Relation between roots and coefficients Questions in English

(A) Given the roots $\alpha, \beta$ of $ax^2 + 2bx + c = 0$,we have $\alpha + \beta = -\frac{2b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Given the roots $\gamma, \delta$ of $px^2 + 2qx + r = 0$,we have $\gamma + \delta = -\frac{2q}{p}$ and $\gamma\delta = \frac{r}{p}$.
Since $\alpha, \beta, \gamma, \delta$ are in $G.P.$,let the common ratio be $k$. Then $\beta = \alpha k$,$\gamma = \alpha k^2$,and $\delta = \alpha k^3$.
From the roots of the first equation: $\alpha + \alpha k = -\frac{2b}{a} \Rightarrow \alpha(1+k) = -\frac{2b}{a}$ and $\alpha^2k = \frac{c}{a}$.
From the roots of the second equation: $\alpha k^2 + \alpha k^3 = -\frac{2q}{p} \Rightarrow \alpha k^2(1+k) = -\frac{2q}{p}$ and $\alpha^2k^5 = \frac{r}{p}$.
Dividing the sum equations: $\frac{\alpha k^2(1+k)}{\alpha(1+k)} = \frac{-2q/p}{-2b/a} \Rightarrow k^2 = \frac{aq}{bp}$.
Dividing the product equations: $\frac{\alpha^2k^5}{\alpha^2k} = \frac{r/p}{c/a} \Rightarrow k^4 = \frac{ar}{pc}$.
Since $k^4 = (k^2)^2$,we have $(\frac{aq}{bp})^2 = \frac{ar}{pc} \Rightarrow \frac{a^2q^2}{b^2p^2} = \frac{ar}{pc}$.
Simplifying,$a^2q^2pc = ar b^2p^2$ $\Rightarrow aq^2c = rb^2p$ $\Rightarrow q^2ac = b^2pr$.

(A) Let the roots of the quadratic equation $x^2 - (a - 2)x + (a - 3) = 0$ be $\alpha$ and $\beta$.
From the properties of roots,we have $\alpha + \beta = a - 2$ and $\alpha \beta = a - 3$.
The sum of the cubes of the roots is given by $S = \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$.
Substituting the values,$S = (a - 2)^3 - 3(a - 3)(a - 2)$.
Expanding this expression: $S = (a^3 - 6a^2 + 12a - 8) - 3(a^2 - 5a + 6) = a^3 - 6a^2 + 12a - 8 - 3a^2 + 15a - 18 = a^3 - 9a^2 + 27a - 26$.
We can rewrite this as $S = (a - 3)^3 + 1$.
Since $a \ge 3$,the term $(a - 3)^3$ is non-negative and its minimum value is $0$ when $a = 3$.
Thus,the sum of the cubes of the roots assumes the least value when $a = 3$.

(A) Given the cubic equation: ${x^3} + px + q = 0$ .....$(i)$
From the properties of roots of a cubic equation $ax^3 + bx^2 + cx + d = 0$:
Sum of roots: $\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{0}{1} = 0$
Sum of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = p$
Product of roots: $\alpha\beta\gamma = -\frac{d}{a} = -q$
We use the algebraic 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 $\alpha + \beta + \gamma = 0$,the right side of the identity becomes $0$.
Therefore,${\alpha^3} + {\beta^3} + {\gamma^3} - 3\alpha\beta\gamma = 0$
${\alpha^3} + {\beta^3} + {\gamma^3} = 3\alpha\beta\gamma$
Substituting $\alpha\beta\gamma = -q$,we get:
${\alpha^3} + {\beta^3} + {\gamma^3} = 3(-q) = -3q$.

(B) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
According to the properties of roots,$\alpha + \beta = p + 3$ and $\alpha \beta = 5p - 2$.
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 $\alpha^2 + \beta^2 = (p + 3)^2 - 2(5p - 2)$.
Expanding this,$\alpha^2 + \beta^2 = p^2 + 6p + 9 - 10p + 4 = p^2 - 4p + 13$.
To find the minimum value,we complete the square: $p^2 - 4p + 13 = (p^2 - 4p + 4) + 9 = (p - 2)^2 + 9$.
The expression $(p - 2)^2 + 9$ attains its minimum value when $(p - 2)^2 = 0$,which implies $p = 2$.

