Mix Examples-Quadratic Equations and Inequations Questions in English

(A) Let the roots be $\alpha_1 = \sqrt{3}+\sqrt{2}i$ and $\alpha_2 = \sqrt{3}-\sqrt{2}$.
Since the coefficients must be rational,the conjugate of $\alpha_1$,which is $\bar{\alpha_1} = \sqrt{3}-\sqrt{2}i$,must also be a root.
Thus,the quadratic factor corresponding to $\alpha_1$ and $\bar{\alpha_1}$ is $(x-(\sqrt{3}+\sqrt{2}i))(x-(\sqrt{3}-\sqrt{2}i)) = ((x-\sqrt{3})-\sqrt{2}i)((x-\sqrt{3})+\sqrt{2}i) = (x-\sqrt{3})^2 + 2 = x^2-2\sqrt{3}x+5$.
Since the coefficients must be rational,we need to eliminate the $\sqrt{3}$ term. The conjugate of $\sqrt{3}-\sqrt{2}$ is $-\sqrt{3}-\sqrt{2}$,and the conjugate of $\sqrt{3}+\sqrt{2}$ is $-\sqrt{3}+\sqrt{2}$.
However,for the root $\alpha_2 = \sqrt{3}-\sqrt{2}$,the minimal polynomial over $\mathbb{Q}$ is $(x^2-3)^2 = 2$,which is $x^4-6x^2+9=2$,or $x^4-6x^2+7=0$. Wait,let's re-evaluate.
For $\alpha_1 = \sqrt{3}+\sqrt{2}i$,$(x-\sqrt{3})^2 = -2 \implies x^2-2\sqrt{3}x+5=0$. To make coefficients rational,we multiply by the conjugate factor $x^2+2\sqrt{3}x+5=0$. Product: $(x^2+5)^2 - (2\sqrt{3}x)^2 = x^4+10x^2+25-12x^2 = x^4-2x^2+25=0$.
For $\alpha_2 = \sqrt{3}-\sqrt{2}$,$(x-\sqrt{3})^2 = 2 \implies x^2-2\sqrt{3}x+1=0$. To make coefficients rational,we multiply by $x^2+2\sqrt{3}x+1=0$. Product: $(x^2+1)^2 - (2\sqrt{3}x)^2 = x^4+2x^2+1-12x^2 = x^4-10x^2+1=0$.
The combined equation is $(x^4-2x^2+25)(x^4-10x^2+1)=0$.

(A) We have $f(x) = 1-2x-5x^2 = -5(x^2+\frac{2}{5}x) + 1 = -5(x+\frac{1}{5})^2 + \frac{1}{5} + 1 = -5(x+\frac{1}{5})^2 + \frac{6}{5}$.
Since the coefficient of $x^2$ is negative,the maximum value is $\alpha = \frac{6}{5}$.
Next,$g(x) = x^2-2x+r = (x-1)^2 + r-1$. The minimum value is $\beta = r-1$.
The given inequality is $5\alpha x^2 + \beta x + 6 > 0$,which becomes $5(\frac{6}{5})x^2 + (r-1)x + 6 > 0$,or $6x^2 + (r-1)x + 6 > 0$.
For this quadratic to be positive for all real $x$,the discriminant $D$ must be less than $0$.
$D = (r-1)^2 - 4(6)(6) < 0$.
$(r-1)^2 - 144 < 0$.
$(r-1-12)(r-1+12) < 0$.
$(r-13)(r+11) < 0$.
Thus,$r \in (-11, 13)$.

(A) Given expression is $f(x) = -3x^2 + 6x + 7$.
Comparing with $ax^2 + bx + c$,we have $a = -3, b = 6, c = 7$.
The extreme value occurs at $x = \alpha = \frac{-b}{2a} = \frac{-6}{2(-3)} = 1$.
The extreme value $\beta = f(\alpha) = f(1) = -3(1)^2 + 6(1) + 7 = 10$.
Now,the equation $x^2 + \alpha x - \beta = 0$ becomes $x^2 + x - 10 = 0$.
Let the roots be $x_1$ and $x_2$.
Then $x_1 + x_2 = -1$ and $x_1x_2 = -10$.
The sum of the squares of the roots is $x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2$.
Substituting the values,$x_1^2 + x_2^2 = (-1)^2 - 2(-10) = 1 + 20 = 21$.

(B) Let $f(x) = \frac{x^2-x+1}{x^2+x+1}$.
We can rewrite the expression as:
$f(x) = \frac{(x^2+x+1) - 2x}{x^2+x+1} = 1 - \frac{2x}{x^2+x+1}$.
To minimize $f(x)$,we need to maximize the term $\frac{2x}{x^2+x+1}$.
For $x \neq 0$,we have $\frac{2x}{x^2+x+1} = \frac{2}{x + \frac{1}{x} + 1}$.
Since $x + \frac{1}{x} \geq 2$ for $x > 0$ and $x + \frac{1}{x} \leq -2$ for $x < 0$,the maximum value of $x + \frac{1}{x} + 1$ is not applicable here directly,but we analyze the range of $\frac{x}{x^2+x+1}$.
Let $y = \frac{x^2-x+1}{x^2+x+1} \implies y(x^2+x+1) = x^2-x+1 \implies (y-1)x^2 + (y+1)x + (y-1) = 0$.
For $x$ to be real,the discriminant $D \geq 0$:
$(y+1)^2 - 4(y-1)^2 \geq 0 \implies (y+1-2y+2)(y+1+2y-2) \geq 0 \implies (3-y)(3y-1) \geq 0$.
This implies $\frac{1}{3} \leq y \leq 3$.
Thus,the minimum value is $\frac{1}{3}$.

(A) Given $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$.
$y(x^2 + 2x + 3) = x^2 + 14x + 9$.
$(y - 1)x^2 + 2(y - 7)x + 3y - 9 = 0$.
Since $x \in R$,the discriminant $D \geq 0$.
$4(y - 7)^2 - 4(y - 1)(3y - 9) \geq 0$.
$(y^2 - 14y + 49) - (3y^2 - 12y + 9) \geq 0$.
$-2y^2 - 2y + 40 \geq 0$.
$y^2 + y - 20 \leq 0$.
$(y + 5)(y - 4) \leq 0$.
Thus,$y \in [-5, 4]$.

