha/10} - \frac{\beta/10}{1-\beta/10} \right) = \frac{1}{10-(\alpha+\beta)+(\alpha\beta/10)} = \frac{10}{100-10-1} = \frac{10}{89}$. Statement $(2)$ is correct.
For $(4)$: $b_n = a_{n-1}+a_{n+1} = \frac{\alpha^{n-1}-\beta^{n-1}+\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta} = \frac{\alpha^{n-1}(1+\alpha^2)-\beta^{n-1}(1+\beta^2)}{\alpha-\beta}$. Since $1+\alpha^2 = \alpha^2-\alpha\beta = \alpha(\alpha-\beta)$,we get $b_n = \alpha^n+\beta^n$. Statement $(4)$ is correct.
Thus,options $1, 2, 4$ are correct.

(B) Given the equation $x^2-ax-b=0$,the roots $\alpha, \beta$ satisfy $\alpha+\beta=a$ and $\alpha\beta=-b$.
Since $P_n = \alpha^n - \beta^n$,we have the recurrence relation $P_n = aP_{n-1} + bP_{n-2}$.
Using $P_6 = aP_5 + bP_4$,we get $45\sqrt{7}i = a(11\sqrt{7}i) + b(-3\sqrt{7}i)$,which simplifies to $11a - 3b = 45$.
Using $P_5 = aP_4 + bP_3$,we get $11\sqrt{7}i = a(-3\sqrt{7}i) + b(-5\sqrt{7}i)$,which simplifies to $-3a - 5b = 11$.
Solving these equations,we find $a=3$ and $b=-4$.
We need to find $|\alpha^4+\beta^4|$. Note that $(\alpha^4+\beta^4)^2 = (\alpha^4-\beta^4)^2 + 4\alpha^4\beta^4 = P_4^2 + 4(\alpha\beta)^4$.
Substituting the values,$P_4^2 = (-3\sqrt{7}i)^2 = 9 \times 7 \times (-1) = -63$.
Also,$4(\alpha\beta)^4 = 4(-b)^4 = 4(-(-4))^4 = 4(256) = 1024$.
Thus,$(\alpha^4+\beta^4)^2 = -63 + 1024 = 961$.
Therefore,$|\alpha^4+\beta^4| = \sqrt{961} = 31$.

(A) Given the sum of the first $11$ terms $S_{11} = 88$ and common difference $d = \frac{3}{2}$.
Using the formula $S_n = \frac{n}{2}(2a + (n-1)d)$:
$88 = \frac{11}{2}(2a + 10 \times \frac{3}{2})$
$8 = \frac{1}{2}(2a + 15) \implies 16 = 2a + 15 \implies 2a = 1 \implies a = \frac{1}{2}$.
The $10^{\text{th}}$ and $11^{\text{th}}$ terms are:
$T_{10} = a + 9d = \frac{1}{2} + 9(\frac{3}{2}) = \frac{1}{2} + \frac{27}{2} = 14$.
$T_{11} = a + 10d = \frac{1}{2} + 10(\frac{3}{2}) = \frac{1}{2} + 15 = \frac{31}{2}$.
For the quadratic equation $3x^2 - px + q = 0$,the sum of roots is $\frac{p}{3} = T_{10} + T_{11} = 14 + \frac{31}{2} = \frac{59}{2} \implies p = \frac{177}{2}$.
The product of roots is $\frac{q}{3} = T_{10} \times T_{11} = 14 \times \frac{31}{2} = 7 \times 31 = 217 \implies q = 651$.
Calculating $q - 2p = 651 - 2(\frac{177}{2}) = 651 - 177 = 474$.

(B) Given $P_n = \alpha^n + \beta^n$. We observe that $P_{10} = P_9 + P_8$ $(123 = 76 + 47)$.
This implies that $\alpha$ and $\beta$ are roots of the quadratic equation $x^2 - x - 1 = 0$.
For this equation,the sum of roots $\alpha + \beta = 1$ and the product of roots $\alpha \beta = -1$.
We need to find the quadratic equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
The sum of the new roots is $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{1}{-1} = -1$.
The product of the new roots is $\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{-1} = -1$.
The required quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - (-1)x + (-1) = 0$,which simplifies to $x^2 + x - 1 = 0$.