(A) The given condition $f(12 - x) = f(x)$ implies that the graph of the quadratic function $f(x)$ is symmetric about the vertical line $x = \frac{12 - x + x}{2} = 6$.
For a quadratic function $f(x) = ax^2 + bx + c$,the axis of symmetry is given by $x = -\frac{b}{2a}$.
Here,$a = \frac{1}{4}$ and the coefficient of $x$ is $b$.
Equating the axis of symmetry to $6$,we get: $-\frac{b}{2(1/4)} = 6$.
$-\frac{b}{1/2} = 6$.
$-2b = 6$.
$b = -3$.

(A) Given $\alpha^2 + \beta^2 = 5$ and $3(\alpha^5 + \beta^5) = 11(\alpha^3 + \beta^3)$.
Let $S_n = \alpha^n + \beta^n$. We have $S_2 = 5$.
Using the identity $\alpha^5 + \beta^5 = (\alpha^2 + \beta^2)(\alpha^3 + \beta^3) - \alpha^2 \beta^2(\alpha + \beta)$ is not direct,so we use the recurrence relation for $x^2 - px + q = 0$ where $p = \alpha + \beta$ and $q = \alpha \beta$.
Since $\alpha^2 + \beta^2 = 5$,we have $p^2 - 2q = 5$.
The relation $3 S_5 = 11 S_3$ can be solved by testing values. If $\alpha \beta = 2$,then $\alpha^2 + \beta^2 = 5$ implies $\alpha, \beta$ are roots of $x^2 - px + 2 = 0$. Since $\alpha^2 + \beta^2 = p^2 - 2(2) = 5$,$p^2 = 9$,so $p = \pm 3$.
If $p = 3$,the roots are $1$ and $2$. Checking the condition: $3(1^5 + 2^5) = 3(1 + 32) = 3(33) = 99$. Also $11(1^3 + 2^3) = 11(1 + 8) = 11(9) = 99$.
Since $99 = 99$,the value $\alpha \beta = 2$ satisfies the given equations.

(A) Given that $\alpha, \beta, \gamma$ are the roots of $x^3 - x - 1 = 0$.
From the properties of roots,$\alpha + \beta + \gamma = 0$.
Therefore,$\beta + \gamma = -\alpha$,$\gamma + \alpha = -\beta$,and $\alpha + \beta = -\gamma$.
The roots of the required equation are $\frac{1}{-\alpha}, \frac{1}{-\beta}, \frac{1}{-\gamma}$,which simplifies to $-\frac{1}{\alpha}, -\frac{1}{\beta}, -\frac{1}{\gamma}$.
Let $y = -\frac{1}{x}$,so $x = -\frac{1}{y}$.
Substitute $x = -\frac{1}{y}$ into the original equation $x^3 - x - 1 = 0$:
$(-\frac{1}{y})^3 - (-\frac{1}{y}) - 1 = 0$
$-\frac{1}{y^3} + \frac{1}{y} - 1 = 0$
Multiply by $-y^3$:
$1 - y^2 + y^3 = 0$
$y^3 - y^2 + 1 = 0$.
Thus,the required equation is $x^3 - x^2 + 1 = 0$.

(A) Let $y = 2\alpha + 1$,then $\alpha = \frac{y - 1}{2}$.
Since $\alpha$ is a root of $x^3 + 2x - 5 = 0$,we substitute $\alpha$ into the equation:
$(\frac{y - 1}{2})^3 + 2(\frac{y - 1}{2}) - 5 = 0$
$\frac{y^3 - 3y^2 + 3y - 1}{8} + y - 1 - 5 = 0$
$y^3 - 3y^2 + 3y - 1 + 8y - 48 = 0$
$y^3 - 3y^2 + 11y - 49 = 0$
Comparing this with $x^3 + bx^2 + cx + d = 0$,we get $b = -3, c = 11, d = -49$.
Therefore,$|b + c + d| = |-3 + 11 - 49| = |-41| = 41$.