(B) The minimum value of the quadratic expression $x^2+bx+5$ occurs at $x = -\frac{b}{2}$. Substituting this,we get $\alpha = (-\frac{b}{2})^2 + b(-\frac{b}{2}) + 5 = 5 - \frac{b^2}{4}$.
The maximum value of the quadratic expression $-x^2+ax+5$ occurs at $x = \frac{a}{2}$. Substituting this,we get $\beta = -(\frac{a}{2})^2 + a(\frac{a}{2}) + 5 = 5 + \frac{a^2}{4}$.
Given the inequality $x^2-10x+24 \leq 0$,we factor it as $(x-4)(x-6) \leq 0$,which implies $4 \leq x \leq 6$.
Comparing this with the interval $[\alpha, \beta]$,we have $\alpha = 4$ and $\beta = 6$.
Equating the expressions:
$5 - \frac{b^2}{4} = 4$ $\Rightarrow \frac{b^2}{4} = 1$ $\Rightarrow b^2 = 4$.
$5 + \frac{a^2}{4} = 6$ $\Rightarrow \frac{a^2}{4} = 1$ $\Rightarrow a^2 = 4$.
Therefore,$a^2b^2 = 4 \times 4 = 16$.

(D) Let $y = \frac{x^2+ax+3}{x^2+x+1}$.
Then $yx^2 + yx + y = x^2 + ax + 3$,which implies $(y-1)x^2 + (y-a)x + (y-3) = 0$.
For $x$ to be real,the discriminant $D \geq 0$.
$(y-a)^2 - 4(y-1)(y-3) \geq 0$.
$y^2 - 2ay + a^2 - 4(y^2 - 4y + 3) \geq 0$.
$-3y^2 + (16-2a)y + (a^2-12) \geq 0$.
$3y^2 + (2a-16)y + (12-a^2) \leq 0$.
For the range of $y$ to be $(-\infty, \infty)$,this quadratic inequality in $y$ must hold for all $y \in R$,which is impossible for a quadratic expression with a positive leading coefficient $(3 > 0)$.
Thus,the expression cannot take all real values for any $a$.

(D) Let $y = \frac{9 \cdot 3^{2x} + 6 \cdot 3^x + 4}{9 \cdot 3^{2x} - 6 \cdot 3^x + 4}$.
Substitute $t = 3^x$,where $t > 0$.
Then $y = \frac{9t^2 + 6t + 4}{9t^2 - 6t + 4}$.
$y(9t^2 - 6t + 4) = 9t^2 + 6t + 4$.
$9t^2(y - 1) - 6t(y + 1) + 4(y - 1) = 0$.
Since $t$ is a real number,the discriminant $D \geq 0$.
$D = [-6(y + 1)]^2 - 4 \cdot 9(y - 1) \cdot 4(y - 1) \geq 0$.
$36(y + 1)^2 - 144(y - 1)^2 \geq 0$.
Divide by $36$: $(y + 1)^2 - 4(y - 1)^2 \geq 0$.
$(y + 1 - 2(y - 1))(y + 1 + 2(y - 1)) \geq 0$.
$(-y + 3)(3y - 1) \geq 0$.
$(y - 3)(3y - 1) \leq 0$.
Thus,$\frac{1}{3} \leq y \leq 3$.
The minimum value is $\frac{1}{3}$.

(A) Let $y = \frac{x^2+14x+9}{x^2+2x+3}$.
Since $x$ is real,the denominator $x^2+2x+3 = (x+1)^2+2$ is always positive.
$y(x^2+2x+3) = x^2+14x+9$
$x^2(y-1) + 2x(y-7) + 3y-9 = 0$.
For $x$ to be real,the discriminant $D \geq 0$.
$D = [2(y-7)]^2 - 4(y-1)(3y-9) \geq 0$
$4(y^2-14y+49) - 4(3y^2-12y+9) \geq 0$
$y^2-14y+49 - 3y^2+12y-9 \geq 0$
$-2y^2-2y+40 \geq 0$
$y^2+y-20 \leq 0$
$(y+5)(y-4) \leq 0$.
Thus,$y \in [-5, 4]$.
The maximum value is $4$ and the minimum value is $-5$.

(A) Given that $1$ is a root of $ax^2+bx+c=0$.
Substituting $x=1$,we get $a(1)^2+b(1)+c=0$,which implies $a+b+c=0$.
Thus,$E_1$ is true.
Given that $\sin \theta$ and $\cos \theta$ are roots of $ax^2+bx+c=0$.
From the sum and product of roots:
$\sin \theta + \cos \theta = -\frac{b}{a}$ and $\sin \theta \cos \theta = \frac{c}{a}$.
Squaring the sum of roots:
$(\sin \theta + \cos \theta)^2 = (-\frac{b}{a})^2$
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = \frac{b^2}{a^2}$
$1 + 2(\frac{c}{a}) = \frac{b^2}{a^2}$
Multiplying by $a^2$ (assuming $a \neq 0$):
$a^2 + 2ac = b^2$
$b^2 - a^2 = 2ac$.
Thus,$E_2$ is true.
Therefore,both $E_1$ and $E_2$ are true.

(B) Let the roots be $\alpha, \beta, \gamma, \delta$ such that $\alpha+\beta=0$,which implies $\beta=-\alpha$.
Since $\alpha$ and $-\alpha$ are roots,the polynomial is divisible by $(x-\alpha)(x+\alpha) = x^2-\alpha^2$.
Let the other two roots be $\gamma$ and $\delta$. Then $(x^2-\alpha^2)(x^2-Sx+P) = x^4-Sx^3+(P-\alpha^2)x^2+S\alpha^2x-P\alpha^2 = x^4-x^3-16x^2+4x+48$.
Comparing coefficients:
$S=1$
$P-\alpha^2=-16$
$S\alpha^2=4$ $\Rightarrow 1 \cdot \alpha^2=4$ $\Rightarrow \alpha^2=4$.
Thus,$\alpha=2, \beta=-2$.
From $P-\alpha^2=-16$,we get $P-4=-16 \Rightarrow P=-12$.
Also,$-P\alpha^2=48 \Rightarrow -(-12)(4)=48$,which is consistent.
Since $S=\gamma+\delta=1$ and $P=\gamma\delta=-12$,we have $\gamma^2+\delta^2 = (\gamma+\delta)^2-2\gamma\delta = 1^2-2(-12) = 1+24=25$.
Then $\gamma^4+\delta^4 = (\gamma^2+\delta^2)^2-2\gamma^2\delta^2 = 25^2-2(-12)^2 = 625-288=337$.
Finally,$\alpha^4+\beta^4+\gamma^4+\delta^4 = 2^4+(-2)^4+337 = 16+16+337 = 369$.