(C) For the equation $x^2+\sqrt{3}x-16=0$,since $\alpha$ and $\beta$ are roots,we have $\alpha^2+\sqrt{3}\alpha-16=0$ and $\beta^2+\sqrt{3}\beta-16=0$. Multiplying by $\alpha^{n-2}$ and $\beta^{n-2}$ respectively and adding,we get $P_n+\sqrt{3}P_{n-1}-16P_{n-2}=0$.
For $n=25$,$P_{25}+\sqrt{3}P_{24}=16P_{23}$.
Thus,$\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$.
For the equation $x^2+3x-1=0$,since $\gamma$ and $\delta$ are roots,we have $\gamma^2+3\gamma-1=0$ and $\delta^2+3\delta-1=0$,which implies $\gamma^2-1=-3\gamma$ and $\delta^2-1=-3\delta$.
Then $Q_{25}-Q_{23} = (\gamma^{25}+\delta^{25})-(\gamma^{23}+\delta^{23}) = \gamma^{23}(\gamma^2-1)+\delta^{23}(\delta^2-1) = \gamma^{23}(-3\gamma)+\delta^{23}(-3\delta) = -3(\gamma^{24}+\delta^{24}) = -3Q_{24}$.
Thus,$\frac{Q_{25}-Q_{23}}{Q_{24}} = -3$.
Finally,$8-3=5$.

(B) Given $f(x) = bx^2 + cx + d$.
Then $f(x+1) = b(x+1)^2 + c(x+1) + d = b(x^2 + 2x + 1) + cx + c + d = bx^2 + 2bx + b + cx + c + d$.
Now,$f(x+1) - f(x) = (bx^2 + 2bx + b + cx + c + d) - (bx^2 + cx + d) = 2bx + b + c$.
We are given the identity $f(x+1) - f(x) = 8x + 3$.
Comparing the coefficients of $x$ and the constant terms:
$2b = 8 \implies b = 4$.
$b + c = 3 \implies 4 + c = 3 \implies c = -1$.
Thus,the values are $b = 4$ and $c = -1$.

(B) Let the roots of the equation $x^2-(a-2)x-a+1=0$ be $\alpha$ and $\beta$.
From the relation between roots and coefficients,we have:
$\alpha + \beta = a-2$
$\alpha \beta = -a+1$
The sum of squares of the roots is $S = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.
Substituting the values:
$S = (a-2)^2 - 2(-a+1)$
$S = a^2 - 4a + 4 + 2a - 2$
$S = a^2 - 2a + 2$
To find the least value,we complete the square:
$S = (a^2 - 2a + 1) + 1 = (a-1)^2 + 1$.
The expression $(a-1)^2 + 1$ assumes its minimum value when $(a-1)^2 = 0$,which implies $a = 1$.
Thus,the value of '$a$' is $1$.

(C) Given $f(x) = 2x^3 + mx^2 - 13x + n = 0$. Since $2$ and $3$ are roots,$f(2) = 0$ and $f(3) = 0$.
For $x = 2$: $2(2)^3 + m(2)^2 - 13(2) + n = 0 \implies 16 + 4m - 26 + n = 0 \implies 4m + n = 10$.
For $x = 3$: $2(3)^3 + m(3)^2 - 13(3) + n = 0 \implies 54 + 9m - 39 + n = 0 \implies 9m + n = -15$.
Subtracting the first equation from the second: $(9m + n) - (4m + n) = -15 - 10 \implies 5m = -25 \implies m = -5$.
Substituting $m = -5$ into $4m + n = 10$: $4(-5) + n = 10 \implies -20 + n = 10 \implies n = 30$.
We need to find $4m + 5n$: $4(-5) + 5(30) = -20 + 150 = 130$.

(D) Given that $\alpha$ and $\beta$ are the roots of $x^2-px+r=0$.
From the relation between roots and coefficients,we have $\alpha+\beta=p$ $(i)$ and $\alpha\beta=r$.
Given that $\frac{\alpha}{2}$ and $2\beta$ are the roots of $x^2-qx+r=0$.
From the relation between roots and coefficients,we have $\frac{\alpha}{2}+2\beta=q$,which implies $\alpha+4\beta=2q$ $(ii)$.
Also,the product of roots is $\frac{\alpha}{2} \times 2\beta = r$,so $\alpha\beta=r$.
Subtracting $(i)$ from $(ii)$,we get $(\alpha+4\beta)-(\alpha+\beta)=2q-p$,which simplifies to $3\beta=2q-p$,so $\beta=\frac{2q-p}{3}$.
Substituting $\beta$ into $(i)$,we get $\alpha=p-\frac{2q-p}{3} = \frac{3p-2q+p}{3} = \frac{4p-2q}{3} = \frac{2(2p-q)}{3}$.
Since $r=\alpha\beta$,we have $r = \left(\frac{2(2p-q)}{3}\right) \left(\frac{2q-p}{3}\right) = \frac{2}{9}(2p-q)(2q-p)$.