(C) Given that $\alpha, \beta$ are the roots of the quadratic equation $x^2 - ax + b = 0$.
By the properties of roots,we have $\alpha + \beta = a$ and $\alpha \beta = b$.
We are given $V_n = \alpha^n + \beta^n$.
Consider the expression $V_{n+1} = \alpha^{n+1} + \beta^{n+1}$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation $x^2 = ax - b$. Thus,$\alpha^2 = a\alpha - b$ and $\beta^2 = a\beta - b$.
Multiplying by $\alpha^{n-1}$ and $\beta^{n-1}$ respectively,we get $\alpha^{n+1} = a\alpha^n - b\alpha^{n-1}$ and $\beta^{n+1} = a\beta^n - b\beta^{n-1}$.
Adding these two equations:
$\alpha^{n+1} + \beta^{n+1} = a(\alpha^n + \beta^n) - b(\alpha^{n-1} + \beta^{n-1})$.
Substituting $V_n$ values,we get $V_{n+1} = aV_n - bV_{n-1}$.

(B) Given the equation $x^3 + qx - r = 0$,we have $\alpha + \beta + \gamma = 0$,$\alpha \beta + \beta \gamma + \gamma \alpha = q$,and $\alpha \beta \gamma = r$.
Since $\alpha \beta \gamma = r$,we have $\beta \gamma = \frac{r}{\alpha}$,$\gamma \alpha = \frac{r}{\beta}$,and $\alpha \beta = \frac{r}{\gamma}$.
The roots of the new equation are $\frac{r}{\alpha} + \frac{1}{\alpha} = \frac{r+1}{\alpha}$,$\frac{r+1}{\beta}$,and $\frac{r+1}{\gamma}$.
Let $y = \frac{r+1}{x}$,which implies $x = \frac{r+1}{y}$.
Substituting this into the original equation: $\left( \frac{r+1}{y} \right)^3 + q \left( \frac{r+1}{y} \right) - r = 0$.
Multiplying by $y^3$: $(r+1)^3 + q(r+1)y^2 - ry^3 = 0$.
Rearranging gives $ry^3 - q(r+1)y^2 - (r+1)^3 = 0$.

(A) Let the three roots of the equation $x^3 + 3x^2 - 1 = 0$ be $\alpha, \beta,$ and $\gamma$.
From the properties of roots,the product of the roots is $\alpha \beta \gamma = -(\frac{-1}{1}) = 1$.
Thus,$\alpha \beta = \frac{1}{\gamma}$.
Since $\gamma$ is a root of the original equation,it satisfies $\gamma^3 + 3\gamma^2 - 1 = 0$.
Let $x = \alpha \beta = \frac{1}{\gamma}$,which implies $\gamma = \frac{1}{x}$.
Substituting $\gamma = \frac{1}{x}$ into the original equation:
$(\frac{1}{x})^3 + 3(\frac{1}{x})^2 - 1 = 0$
$\frac{1}{x^3} + \frac{3}{x^2} - 1 = 0$
Multiplying by $x^3$,we get $1 + 3x - x^3 = 0$,which simplifies to $x^3 - 3x - 1 = 0$.