(B) Given the equation: $e^{4t} - 10e^{3t} + 29e^{2t} - 20e^t + 4 = 0$.
Let $x = e^t$. Then the equation becomes $x^4 - 10x^3 + 29x^2 - 20x + 4 = 0$.
Let the roots of the equation in $t$ be $t_1, t_2, t_3, t_4$.
Then the roots of the equation in $x$ are $x_1 = e^{t_1}, x_2 = e^{t_2}, x_3 = e^{t_3}, x_4 = e^{t_4}$.
By Vieta's formulas,the product of the roots of the polynomial $x^4 - 10x^3 + 29x^2 - 20x + 4 = 0$ is given by the constant term:
$x_1 \cdot x_2 \cdot x_3 \cdot x_4 = 4$.
Substituting back $e^t$:
$e^{t_1} \cdot e^{t_2} \cdot e^{t_3} \cdot e^{t_4} = 4$.
$e^{(t_1 + t_2 + t_3 + t_4)} = 4$.
Taking the natural logarithm on both sides:
$t_1 + t_2 + t_3 + t_4 = \log_e 4$.
Since $4 = 2^2$,we have:
$t_1 + t_2 + t_3 + t_4 = \log_e 2^2 = 2\log_e 2$.

(B) Let the roots of the given equation $f(x) = x^4-2ax^3+4bx^2+8ax+16=0$ be $\alpha_1, \alpha_2, \alpha_3, \alpha_4$.
The roots of the new equation are $p\alpha_1, p\alpha_2, p\alpha_3, p\alpha_4$.
Thus,the new equation is $f(\frac{x}{p}) = 0$.
Substituting $\frac{x}{p}$ into the equation:
$(\frac{x}{p})^4 - 2a(\frac{x}{p})^3 + 4b(\frac{x}{p})^2 + 8a(\frac{x}{p}) + 16 = 0$
Multiplying by $p^4$:
$x^4 - 2apx^3 + 4bp^2x^2 + 8ap^3x + 16p^4 = 0$
Since this is a reciprocal equation,the coefficient of $x^4$ must equal the constant term:
$1 = 16p^4$
$p^4 = \frac{1}{16}$
$p^2 = \frac{1}{4} \implies |p| = \frac{1}{2}$

(B) Given $P(x) = 2x^4 + ax^3 + bx^2 + cx + d$ is a polynomial of degree $4$ with leading coefficient $2$.
We observe that for $x = 1, 2, 3, 4$,the values are $P(x) = x^2 + 3$.
Let $Q(x) = P(x) - (x^2 + 3)$.
Since $P(x)$ is a polynomial of degree $4$ with leading coefficient $2$,$Q(x)$ is also a polynomial of degree $4$ with leading coefficient $2$.
Since $P(1)=4, P(2)=7, P(3)=12, P(4)=19$,we have $Q(1)=0, Q(2)=0, Q(3)=0, Q(4)=0$.
Thus,$Q(x) = 2(x-1)(x-2)(x-3)(x-4)$.
Therefore,$P(x) = 2(x-1)(x-2)(x-3)(x-4) + x^2 + 3$.
To find $P(5)$,substitute $x = 5$:
$P(5) = 2(5-1)(5-2)(5-3)(5-4) + (5^2 + 3)$
$P(5) = 2(4)(3)(2)(1) + (25 + 3)$
$P(5) = 2(24) + 28 = 48 + 28 = 76$.

(C) Given the equation $2x^4-8x^3+3x^2-1=0$.
To remove the $x^3$ term,we use the transformation $x = y - \frac{a_1}{n a_0}$,where $a_0=2$ and $a_1=-8$ for a quartic equation $a_0x^4+a_1x^3+a_2x^2+a_3x+a_4=0$.
Here,$h = -\frac{-8}{4 \times 2} = \frac{8}{8} = 1$.
Substitute $x = y+1$ into the equation:
$2(y+1)^4 - 8(y+1)^3 + 3(y+1)^2 - 1 = 0$.
Expanding the terms:
$2(y^4+4y^3+6y^2+4y+1) - 8(y^3+3y^2+3y+1) + 3(y^2+2y+1) - 1 = 0$.
$2y^4 + 8y^3 + 12y^2 + 8y + 2 - 8y^3 - 24y^2 - 24y - 8 + 3y^2 + 6y + 3 - 1 = 0$.
Combining like terms:
$2y^4 + (8-8)y^3 + (12-24+3)y^2 + (8-24+6)y + (2-8+3-1) = 0$.
$2y^4 - 9y^2 - 10y - 4 = 0$.
Comparing this with $2x^4+bx^2+cx+d=0$,we find $b = -9$.

(C) The given equation is $6x^6-25x^5+31x^4-31x^2+25x-6=0$.
This is a reciprocal equation of the first kind.
By testing $x=1$ and $x=-1$,we find that $x=1$ and $x=-1$ are roots of the equation.
Let the roots be $a_1, a_2, a_3, a_4, a_5, a_6$.
Given $a_1+a_2 = \frac{5}{2}$.
Since $1$ and $-1$ are roots,let $a_3=1$ and $a_4=-1$.
The sum of all roots is given by $-\frac{\text{coefficient of } x^5}{\text{coefficient of } x^6} = -\frac{-25}{6} = \frac{25}{6}$.
Thus,$a_1+a_2+a_3+a_4+a_5+a_6 = \frac{25}{6}$.
Substituting the known values: $\frac{5}{2} + 1 - 1 + a_5 + a_6 = \frac{25}{6}$.
$a_5+a_6 = \frac{25}{6} - \frac{5}{2} = \frac{25-15}{6} = \frac{10}{6} = \frac{5}{3}$.
Since the equation is a reciprocal equation of the first kind,the roots occur in pairs $(r, 1/r)$. The roots $1$ and $-1$ are real. The remaining roots $a_1, a_2, a_5, a_6$ are non-real.
The sum of all non-real roots is $(a_1+a_2) + (a_5+a_6) = \frac{5}{2} + \frac{5}{3} = \frac{15+10}{6} = \frac{25}{6}$.