(D) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
Given that the Arithmetic Mean ($A$.$M$.) of the roots is $p$,we have $\frac{\alpha+\beta}{2} = p$,which implies $\alpha+\beta = 2p$.
Given that the Geometric Mean ($G$.$M$.) of the roots is $q$,we have $\sqrt{\alpha\beta} = q$,which implies $\alpha\beta = q^{2}$.
$A$ quadratic equation with roots $\alpha$ and $\beta$ is given by $x^{2} - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^{2} - (2p)x + q^{2} = 0$,which simplifies to $x^{2} - 2px + q^{2} = 0$.

(B) The given equation is $x^{2}+x+1=0$.
Since the roots are $\alpha$ and $\beta$,we know that $\alpha + \beta = -1$ and $\alpha \beta = 1$.
We need to find $\alpha^{2}+\beta^{2}$.
Using the identity $\alpha^{2}+\beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta$,we substitute the values:
$\alpha^{2}+\beta^{2} = (-1)^{2} - 2(1) = 1 - 2 = -1$.
Alternatively,the roots of $x^{2}+x+1=0$ are the complex cube roots of unity,$\omega$ and $\omega^{2}$.
Thus,$\alpha^{2}+\beta^{2} = \omega^{2} + (\omega^{2})^{2} = \omega^{2} + \omega^{4} = \omega^{2} + \omega = -1$ (since $1+\omega+\omega^{2}=0$).

(D) Given the cubic equation $x^{3}+4x+2=0$.
Let the roots be $\alpha, \beta, \gamma$.
From the properties of roots of a cubic equation $ax^{3}+bx^{2}+cx+d=0$:
Sum of roots $\Sigma \alpha = -\frac{b}{a} = 0$.
Sum of roots taken two at a time $\Sigma \alpha \beta = \frac{c}{a} = 4$.
Product of roots $\alpha \beta \gamma = -\frac{d}{a} = -2$.
We know the identity $\alpha^{3}+\beta^{3}+\gamma^{3}-3 \alpha \beta \gamma = (\alpha+\beta+\gamma)(\alpha^{2}+\beta^{2}+\gamma^{2}-\alpha \beta-\beta \gamma-\gamma \alpha)$.
Since $\Sigma \alpha = 0$,the right side becomes $0$.
Therefore,$\alpha^{3}+\beta^{3}+\gamma^{3} = 3 \alpha \beta \gamma$.
Substituting the value of the product of roots:
$\alpha^{3}+\beta^{3}+\gamma^{3} = 3(-2) = -6$.

(B) Given the cubic equation $x^{3}-2x+1=0$ with roots $\alpha, \beta, \gamma$.
From Vieta's formulas,we have:
$\alpha+\beta+\gamma = 0$
$\alpha\beta+\beta\gamma+\gamma\alpha = -2$
$\alpha\beta\gamma = -1$
We need to evaluate $\sum\left(\frac{1}{\alpha+\beta-\gamma}\right)$.
Since $\alpha+\beta+\gamma = 0$,we have $\alpha+\beta = -\gamma$.
Substituting this into the expression:
$\sum\left(\frac{1}{-\gamma-\gamma}\right) = \sum\left(\frac{1}{-2\gamma}\right) = -\frac{1}{2} \sum\left(\frac{1}{\gamma}\right)$
$= -\frac{1}{2} \left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right)$
$= -\frac{1}{2} \left(\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}\right)$
$= -\frac{1}{2} \left(\frac{-2}{-1}\right) = -\frac{1}{2} \times 2 = -1$.

(C) Given the cubic equation $x^{3}-5x^{2}-x+5=0$.
Let the roots be $a, -a, b$.
From the relationship between roots and coefficients,the sum of the roots is given by:
$a + (-a) + b = -(-5)/1 = 5$.
This simplifies to $b = 5$.
Now,we check which of the given quadratic equations is satisfied by $x = 5$:
For option $C$,$x^{2}-3x-10=0$:
Substituting $x = 5$,we get $(5)^{2}-3(5)-10 = 25-15-10 = 0$.
Since the equation is satisfied,$b=5$ is a root of $x^{2}-3x-10=0$.