(A) Given the equation $x^3 - 2x^2 + 3x - 2 = 0$.
By inspection,$x = 1$ is a root.
Dividing by $(x - 1)$,we get $(x - 1)(x^2 - x + 2) = 0$.
Thus,$\alpha = 1$ and $\beta, \gamma$ are roots of $x^2 - x + 2 = 0$.
From the quadratic equation,$\beta + \gamma = 1$ and $\beta \gamma = 2$.
We need to evaluate $S = \frac{\alpha \beta}{\alpha + \beta} + \frac{\alpha \gamma}{\alpha + \gamma} + \frac{\beta \gamma}{\beta + \gamma}$.
Substituting $\alpha = 1$: $S = \frac{\beta}{1 + \beta} + \frac{\gamma}{1 + \gamma} + \frac{2}{1} = \frac{\beta(1 + \gamma) + \gamma(1 + \beta)}{(1 + \beta)(1 + \gamma)} + 2$.
$S = \frac{\beta + \beta \gamma + \gamma + \beta \gamma}{1 + (\beta + \gamma) + \beta \gamma} + 2 = \frac{(\beta + \gamma) + 2(\beta \gamma)}{1 + (\beta + \gamma) + \beta \gamma} + 2$.
Substituting values: $S = \frac{1 + 2(2)}{1 + 1 + 2} + 2 = \frac{5}{4} + 2 = \frac{13}{4}$.

(D) Given $P(x) = x^2 + ax + b = (x - a)(x - b)$.
Expanding the right side: $x^2 - (a + b)x + ab = x^2 + ax + b$.
Comparing coefficients:
$a = -(a + b)$ $\Rightarrow 2a + b = 0$ $\Rightarrow b = -2a$.
$b = ab$.
Case $1$: If $b \neq 0$,then $a = 1$. Substituting into $b = -2a$,we get $b = -2$.
Then $P(x) = x^2 + x - 2$. $P(2) = 2^2 + 2 - 2 = 4$.
Case $2$: If $b = 0$,then $2a + 0 = 0 \Rightarrow a = 0$.
Then $P(x) = x^2$. $P(2) = 2^2 = 4$.
In both cases,$P(2) = 4$.

(D) Since $\alpha, \beta, \gamma$ are roots of $x^3 - x - 2 = 0$,we have $\alpha^3 = \alpha + 2$,$\beta^3 = \beta + 2$,and $\gamma^3 = \gamma + 2$.
Multiplying by $\alpha^2$,we get $\alpha^5 = \alpha^3 + 2\alpha^2 = (\alpha + 2) + 2\alpha^2 = 2\alpha^2 + \alpha + 2$.
Similarly,$\beta^5 = 2\beta^2 + \beta + 2$ and $\gamma^5 = 2\gamma^2 + \gamma + 2$.
Summing these,$\alpha^5 + \beta^5 + \gamma^5 = 2(\alpha^2 + \beta^2 + \gamma^2) + (\alpha + \beta + \gamma) + 6$.
From the equation $x^3 + 0x^2 - x - 2 = 0$,we have $\sum \alpha = 0$ and $\sum \alpha\beta = -1$.
Using the identity $\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2(\sum \alpha\beta) = 0^2 - 2(-1) = 2$.
Substituting these values: $\alpha^5 + \beta^5 + \gamma^5 = 2(2) + 0 + 6 = 10$.

(C) Let $a^2(a + p) = b^2(b + p) = c^2(c + p) = k$.
Then $a, b, c$ are the roots of the cubic equation $x^3 + px^2 - k = 0$.
Comparing this with the standard cubic equation $x^3 + Ax^2 + Bx + C = 0$,we have $A = p$,$B = 0$,and $C = -k$.
By Vieta's formulas,the sum of the roots taken two at a time is given by the coefficient of $x$,which is $B$.
Therefore,$ab + bc + ca = 0$.

(A) Let the roots of the equation $x^2 + (k + 1)x + \lambda = 0$ be $\alpha$ and $\alpha^2$.
From the properties of roots,we have:
$\alpha + \alpha^2 = -(k + 1)$
$\alpha \cdot \alpha^2 = \alpha^3 = \lambda$
If the roots are squares of each other,we consider the following cases:
$1$. If $\alpha = 0$,then $\alpha^2 = 0$. The equation becomes $x^2 = 0$,so $k+1 = 0 \implies k = -1$.
$2$. If $\alpha = 1$,then $\alpha^2 = 1$. The equation becomes $(x-1)^2 = x^2 - 2x + 1 = 0$. Comparing with $x^2 + (k+1)x + \lambda = 0$,we get $k+1 = -2 \implies k = -3$.
$3$. If $\alpha = \omega$ (where $\omega$ is a complex cube root of unity),then $\alpha^2 = \omega^2$. The equation is $x^2 + x + 1 = 0$. Comparing with $x^2 + (k+1)x + \lambda = 0$,we get $k+1 = 1 \implies k = 0$.
The possible values of $k$ are $-1, -3, 0$.
The sum of these values is $(-1) + (-3) + 0 = -4$.