(B) Given equation is $\left(x^4+1\right)=\frac{1}{a}(x+1)^4$.
Multiplying by $a$,we get $a(x^4+1) = (x+1)^4$.
Expanding the right side: $a(x^4+1) = x^4+4x^3+6x^2+4x+1$.
Rearranging the terms: $(a-1)x^4 - 4x^3 - 6x^2 - 4x + (a-1) = 0$.
For an equation to be a reciprocal equation,the coefficients of $x^k$ and $x^{n-k}$ must be equal (or proportional).
Here,the coefficient of $x^4$ is $(a-1)$ and the constant term is $(a-1)$.
For the equation to be of degree $4$,the coefficient of $x^4$ must not be zero,so $a-1 \neq 0$,which means $a \neq 1$.
Thus,the equation is a reciprocal equation for all $a \in R - \{1\}$.

(B) Given $x=2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$.
Subtracting $2$ from both sides,we get $x-2=2^{\frac{2}{3}}+2^{\frac{1}{3}}$.
Cubing both sides,we have $(x-2)^3 = (2^{\frac{2}{3}}+2^{\frac{1}{3}})^3$.
Using the identity $(a+b)^3 = a^3+b^3+3ab(a+b)$,we get:
$x^3-6x^2+12x-8 = (2^{\frac{2}{3}})^3 + (2^{\frac{1}{3}})^3 + 3(2^{\frac{2}{3}})(2^{\frac{1}{3}})(2^{\frac{2}{3}}+2^{\frac{1}{3}})$.
$x^3-6x^2+12x-8 = 4 + 2 + 3(2^1)(x-2)$.
$x^3-6x^2+12x-8 = 6 + 6(x-2)$.
$x^3-6x^2+12x-8 = 6 + 6x - 12$.
$x^3-6x^2+12x-8 = 6x - 6$.
Rearranging the terms,we get $x^3-6x^2+6x = 8-6 = 2$.

(D) Given that $\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of $a x^4+b x^3+c x^2+d x+e=0$.
The roots of $e x^4+d x^3+c x^2+b x+a=0$ are the reciprocals of the roots of the first equation.
Thus,$\alpha_2 = \frac{1}{\delta_1}, \beta_2 = \frac{1}{\gamma_1}, \gamma_2 = \frac{1}{\beta_1}, \delta_2 = \frac{1}{\alpha_1}$.
Given $\alpha_1 - \delta_2 = 2 \implies \alpha_1 - \frac{1}{\alpha_1} = 2 \implies \alpha_1^2 - 2\alpha_1 - 1 = 0$.
Given $\delta_1 - \alpha_2 = 4 \implies \delta_1 - \frac{1}{\delta_1} = 4 \implies \delta_1^2 - 4\delta_1 - 1 = 0$.
Since $\alpha_1$ and $\delta_1$ are roots of the quartic equation,the quadratic factors are $(x^2 - 2x - 1)$ and $(x^2 - 4x - 1)$.
Thus,$a x^4+b x^3+c x^2+d x+e = (x^2 - 2x - 1)(x^2 - 4x - 1)$.
To find $a+b+c+d+e$,we substitute $x=1$:
$a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = (1^2 - 2(1) - 1)(1^2 - 4(1) - 1)$.
$a+b+c+d+e = (-2)(-4) = 8$.

(C) Let $\alpha, \beta, \gamma$ be the roots of the equation $x^3-ax^2+ax-1=0$. \\ By Vieta's formulas: \\ $\alpha+\beta+\gamma = a$ \\ $\alpha\beta+\beta\gamma+\gamma\alpha = a$ \\ $\alpha\beta\gamma = 1$ (Note: The constant term is $-1$,so product is $-(-1)/1 = 1$). \\ The cubic equation whose roots are $\alpha^2, \beta^2, \gamma^2$ is $x^3 - (\alpha^2+\beta^2+\gamma^2)x^2 + (\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)x - (\alpha\beta\gamma)^2 = 0$. \\ Comparing this with $x^3-ax^2+ax-1=0$,we get: \\ $a = \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = a^2 - 2a$. \\ Thus,$a^2 - 3a = 0$,which implies $a(a-3) = 0$. \\ Since $a$ is non-zero,$a = 3$.

(D) Given equation is $x^5+15x^4+94x^3+305x^2+507x+353=0$.
If all roots of the equation are increased by $k$,the transformed equation is obtained by replacing $x$ with $(x-k)$.
The transformed equation is $(x-k)^5+15(x-k)^4+94(x-k)^3+305(x-k)^2+507(x-k)+353=0$.
To eliminate the $x^4$ term,the coefficient of $x^4$ must be zero.
The coefficient of $x^4$ is given by $\binom{5}{1}(-k) + 15 = -5k + 15$.
Setting $-5k + 15 = 0$,we get $k = 3$.
Now,we substitute $k=3$ into the transformed equation $(x-3)^5+15(x-3)^4+94(x-3)^3+305(x-3)^2+507(x-3)+353=0$.
The coefficient of $x$ is given by:
$\binom{5}{4}(-3)^4 + 15 \times \binom{4}{3}(-3)^3 + 94 \times \binom{3}{2}(-3)^2 + 305 \times \binom{2}{1}(-3)^1 + 507$.
$= 5(81) + 15(4 \times -27) + 94(3 \times 9) + 305(2 \times -3) + 507$.
$= 405 - 1620 + 2538 - 1830 + 507 = 0$.
Thus,the coefficient of $x$ is $0$.

(C) Given the equation $x^3-7x^2+14x-8=0$. Let the roots be diminished by $k$,so $y = x - k$,which implies $x = y + k$.
Substituting $x = y + k$ into the original equation:
$(y+k)^3 - 7(y+k)^2 + 14(y+k) - 8 = 0$
Expanding the terms:
$(y^3 + 3y^2k + 3yk^2 + k^3) - 7(y^2 + 2yk + k^2) + 14(y + k) - 8 = 0$
$y^3 + y^2(3k - 7) + y(3k^2 - 14k + 14) + (k^3 - 7k^2 + 14k - 8) = 0$
Comparing this with $y^3 + py - \frac{20}{27} = 0$,the coefficient of $y^2$ must be zero:
$3k - 7 = 0 \Rightarrow k = \frac{7}{3}$
Now,find $p$ as the coefficient of $y$:
$p = 3k^2 - 14k + 14$
$p = 3(\frac{7}{3})^2 - 14(\frac{7}{3}) + 14$
$p = 3(\frac{49}{9}) - \frac{98}{3} + 14$
$p = \frac{49}{3} - \frac{98}{3} + \frac{42}{3} = -\frac{7}{3}$

(B) To find the polynomial whose roots are translated by $-2$,we replace $x$ with $(x+2)$ in the original equation $x^5-2x^4+3x^3-4x^2+5x-6=0$.
Substituting $x \to x+2$:
$(x+2)^5 - 2(x+2)^4 + 3(x+2)^3 - 4(x+2)^2 + 5(x+2) - 6 = 0$.
Expanding each term:
$(x^5+10x^4+40x^3+80x^2+80x+32) - 2(x^4+8x^3+24x^2+32x+16) + 3(x^3+6x^2+12x+8) - 4(x^2+4x+4) + 5(x+2) - 6 = 0$.
Grouping like terms:
$x^5 + (10-2)x^4 + (40-16+3)x^3 + (80-48+18-4)x^2 + (80-64+36-16+5)x + (32-32+24-16+10-6) = 0$.
Simplifying the coefficients:
$x^5 + 8x^4 + 27x^3 + 46x^2 + 41x + 12 = 0$.