(A) Given the quadratic equation $x^{2}-ax+b^{2}=0$.
For a quadratic equation of the form $Ax^{2}+Bx+C=0$,the sum of the roots $\alpha+\beta = -B/A$ and the product of the roots $\alpha\beta = C/A$.
Here,$A=1$,$B=-a$,and $C=b^{2}$.
Therefore,$\alpha+\beta = -(-a)/1 = a$ and $\alpha\beta = b^{2}/1 = b^{2}$.
We need to find the value of $\alpha^{2}+\beta^{2}$.
Using the algebraic identity $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2}-2\alpha\beta$.
Substituting the values,we get $\alpha^{2}+\beta^{2} = (a)^{2}-2(b^{2}) = a^{2}-2b^{2}$.

(C) Given the quadratic equation $x^2+bx+c=0$,the sum of roots is $\alpha+\beta = -b = 5$,which implies $b = -5$.
Using the identity $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$,we substitute the given values:
$60 = (5)^3 - 3\alpha\beta(5)$.
$60 = 125 - 15\alpha\beta$.
$15\alpha\beta = 125 - 60 = 65$.
$\alpha\beta = \frac{65}{15} = \frac{13}{3}$.
Since $\alpha\beta = c$,we have $c = \frac{13}{3}$.
Now,calculate $3c+2$:
$3(\frac{13}{3}) + 2 = 13 + 2 = 15$.
Since $b = -5$,we check the options:
$2b = 2(-5) = -10$.
$3b = 3(-5) = -15$.
$-3b = -3(-5) = 15$.
$-2b = -2(-5) = 10$.
Thus,$3c+2 = -3b$.

(C) Given the equation $x^3+a x^2+b x+c=0$,the sum of the roots is $\alpha+\beta+\gamma = -a$.
We can rewrite the terms in the expression as follows:
$\alpha+\beta-2 \gamma = (\alpha+\beta+\gamma) - 3 \gamma = -a - 3 \gamma$.
Similarly,$\beta+\gamma-2 \alpha = -a - 3 \alpha$ and $\gamma+\alpha-2 \beta = -a - 3 \beta$.
The product is $(-a-3 \alpha)(-a-3 \beta)(-a-3 \gamma) = -(a+3 \alpha)(a+3 \beta)(a+3 \gamma)$.
Let $f(x) = x^3+a x^2+b x+c = (x-\alpha)(x-\beta)(x-\gamma)$.
Then $f(-a/3) = (-a/3-\alpha)(-a/3-\beta)(-a/3-\gamma) = (-1/27)(a+3 \alpha)(a+3 \beta)(a+3 \gamma)$.
Thus,$(a+3 \alpha)(a+3 \beta)(a+3 \gamma) = -27 f(-a/3)$.
The original expression is $-(-27 f(-a/3)) = 27 f(-a/3)$.
$f(-a/3) = (-a/3)^3 + a(-a/3)^2 + b(-a/3) + c = -a^3/27 + a^3/9 - ab/3 + c = (2a^3 - 9ab + 27c)/27$.
Multiplying by $27$,we get $2a^3 - 9ab + 27c$.

(C) Given the cubic equation $x^3+px^2+qx+r=0$,the roots are $\alpha, \beta, \gamma$.
From Vieta's formulas:
$\alpha+\beta+\gamma = -p$
$\alpha\beta+\beta\gamma+\gamma\alpha = q$
$\alpha\beta\gamma = -r$
We use the identity:
$\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)$
We know that $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (-p)^2 - 2q = p^2-2q$.
Substituting these into the identity:
$\alpha^3+\beta^3+\gamma^3 - 3(-r) = (-p)((p^2-2q) - q)$
$\alpha^3+\beta^3+\gamma^3 + 3r = -p(p^2-3q)$
$\alpha^3+\beta^3+\gamma^3 = -p^3+3pq-3r$
$\alpha^3+\beta^3+\gamma^3 = 3pq-3r-p^3$.