(C) Given $|f(a)| + |f(b)| + |f(c)| = 0$. Since the absolute value is always non-negative,this implies $f(a) = 0, f(b) = 0$,and $f(c) = 0$.
Thus,$a, b, c$ are the roots of the cubic equation $x^3 - 2x + 2 = 0$.
For any root $x$,we have $x^3 - 2x + 2 = 0$,which implies $x^3 = 2x - 2$.
Dividing by $x$ (assuming $x \neq 0$,which is true as $f(0) = 2 \neq 0$),we get $x^2 = 2 - \frac{2}{x}$,or $x^2 + \frac{2}{x} = 2$.
Since $a, b, c$ are roots,$a^2 + \frac{2}{a} = 2$,$b^2 + \frac{2}{b} = 2$,and $c^2 + \frac{2}{c} = 2$.
We need to evaluate $f^2(a^2 + \frac{2}{a}) + f^2(b^2 + \frac{2}{b}) - f^2(c^2 + \frac{2}{c}) = f^2(2) + f^2(2) - f^2(2) = f^2(2)$.
Calculating $f(2) = 2^3 - 2(2) + 2 = 8 - 4 + 2 = 6$.
Therefore,$f^2(2) = 6^2 = 36$.

(A) Given $\alpha + \beta = 3$ and $|\alpha - \beta| = 4$.
Squaring the second equation,we get $(\alpha - \beta)^2 = 16$.
We know the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.
Substituting the values: $16 = (3)^2 - 4\alpha\beta$.
$16 = 9 - 4\alpha\beta \Rightarrow 4\alpha\beta = 9 - 16 = -7$.
Thus,$\alpha\beta = -\frac{7}{4}$.
The quadratic equation with roots $\alpha$ and $\beta$ is given by $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.
Substituting the values: $x^2 - 3x - \frac{7}{4} = 0$.
Multiplying by $4$,we get $4x^2 - 12x - 7 = 0$.

(A) Given expression: $x^3 - 3x^2 - 9x + c = (x - a)^2 (x - b)$
Expanding the right side: $(x^2 - 2ax + a^2)(x - b) = x^3 - bx^2 - 2ax^2 + 2abx + a^2x - a^2b = x^3 - (2a + b)x^2 + (a^2 + 2ab)x - a^2b$
Comparing coefficients with $x^3 - 3x^2 - 9x + c$:
$2a + b = 3$ $(1)$
$a^2 + 2ab = -9$ $(2)$
$c = -a^2b$ $(3)$
From $(1)$,$b = 3 - 2a$. Substituting into $(2)$:
$a^2 + 2a(3 - 2a) = -9$
$a^2 + 6a - 4a^2 = -9$
$-3a^2 + 6a + 9 = 0$
$a^2 - 2a - 3 = 0$
$(a - 3)(a + 1) = 0$
So,$a = 3$ or $a = -1$.
If $a = 3$,$b = 3 - 2(3) = -3$. Then $c = -(3)^2(-3) = 27$.
If $a = -1$,$b = 3 - 2(-1) = 5$. Then $c = -(-1)^2(5) = -5$.
Thus,$c = -5$ or $27$.

(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Dividing by $a$,we get $\frac{2b}{a} = 1 + \frac{c}{a}$ $(i)$.
Given $\alpha + \beta = -\frac{b}{a} = 15$,so $\frac{b}{a} = -15$.
Substituting this into $(i)$,we get $2(-15) = 1 + \frac{c}{a}$,which implies $-30 = 1 + \frac{c}{a}$,so $\frac{c}{a} = -31$.
Since $\alpha \beta = \frac{c}{a}$,we have $\alpha \beta = -31$.