(B) Given $f(x)=a x^2+b x+c$ and $GCD(a, b, c)=1$.
Since $\frac{-7+\sqrt{11} i}{6}$ is a root of $f(x)=0$,its conjugate $\frac{-7-\sqrt{11} i}{6}$ must also be a root.
Sum of roots $= \frac{-7+\sqrt{11} i}{6} + \frac{-7-\sqrt{11} i}{6} = \frac{-14}{6} = -\frac{7}{3} = -\frac{b}{a} \implies \frac{b}{a} = \frac{7}{3}$.
Product of roots $= \left(\frac{-7+\sqrt{11} i}{6}\right) \left(\frac{-7-\sqrt{11} i}{6}\right) = \frac{(-7)^2 - (\sqrt{11} i)^2}{36} = \frac{49+11}{36} = \frac{60}{36} = \frac{5}{3} = \frac{c}{a}$.
Thus,$a=3, b=7, c=5$,which satisfies $GCD(3, 7, 5)=1$.
So,$f(x) = 3x^2+7x+5$.
Given $f\left(\frac{x}{k}\right)-L = (x+4)(3x-5) = 3x^2+7x-20$.
Substituting $f\left(\frac{x}{k}\right) = 3\left(\frac{x}{k}\right)^2 + 7\left(\frac{x}{k}\right) + 5$,we get:
$3\frac{x^2}{k^2} + \frac{7x}{k} + 5 - L = 3x^2 + 7x - 20$.
Comparing coefficients:
$\frac{3}{k^2} = 3 \implies k^2 = 1 \implies k = 1$ (taking positive root).
$\frac{7}{k} = 7 \implies k = 1$.
$5 - L = -20 \implies L = 25$.
Therefore,$k=1$ and $L=25$.

(C) Given the equation $x^2-x+1=0$. The roots are $\alpha = -\omega$ and $\alpha = -\omega^2$,where $\omega$ is the complex cube root of unity.
Since $\alpha^2-\alpha+1=0$,we have $\alpha+\frac{1}{\alpha} = 1$.
Let $S_n = \alpha^n + \frac{1}{\alpha^n}$.
For $n=1$,$S_1 = \alpha + \frac{1}{\alpha} = 1$.
For $n=2$,$S_2 = \alpha^2 + \frac{1}{\alpha^2} = (\alpha+\frac{1}{\alpha})^2 - 2 = 1^2 - 2 = -1$.
For $n=3$,$S_3 = \alpha^3 + \frac{1}{\alpha^3} = (-1)^3 + \frac{1}{(-1)^3} = -1 - 1 = -2$.
For $n=4$,$S_4 = \alpha^4 + \frac{1}{\alpha^4} = \alpha + \frac{1}{\alpha} = 1$.
For $n=5$,$S_5 = \alpha^5 + \frac{1}{\alpha^5} = \alpha^2 + \frac{1}{\alpha^2} = -1$.
For $n=6$,$S_6 = \alpha^6 + \frac{1}{\alpha^6} = 1 + 1 = 2$.
The terms are $S_n^3$.
$S_1^3 = 1^3 = 1$,$S_2^3 = (-1)^3 = -1$,$S_3^3 = (-2)^3 = -8$,$S_4^3 = 1^3 = 1$,$S_5^3 = (-1)^3 = -1$,$S_6^3 = 2^3 = 8$.
The sequence of terms is $1, -1, -8, 1, -1, 8, 1, -1, -8, 1, -1, 8$.
Summing these $12$ terms: $(1-1-8) + (1-1+8) + (1-1-8) + (1-1+8) = -8 + 8 - 8 + 8 = 0$.

(D) Given the equation $x^2-x+1=0$. Since $\alpha$ is a root,$\alpha^2-\alpha+1=0$.
Dividing by $\alpha$,we get $\alpha-1+\frac{1}{\alpha}=0$,so $\alpha+\frac{1}{\alpha}=1$.
Now,$\left(\alpha+\frac{1}{\alpha}\right)^2 = \alpha^2+\frac{1}{\alpha^2}+2 = 1^2 = 1$,which implies $\alpha^2+\frac{1}{\alpha^2} = -1$.
Also,$\alpha^3+\frac{1}{\alpha^3} = (\alpha+\frac{1}{\alpha})(\alpha^2+\frac{1}{\alpha^2}-1) = (1)(-1-1) = -2$.
For $\alpha^4+\frac{1}{\alpha^4}$,we use $(\alpha^2+\frac{1}{\alpha^2})^2 = \alpha^4+\frac{1}{\alpha^4}+2 = (-1)^2 = 1$,so $\alpha^4+\frac{1}{\alpha^4} = -1$.
Substituting these values into the expression:
$(1)^3 + (-1)^3 + (-2)^3 + (-1)^3 = 1 - 1 - 8 - 1 = -9$.

(B) Given the inequality $\sqrt{2} \sin^2 x + (3\sqrt{2} + 1) \sin x + 3 > 0$.
Factoring the quadratic expression: $(\sqrt{2} \sin x + 1)(\sin x + 3) > 0$.
Since $\sin x + 3 > 0$ for all real $x$,we must have $\sqrt{2} \sin x + 1 > 0$,which implies $\sin x > -\frac{1}{\sqrt{2}}$.
This inequality holds for $x \in \left(-\frac{\pi}{4} + 2n\pi, \frac{5\pi}{4} + 2n\pi\right)$.
For the second inequality $x^2 - 7x + 10 < 0$,we factor it as $(x - 2)(x - 5) < 0$,which gives $x \in (2, 5)$.
Since $2 \approx 2$ and $\frac{5\pi}{4} \approx 3.927$,the intersection of $x \in (2, 5)$ and the solution set of the trigonometric inequality is $x \in \left(2, \frac{5\pi}{4}\right)$.