(C) Given the cubic equation $5x^3-4x^2+3x-2=0$.
Let the roots be $\alpha, \beta, \gamma$.
From Vieta's formulas:
$e_1 = \alpha+\beta+\gamma = -(\frac{-4}{5}) = \frac{4}{5}$
$e_2 = \alpha\beta+\beta\gamma+\gamma\alpha = \frac{3}{5}$
$e_3 = \alpha\beta\gamma = -(\frac{-2}{5}) = \frac{2}{5}$
Since $\alpha, \beta, \gamma$ are roots,they satisfy $5x^3 = 4x^2-3x+2$,or $x^3 = \frac{4}{5}x^2-\frac{3}{5}x+\frac{2}{5}$.
Summing for all roots:
$\sum \alpha^3 = \frac{4}{5}(\sum \alpha^2) - \frac{3}{5}(\sum \alpha) + 3(\frac{2}{5})$
We know $\sum \alpha^2 = (\sum \alpha)^2 - 2(\sum \alpha\beta) = (\frac{4}{5})^2 - 2(\frac{3}{5}) = \frac{16}{25} - \frac{6}{5} = \frac{16-30}{25} = -\frac{14}{25}$.
Substituting back:
$\sum \alpha^3 = \frac{4}{5}(-\frac{14}{25}) - \frac{3}{5}(\frac{4}{5}) + \frac{6}{5} = -\frac{56}{125} - \frac{12}{25} + \frac{6}{5} = \frac{-56-60+150}{125} = \frac{34}{125}$.

(D) Given the cubic equation $x^3+px^2+qx+r=0$,the roots are $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha+\beta+\gamma = -p$
$\alpha\beta+\beta\gamma+\gamma\alpha = q$
$\alpha\beta\gamma = -r$
We know that $\alpha+\beta = -p-\gamma$,$\beta+\gamma = -p-\alpha$,and $\gamma+\alpha = -p-\beta$.
Therefore,$(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = (-p-\gamma)(-p-\alpha)(-p-\beta) = -(p+\gamma)(p+\alpha)(p+\beta)$.
Let $f(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3+px^2+qx+r$.
Then $f(-p) = (-p-\alpha)(-p-\beta)(-p-\gamma) = (-p)^3+p(-p)^2+q(-p)+r = -p^3+p^3-pq+r = r-pq$.
Since $f(-p) = -(p+\alpha)(p+\beta)(p+\gamma)$,we have $-(p+\alpha)(p+\beta)(p+\gamma) = r-pq$.
Thus,$(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = pq-r$.

(A) Let the roots of the quadratic equation $x^2-35x+c=0$ be $2t$ and $3t$.
Sum of the roots $= 2t + 3t = -(-35)/1 = 35$.
$5t = 35 \Rightarrow t = 7$.
Product of the roots $= (2t)(3t) = c/1 = c$.
$6t^2 = c$.
Since $t = 7$,$c = 6(7^2) = 6 \times 49$.
Given $c = 6K$,we have $6K = 6 \times 49$.
Therefore,$K = 49$.

(A) Given $\alpha+\beta=a$ and $\alpha\beta=b$.
We know that $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = a^2-2b$.
Also,$\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2) = a(a^2-2b-b) = a(a^2-3b) = a^3-3ab$.
The quadratic equation with roots $p$ and $q$ is given by $x^2 - (p+q)x + pq = 0$.
Here,the roots are $p = a^2-2b$ and $q = a^3-3ab$.
The constant term $C$ is the product of the roots: $C = pq = (a^2-2b)(a^3-3ab)$.
Expanding this: $C = a^2(a^3) - a^2(3ab) - 2b(a^3) + 2b(3ab) = a^5 - 3a^3b - 2a^3b + 6ab^2 = a^5 - 5a^3b + 6ab^2$.

(C) Let the roots of the cubic equation $ax^3+bx^2+cx+1=0$ be $-1, -1, \alpha$.
From the relation between roots and coefficients,the product of roots is $\alpha \times (-1) \times (-1) = -\frac{1}{a}$,which implies $\alpha = -\frac{1}{a}$.
The sum of roots is $(-1) + (-1) + \alpha = -\frac{b}{a}$.
Substituting $\alpha = -\frac{1}{a}$,we get $-2 - \frac{1}{a} = -\frac{b}{a}$.
Multiplying by $-a$,we get $2a + 1 = b$,so $b = 2a + 1$.
Since $-1$ is a root,$a(-1)^3 + b(-1)^2 + c(-1) + 1 = 0$,which simplifies to $-a + b - c + 1 = 0$.
Substituting $b = 2a + 1$,we get $-a + (2a + 1) - c + 1 = 0$,which simplifies to $a + 2 - c = 0$,hence $c = a + 2$.