(B) Let the roots of $x^2 + ax + b = 0$ be $\alpha, \beta$ and the roots of $x^2 + bx + a = 0$ be $\alpha', \beta'$.
Given that $|\alpha - \beta| = |\alpha' - \beta'|$.
Squaring both sides,we get $(\alpha - \beta)^2 = (\alpha' - \beta')^2$.
Using the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$,we have:
$a^2 - 4b = b^2 - 4a$.
Rearranging the terms,we get $a^2 - b^2 = 4b - 4a$.
$(a - b)(a + b) = -4(a - b)$.
Since $a \neq b$,we can divide both sides by $(a - b)$:
$a + b = -4$.

(C) Given $x^3 + 8x + 1 = 0$,the roots are $a, b, c$. Thus,$x^3 = -(8x + 1)$.
For any root $r$,$r^3 = -(8r + 1)$,which implies $8r + 1 = -r^3$.
Substituting this into the expression:
$\frac{bc}{(-b^3)(-c^3)} + \frac{ac}{(-a^3)(-c^3)} + \frac{ab}{(-a^3)(-b^3)}$
$= \frac{bc}{b^3 c^3} + \frac{ac}{a^3 c^3} + \frac{ab}{a^3 b^3} = \frac{1}{b^2 c^2} + \frac{1}{a^2 c^2} + \frac{1}{a^2 b^2}$
$= \frac{a^2 + b^2 + c^2}{a^2 b^2 c^2}$
From the equation $x^3 + 0x^2 + 8x + 1 = 0$,we have $\sum a = 0$,$\sum ab = 8$,and $abc = -1$.
$a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + bc + ca) = (0)^2 - 2(8) = -16$.
$(abc)^2 = (-1)^2 = 1$.
Therefore,the value is $\frac{-16}{1} = -16$.

(B) Let $P(x) = x^3 - 2x^2 + 4x + 5074$. Since $r_1, r_2, r_3$ are the roots of $P(x) = 0$,we can write the polynomial as $P(x) = (x - r_1)(x - r_2)(x - r_3)$.
We want to find the value of $(r_1 + 2)(r_2 + 2)(r_3 + 2)$.
Note that $(r_1 + 2)(r_2 + 2)(r_3 + 2) = -(-r_1 - 2)(-r_2 - 2)(-r_3 - 2) = -((-2 - r_1)(-2 - r_2)(-2 - r_3))$.
This is equivalent to $-P(-2)$.
Substitute $x = -2$ into the polynomial $P(x)$:
$P(-2) = (-2)^3 - 2(-2)^2 + 4(-2) + 5074$
$P(-2) = -8 - 2(4) - 8 + 5074$
$P(-2) = -8 - 8 - 8 + 5074 = -24 + 5074 = 5050$.
Therefore,$(r_1 + 2)(r_2 + 2)(r_3 + 2) = -P(-2) = -5050$.

(A) Given that $\sin \alpha$ and $\cos \alpha$ are roots of $ax^2 + bx + c = 0$.
From the properties of quadratic equations,the sum of roots is $\sin \alpha + \cos \alpha = -\frac{b}{a}$ and the product of roots is $\sin \alpha \cos \alpha = \frac{c}{a}$.
We know the identity $\sin^2 \alpha + \cos^2 \alpha = 1$.
This can be written as $(\sin \alpha + \cos \alpha)^2 - 2 \sin \alpha \cos \alpha = 1$.
Substituting the values of the sum and product of roots:
$(-\frac{b}{a})^2 - 2(\frac{c}{a}) = 1$
$\frac{b^2}{a^2} - \frac{2c}{a} = 1$
Multiplying both sides by $a^2$ (since $a \neq 0$):
$b^2 - 2ac = a^2$
Rearranging the terms,we get $a^2 - b^2 + 2ac = 0$.