(A) Let $y = \frac{x^2-2x+1}{x^2+x-1}$.
$y(x^2+x-1) = x^2-2x+1$
$yx^2 + yx - y = x^2 - 2x + 1$
$(y-1)x^2 + (y+2)x - (y+1) = 0$.
Since $x \in \mathbb{R}$,the discriminant $D \geq 0$.
$D = (y+2)^2 - 4(y-1)(-(y+1)) \geq 0$
$(y+2)^2 + 4(y-1)(y+1) \geq 0$
$y^2 + 4y + 4 + 4(y^2-1) \geq 0$
$y^2 + 4y + 4 + 4y^2 - 4 \geq 0$
$5y^2 + 4y \geq 0$
$y(5y+4) \geq 0$.
The solution to this inequality is $y \in (-\infty, -\frac{4}{5}] \cup [0, \infty)$.
Therefore,the values of $y$ do not lie in the interval $\left(-\frac{4}{5}, 0\right)$.

(C) The expression $2x^2 + 4x + 5 = 2(x^2 + 2x + 1) + 3 = 2(x+1)^2 + 3$. Since $(x+1)^2 \geq 0$,the minimum value is $3$ $(IV)$.
$(B)$ Let $y = \frac{x^2 + 4x + 1}{x^2 + x + 1}$. Then $y(x^2 + x + 1) = x^2 + 4x + 1 \implies (y-1)x^2 + (y-4)x + (y-1) = 0$. For $x$ to be real,$D \geq 0 \implies (y-4)^2 - 4(y-1)^2 \geq 0 \implies (y-4-2y+2)(y-4+2y-2) \geq 0 \implies (-y-2)(3y-6) \geq 0 \implies (y+2)(y-2) \leq 0 \implies -2 \leq y \leq 2$. The maximum value is $2$ $(III)$.
$(C)$ and $(D)$ Given $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$.
$x^2 + 1 \leq 3x^2 - 5x + 6 \implies 2x^2 - 5x + 5 \geq 0$ (Always true as $D = 25 - 40 < 0$).
$3x^2 - 5x + 6 \leq 2x^2 + 2 \implies x^2 - 5x + 4 \leq 0 \implies (x-1)(x-4) \leq 0 \implies 1 \leq x \leq 4$.
Thus $a = 1$ $(II)$ and $b = 4$ $(V)$.
Matching: $(A)$ $\rightarrow IV, (B)$ $\rightarrow III, (C)$ $\rightarrow V, (D)$ $\rightarrow II$. Correct option is $(C)$.

(C) Let $y = \frac{x^2-x+2}{x^2+x-2}$.
Rearranging the terms,we get $y(x^2+x-2) = x^2-x+2$.
$yx^2 + yx - 2y = x^2 - x + 2$.
$(y-1)x^2 + (y+1)x - (2y+2) = 0$.
For $x$ to be a real number,the discriminant $D \ge 0$.
$D = (y+1)^2 - 4(y-1)(-2y-2) \ge 0$.
$(y+1)^2 + 8(y-1)(y+1) \ge 0$.
$(y+1)[(y+1) + 8(y-1)] \ge 0$.
$(y+1)(y+1+8y-8) \ge 0$.
$(y+1)(9y-7) \ge 0$.
The critical points are $y = -1$ and $y = \frac{7}{9}$.
Testing the intervals,the inequality holds for $y \in (-\infty, -1] \cup \left[\frac{7}{9}, \infty\right)$.
Thus,the correct option is $C$.

(D) Let $y = \frac{x^2-3x+4}{x^2+3x+4}$.
$y(x^2+3x+4) = x^2-3x+4$
$yx^2 + 3yx + 4y = x^2 - 3x + 4$
$x^2(y-1) + x(3y+3) + (4y-4) = 0$.
Since $x$ is real,the discriminant $D \geq 0$.
$D = (3y+3)^2 - 4(y-1)(4y-4) \geq 0$
$9(y+1)^2 - 16(y-1)^2 \geq 0$
$9(y^2+2y+1) - 16(y^2-2y+1) \geq 0$
$9y^2 + 18y + 9 - 16y^2 + 32y - 16 \geq 0$
$-7y^2 + 50y - 7 \geq 0$
$7y^2 - 50y + 7 \leq 0$
$(7y-1)(y-7) \leq 0$.
Thus,the interval is $[\frac{1}{7}, 7]$.

(D) Let $f(x) = \frac{x^2-6x+5}{x^2+2x+1}$ for $x \in R$.
Set $y = \frac{x^2-6x+5}{x^2+2x+1}$.
Then $y(x^2+2x+1) = x^2-6x+5$,which simplifies to $(y-1)x^2 + (2y+6)x + (y-5) = 0$.
Since $x$ is a real number,the discriminant $D$ must be greater than or equal to $0$.
$D = (2y+6)^2 - 4(y-1)(y-5) \geq 0$.
$4(y+3)^2 - 4(y^2-6y+5) \geq 0$.
$(y^2+6y+9) - (y^2-6y+5) \geq 0$.
$12y + 4 \geq 0$.
$12y \geq -4$.
$y \geq -\frac{1}{3}$.
Thus,the least value of the expression is $-\frac{1}{3}$.

(A) Given $x = \frac{1}{2} \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right)$.
First,calculate $x^2$:
$x^2 = \frac{1}{4} \left( 3 + \frac{1}{3} + 2 \right) = \frac{1}{4} \left( \frac{9+1+6}{3} \right) = \frac{1}{4} \left( \frac{16}{3} \right) = \frac{4}{3}$.
Now,calculate $x^2 - 1$:
$x^2 - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.
Therefore,$\sqrt{x^2 - 1} = \frac{1}{\sqrt{3}}$.
Substitute these values into the expression $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$:
$= \frac{\frac{1}{\sqrt{3}}}{\frac{1}{2} \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) - \frac{1}{\sqrt{3}}}$
$= \frac{\frac{1}{\sqrt{3}}}{\frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}} - \frac{1}{\sqrt{3}}}$
$= \frac{\frac{1}{\sqrt{3}}}{\frac{\sqrt{3}}{2} - \frac{1}{2\sqrt{3}}}$
$= \frac{\frac{1}{\sqrt{3}}}{\frac{3 - 1}{2\sqrt{3}}} = \frac{\frac{1}{\sqrt{3}}}{\frac{2}{2\sqrt{3}}} = \frac{\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = 1$.