(B) Given that $\alpha$ and $\beta$ are roots of $2x^2-4x+3=0$.
Sum of roots: $\alpha+\beta = -(\frac{-4}{2}) = 2$.
Product of roots: $\alpha\beta = \frac{3}{2}$.
Calculate $\alpha^2+\beta^2$:
$\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (2)^2 - 2(\frac{3}{2}) = 4-3 = 1$.
Calculate $\alpha^4+\beta^4$:
$(\alpha^2+\beta^2)^2 = \alpha^4+\beta^4+2\alpha^2\beta^2$.
$1^2 = \alpha^4+\beta^4 + 2(\frac{3}{2})^2$.
$1 = \alpha^4+\beta^4 + 2(\frac{9}{4}) = \alpha^4+\beta^4 + \frac{9}{2}$.
$\alpha^4+\beta^4 = 1 - \frac{9}{2} = -\frac{7}{2}$.
Now,substitute these values into the expression:
$\frac{2(\alpha^4+\beta^4)+3(\alpha^2+\beta^2)}{\alpha+\beta} = \frac{2(-\frac{7}{2}) + 3(1)}{2} = \frac{-7+3}{2} = \frac{-4}{2} = -2$.

(C) Given: $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$.
Sum of roots: $\alpha+\beta = -\frac{b}{a}$.
Product of roots: $\alpha\beta = \frac{c}{a}$.
New roots are $S_1 = \alpha+\beta = -\frac{b}{a}$ and $S_2 = \frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c}$.
The required equation is $(x-S_1)(x-S_2) = 0$.
$(x - (-\frac{b}{a}))(x - (-\frac{b}{c})) = 0$.
$(x + \frac{b}{a})(x + \frac{b}{c}) = 0$.
Multiplying by $ac$: $(ax+b)(cx+b) = 0$.
$acx^2 + abx + bcx + b^2 = 0$.
$acx^2 + (ab+bc)x + b^2 = 0$.

(B) Given the equation $y^2+y+1=0$.
Since $a$ and $b$ are the roots,by Vieta's formulas,$a+b = -1$ and $ab = 1$.
We need to find the value of $a^4+b^4+a^{-1}b^{-1} = a^4+b^4+\frac{1}{ab}$.
First,calculate $a^2+b^2 = (a+b)^2 - 2ab = (-1)^2 - 2(1) = 1 - 2 = -1$.
Then,$a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2 = (-1)^2 - 2(1)^2 = 1 - 2 = -1$.
Substituting these values into the expression:
$a^4+b^4+\frac{1}{ab} = -1 + \frac{1}{1} = -1 + 1 = 0$.

(D) Given that $c$ and $d$ are the roots of $x^2+ax+b=0$.
From the relation between roots and coefficients,we have $c+d = -a$ and $cd = b$.
Now,consider the equation $x^2+(4c+a)x+(b+2ac+4c^2)=0$.
Substitute $a = -(c+d)$ and $b = cd$ into the equation:
$x^2+(4c-(c+d))x+(cd+2c(-(c+d))+4c^2)=0$
$x^2+(3c-d)x+(cd-2c^2-2cd+4c^2)=0$
$x^2+(3c-d)x+(2c^2-cd)=0$
$x^2+(3c-d)x+c(2c-d)=0$
We can factor this as $(x+c)(x+2c-d)=0$.
Thus,the roots are $x = -c$ and $x = d-2c$.
Comparing with the given options,$d-2c$ is a root.

(C) Given that $\alpha, \beta, \gamma$ are the roots of $x^3+ax^2+bx+c=0$.
By Vieta's formulas:
$\alpha+\beta+\gamma = -a$
$\alpha\beta+\beta\gamma+\gamma\alpha = b$
$\alpha\beta\gamma = -c$
Let the roots of the new equation $x^3+(2b-a^2)x^2+(b^2-2ac)x-c^2=0$ be $\alpha', \beta', \gamma'$.
Comparing coefficients:
$\alpha'+\beta'+\gamma' = -(2b-a^2) = a^2-2b = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = \alpha^2+\beta^2+\gamma^2$.
$\alpha'\beta'+\beta'\gamma'+\gamma'\alpha' = b^2-2ac = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2(\alpha+\beta+\gamma)(\alpha\beta\gamma) = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$.
$\alpha'\beta'\gamma' = c^2 = (\alpha\beta\gamma)^2 = \alpha^2\beta^2\gamma^2$.
Thus,the roots are $\alpha^2, \beta^2, \gamma^2$.