(C) Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.
Given $\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$.
Substituting the values: $-b/a = \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} = \frac{b^2 - 2ac}{c^2}$.
$-b/a = \frac{b^2 - 2ac}{c^2} \Rightarrow -bc^2 = ab^2 - 2a^2c$.
$2a^2c = ab^2 + bc^2$.
Dividing by $abc$: $\frac{2a}{b} = \frac{b}{c} + \frac{c}{a}$.
This implies $\frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in $A.P.$
Therefore,their reciprocals $\frac{a}{c}, \frac{b}{a}, \frac{c}{b}$ are in $H.P.$

(D) Given the equation $x^4 - 100x^3 + 2x^2 + 4x + 10 = 0$ with roots $\alpha, \beta, \gamma, \delta$.
To find the sum of the reciprocals $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}$,we can substitute $x = \frac{1}{y}$.
Substituting into the equation: $(\frac{1}{y})^4 - 100(\frac{1}{y})^3 + 2(\frac{1}{y})^2 + 4(\frac{1}{y}) + 10 = 0$.
Multiplying by $y^4$: $1 - 100y + 2y^2 + 4y^3 + 10y^4 = 0$,which is $10y^4 + 4y^3 + 2y^2 - 100y + 1 = 0$.
The roots of this new equation are $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta}$.
The sum of the roots is given by $-\frac{\text{coefficient of } y^3}{\text{coefficient of } y^4} = -\frac{4}{10} = -\frac{2}{5}$.

(C) The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
For the given vertices,the centroid is $\left(\frac{\alpha+\beta+\gamma}{3}, \frac{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}}{3}\right)$.
From the equation $x^3 - 3px^2 + 3qx - 1 = 0$,by Vieta's formulas:
$\alpha+\beta+\gamma = 3p$
$\alpha\beta + \beta\gamma + \gamma\alpha = 3q$
$\alpha\beta\gamma = 1$
Substituting these values into the centroid formula:
$x$-coordinate $= \frac{3p}{3} = p$
$y$-coordinate $= \frac{\frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma}}{3} = \frac{3q}{3(1)} = q$
Thus,the centroid is $(p, q)$.

(C) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$.
The harmonic mean $H$ of the roots is given by the formula $H = \frac{2\alpha\beta}{\alpha + \beta}$.
From the given equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + (8 + 2\sqrt{5}) = 0$,we have:
Sum of roots $\alpha + \beta = -\frac{b}{a} = \frac{4 + \sqrt{5}}{5 + \sqrt{2}}$
Product of roots $\alpha\beta = \frac{c}{a} = \frac{8 + 2\sqrt{5}}{5 + \sqrt{2}}$
Substituting these values into the formula for $H$:
$H = 2 \cdot \frac{\frac{8 + 2\sqrt{5}}{5 + \sqrt{2}}}{\frac{4 + \sqrt{5}}{5 + \sqrt{2}}}$
$H = 2 \cdot \frac{8 + 2\sqrt{5}}{4 + \sqrt{5}}$
$H = 2 \cdot \frac{2(4 + \sqrt{5})}{4 + \sqrt{5}}$
$H = 2 \cdot 2 = 4$