(C) Let $f(x) = -2x^2 + 3x + 1$.
To find the extreme value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = -4x + 3$.
Setting $f'(x) = 0$,we get $-4x + 3 = 0$,which implies $x = \frac{3}{4}$.
The extreme value $k$ is $f\left(\frac{3}{4}\right) = -2\left(\frac{3}{4}\right)^2 + 3\left(\frac{3}{4}\right) + 1 = -2\left(\frac{9}{16}\right) + \frac{9}{4} + 1 = -\frac{9}{8} + \frac{18}{8} + \frac{8}{8} = \frac{17}{8}$.
Now,we need to find the set of $x$ such that $kx^2 + 2x + 1 > 0$,where $k = \frac{17}{8}$.
Substituting $k$,we get $\frac{17}{8}x^2 + 2x + 1 > 0$.
Multiplying by $8$,we get $17x^2 + 16x + 8 > 0$.
The discriminant $D$ of this quadratic is $D = b^2 - 4ac = (16)^2 - 4(17)(8) = 256 - 544 = -288$.
Since $D < 0$ and the coefficient of $x^2$ $(17)$ is positive,the quadratic $17x^2 + 16x + 8$ is always positive for all real $x$.
Thus,the solution set is $(-\infty, \infty)$.

(B) The minimum value of a quadratic function $f(x)=ax^2+bx+c$ (where $a>0$) is given by $-\frac{D}{4a} = \frac{4ac-b^2}{4a}$.
For $f(x)=2x^2+\alpha x+8$,we have $a=2, b=\alpha, c=8$. The minimum value is $\frac{4(2)(8)-\alpha^2}{4(2)} = \frac{64-\alpha^2}{8}$.
The maximum value of a quadratic function $g(x)=ax^2+bx+c$ (where $a < 0$) is given by $-\frac{D}{4a} = \frac{4ac-b^2}{4a}$.
For $g(x)=-3x^2-4x+\alpha^2$,we have $a=-3, b=-4, c=\alpha^2$. The maximum value is $\frac{4(-3)(\alpha^2)-(-4)^2}{4(-3)} = \frac{-12\alpha^2-16}{-12} = \frac{12\alpha^2+16}{12}$.
Equating the two values:
$\frac{64-\alpha^2}{8} = \frac{12\alpha^2+16}{12}$
Multiplying both sides by $24$:
$3(64-\alpha^2) = 2(12\alpha^2+16)$
$192-3\alpha^2 = 24\alpha^2+32$
$160 = 27\alpha^2$
$\alpha^2 = \frac{160}{27}$.
Thus,option $(B)$ is correct.

(C) Given that,$x = \frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right) = \frac{1}{2} \left( \frac{7+1}{\sqrt{7}} \right) = \frac{4}{\sqrt{7}}$.
Then,$x^2 = \frac{16}{7}$.
So,$x^2 - 1 = \frac{16}{7} - 1 = \frac{9}{7}$.
Thus,$\sqrt{x^2 - 1} = \sqrt{\frac{9}{7}} = \frac{3}{\sqrt{7}}$.
Now,substitute these values into the expression:
$\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}} = \frac{\frac{3}{\sqrt{7}}}{\frac{4}{\sqrt{7}} - \frac{3}{\sqrt{7}}} = \frac{\frac{3}{\sqrt{7}}}{\frac{1}{\sqrt{7}}} = 3$.

(D) Given that $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$,we have $\alpha^{2}+3 p^{2} \alpha+5 q^{2}=0$.
Since $\beta$ is a root of $x^{2}+9 p^{2} x+15 q^{2}=0$,we have $\beta^{2}+9 p^{2} \beta+15 q^{2}=0$.
Let $f(x)=x^{2}+6 p^{2} x+10 q^{2}$.
Evaluating $f(\alpha)$:
$f(\alpha)=\alpha^{2}+6 p^{2} \alpha+10 q^{2} = (\alpha^{2}+3 p^{2} \alpha+5 q^{2}) + 3 p^{2} \alpha + 5 q^{2} = 0 + 3 p^{2} \alpha + 5 q^{2}$.
Since $\alpha > 0$ and $p, q$ are real numbers (assuming $p, q \neq 0$ for non-trivial roots),$f(\alpha) > 0$.
Evaluating $f(\beta)$:
$f(\beta)=\beta^{2}+6 p^{2} \beta+10 q^{2} = (\beta^{2}+9 p^{2} \beta+15 q^{2}) - (3 p^{2} \beta + 5 q^{2}) = 0 - (3 p^{2} \beta + 5 q^{2})$.
Since $\beta > \alpha > 0$,$f(\beta) < 0$.
Since $f(x)$ is a continuous polynomial function and $f(\alpha) > 0$ while $f(\beta) < 0$,by the Intermediate Value Theorem,there must exist a root $\gamma$ such that $\alpha < \gamma < \beta$.

(C) Given the quadratic equation $x^2 - a(x-1) + b = 0$,which simplifies to $x^2 - ax + a + b = 0$.
Since $\alpha$ and $\beta$ are the roots,they satisfy the equation:
$\alpha^2 - a\alpha + a + b = 0 \implies \alpha^2 - a\alpha = -(a+b)$
$\beta^2 - a\beta + a + b = 0 \implies \beta^2 - a\beta = -(a+b)$
Substituting these into the expression:
$\frac{1}{\alpha^2 - a\alpha} + \frac{1}{\beta^2 - a\beta} + \frac{2}{a+b} = \frac{1}{-(a+b)} + \frac{1}{-(a+b)} + \frac{2}{a+b}$
$= -\frac{1}{a+b} - \frac{1}{a+b} + \frac{2}{a+b} = -\frac{2}{a+b} + \frac{2}{a+b} = 0$.

(C) Let $y = \frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$.
$y(2 x^{2}+4 x+9) = x^{2}+2 x+4$.
$2 y x^{2} + 4 y x + 9 y = x^{2} + 2 x + 4$.
$(2 y-1) x^{2} + (4 y-2) x + (9 y-4) = 0$.
Since $x$ is a real number,the discriminant $D \geq 0$.
$D = (4 y-2)^{2} - 4(2 y-1)(9 y-4) \geq 0$.
$4(2 y-1)^{2} - 4(2 y-1)(9 y-4) \geq 0$.
$4(2 y-1) [ (2 y-1) - (9 y-4) ] \geq 0$.
$4(2 y-1) (-7 y + 3) \geq 0$.
$(2 y-1) (7 y - 3) \leq 0$.
This inequality holds for $\frac{3}{7} \leq y \leq \frac{1}{2}$.
Therefore,the maximum value of $y$ is $\frac{1}{2}$.

(A) To find the least value of the expression $\frac{6a}{5b} + \frac{10b}{3a}$,we use the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality.
For any two positive real numbers $x$ and $y$,the inequality states that $\frac{x+y}{2} \geq \sqrt{xy}$,or $x+y \geq 2\sqrt{xy}$.
Let $x = \frac{6a}{5b}$ and $y = \frac{10b}{3a}$.
Then,$\frac{6a}{5b} + \frac{10b}{3a} \geq 2 \sqrt{\frac{6a}{5b} \times \frac{10b}{3a}}$.
Simplifying the expression inside the square root:
$\frac{6a}{5b} \times \frac{10b}{3a} = \frac{6 \times 10}{5 \times 3} \times \frac{a}{a} \times \frac{b}{b} = \frac{60}{15} = 4$.
Therefore,$\frac{6a}{5b} + \frac{10b}{3a} \geq 2 \sqrt{4} = 2 \times 2 = 4$.
The least possible value is $4$.

(C) Given $a^{n} + b^{n} = c^{n} + d^{n}$ for all $n \in \mathbb{N}$.
For $n = 1$,we have $a + b = c + d$.
For $n = 2$,we have $a^{2} + b^{2} = c^{2} + d^{2}$.
From $a + b = c + d$,we get $a - c = d - b$.
From $a^{2} + b^{2} = c^{2} + d^{2}$,we get $a^{2} - c^{2} = d^{2} - b^{2}$,which implies $(a - c)(a + c) = (d - b)(d + b)$.
If $a \neq c$,then $a + c = d + b$. Since $a - c = d - b$,adding these gives $2a = 2d \Rightarrow a = d$,which implies $b = c$.
If $a = c$,then $b = d$.
In both cases,the set of values ${a, b}$ is the same as ${c, d}$.
Option $(C)$ provides the conditions $a + b = c + d$ and $a^{2} + b^{2} = c^{2} + d^{2}$,which are necessary and sufficient for the equality to hold for all $n$.

(D) $1$) From the equation $x^2 - 3x + r = 0$,we have $\alpha + \beta = 3$ and $\alpha\beta = r$.
$2$) From the equation $x^2 + 3x + r = 0$,the roots are $\frac{\alpha}{2}$ and $2\beta$. Thus,$\frac{\alpha}{2} + 2\beta = -3$ and $(\frac{\alpha}{2})(2\beta) = r$,which simplifies to $\alpha\beta = r$. This is consistent.
$3$) Multiplying the first equation by $2$,we get $\alpha + 4\beta = -6$. Subtracting $\alpha + \beta = 3$ from this gives $3\beta = -9$,so $\beta = -3$. Substituting back,$\alpha = 6$.
$4$) Then $r = \alpha\beta = 6(-3) = -18$.
$5$) The roots of $x^2 + 6x - m = 0$ are $x_1 = 2\alpha + \beta + 2r = 2(6) - 3 + 2(-18) = 12 - 3 - 36 = -27$ and $x_2 = \alpha - 2\beta - \frac{r}{2} = 6 - 2(-3) - \frac{-18}{2} = 6 + 6 + 9 = 21$.
$6$) The product of the roots is $x_1 x_2 = (-27)(21) = -567$. Since the equation is $x^2 + 6x - m = 0$,the product of the roots is $-m$. Therefore,$-m = -567$,which implies $m = 567$.

(C) Let $x_1 = \tan A$ and $x_2 = \tan B$ be the roots of the equation $x^2 - 2x - 5 = 0$.
From the properties of roots,$x_1 + x_2 = 2$ and $x_1 x_2 = -5$.
Using the formula for $\tan(A+B)$,we have $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2}{1 - (-5)} = \frac{2}{6} = \frac{1}{3}$.
We know that $\sin^2(\frac{A+B}{2}) = \frac{1 - \cos(A+B)}{2}$.
Given $\tan(A+B) = \frac{1}{3}$,we can find $\cos(A+B)$. Since $\tan(A+B) > 0$,$A+B$ lies in the first or third quadrant. However,since $A, B \in (-\frac{\pi}{2}, \frac{\pi}{2})$,$A+B \in (-\pi, \pi)$. For $\tan(A+B) = 1/3$,$\cos(A+B) = \frac{3}{\sqrt{1^2 + 3^2}} = \frac{3}{\sqrt{10}}$.
Substituting this into the identity: $\sin^2(\frac{A+B}{2}) = \frac{1 - 3/\sqrt{10}}{2} = \frac{\sqrt{10} - 3}{2\sqrt{10}}$.
Finally,$20 \sin^2(\frac{A+B}{2}) = 20 \times \frac{\sqrt{10} - 3}{2\sqrt{10}} = \frac{10(\sqrt{10} - 3)}{\sqrt{10}} = 10 - \frac{30}{\sqrt{10}} = 10 - 3\sqrt{10}$.

(C) Let the roots be $a, ar, ar^2, ar^3$ in $G$.$P$.
From the first equation $x^2 - x + p = 0$,we have $\alpha + \beta = a + ar = 1$ and $\alpha\beta = a(ar) = a^2r = p$.
From the second equation $x^2 - 4x + q = 0$,we have $\gamma + \delta = ar^2 + ar^3 = r^2(a + ar) = 4$ and $\gamma\delta = (ar^2)(ar^3) = a^2r^5 = q$.
Since $a + ar = 1$,we substitute this into the second sum equation: $r^2(1) = 4 \implies r^2 = 4 \implies r = \pm 2$.
Case $1$: If $r = 2$,then $a(1 + 2) = 1 \implies a = 1/3$. Then $p = a^2r = (1/9)(2) = 2/9$,which is not an integer. This case is rejected.
Case $2$: If $r = -2$,then $a(1 - 2) = 1 \implies -a = 1 \implies a = -1$.
Now calculate $p$ and $q$: $p = a^2r = (-1)^2(-2) = -2$.
$q = a^2r^5 = (-1)^2(-2)^5 = 1 \times (-32) = -32$.
Finally,$|p + q| = |-2 + (-32)| = |-34| = 34$.