(B) Let the roots of the equation $x^{2} + (2 - \lambda)x + (10 - \lambda) = 0$ be $\alpha$ and $\beta$.
From the properties of roots,$\alpha + \beta = \lambda - 2$ and $\alpha\beta = 10 - \lambda$.
The sum of the cubes of the roots is $S = \alpha^{3} + \beta^{3} = (\alpha + \beta)^{3} - 3\alpha\beta(\alpha + \beta)$.
Substituting the values,$S = (\lambda - 2)^{3} - 3(10 - \lambda)(\lambda - 2)$.
$S = (\lambda - 2)[(\lambda - 2)^{2} - 3(10 - \lambda)] = (\lambda - 2)(\lambda^{2} - 4\lambda + 4 - 30 + 3\lambda) = (\lambda - 2)(\lambda^{2} - \lambda - 26)$.
$S = \lambda^{3} - \lambda^{2} - 26\lambda - 2\lambda^{2} + 2\lambda + 52 = \lambda^{3} - 3\lambda^{2} - 24\lambda + 52$.
To find the minimum,set $\frac{dS}{d\lambda} = 3\lambda^{2} - 6\lambda - 24 = 0$.
$3(\lambda^{2} - 2\lambda - 8) = 0 \implies 3(\lambda - 4)(\lambda + 2) = 0$.
For minimum,$\frac{d^{2}S}{d\lambda^{2}} = 6\lambda - 6$. At $\lambda = 4$,$6(4) - 6 = 18 > 0$,so it is a minimum.
The difference of the roots is $|\alpha - \beta| = \sqrt{(\alpha + \beta)^{2} - 4\alpha\beta} = \sqrt{(\lambda - 2)^{2} - 4(10 - \lambda)}$.
For $\lambda = 4$,$|\alpha - \beta| = \sqrt{(4 - 2)^{2} - 4(10 - 4)} = \sqrt{4 - 24} = \sqrt{-20}$.
The magnitude of the difference is $|\sqrt{-20}| = \sqrt{20} = 2\sqrt{5}$.

(A) Given that $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$ are the roots of $ax^2 + bx + 1 = 0$.
Sum of roots: $\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = -\frac{b}{a} \Rightarrow \frac{\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha \beta}} = -\frac{b}{a}$.
Product of roots: $\frac{1}{\sqrt{\alpha \beta}} = \frac{1}{a} \Rightarrow a = \sqrt{\alpha \beta}$.
Substituting $a$ into the sum equation: $\frac{\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha \beta}} = -\frac{b}{\sqrt{\alpha \beta}} \Rightarrow b = -(\sqrt{\alpha} + \sqrt{\beta})$.
The given equation is $x^2 + (b^3 - 3ab)x + a^3 = 0$.
Note that $b^3 - 3ab = -(\sqrt{\alpha} + \sqrt{\beta})^3 - 3(\sqrt{\alpha \beta})(-(\sqrt{\alpha} + \sqrt{\beta})) = -(\alpha^{3/2} + \beta^{3/2} + 3\sqrt{\alpha \beta}(\sqrt{\alpha} + \sqrt{\beta})) + 3\sqrt{\alpha \beta}(\sqrt{\alpha} + \sqrt{\beta}) = -(\alpha^{3/2} + \beta^{3/2})$.
Also,$a^3 = (\sqrt{\alpha \beta})^3 = \alpha^{3/2} \beta^{3/2}$.
The equation becomes $x^2 - (\alpha^{3/2} + \beta^{3/2})x + \alpha^{3/2} \beta^{3/2} = 0$.
This is a quadratic equation with roots $\alpha^{3/2}$ and $\beta^{3/2}$.

(D) Given the equation $x^2 - 4\sqrt{2}kx + 2e^{4\ln k} - 1 = 0$.
Since $e^{4\ln k} = k^4$,the equation becomes $x^2 - 4\sqrt{2}kx + 2k^4 - 1 = 0$.
From the properties of roots,$\alpha + \beta = 4\sqrt{2}k$ and $\alpha\beta = 2k^4 - 1$.
We are given $\alpha^2 + \beta^2 = 66$.
Using the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$,we have:
$(4\sqrt{2}k)^2 = 66 + 2(2k^4 - 1)$
$32k^2 = 66 + 4k^4 - 2$
$4k^4 - 32k^2 + 64 = 0$
Dividing by $4$,we get $k^4 - 8k^2 + 16 = 0$,which is $(k^2 - 4)^2 = 0$.
Thus,$k^2 = 4$,so $k = 2$ or $k = -2$.
Now,$\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = (4\sqrt{2}k)(66 - (2k^4 - 1)) = (4\sqrt{2}k)(67 - 2k^4)$.
If $k = 2$,$\alpha^3 + \beta^3 = (4\sqrt{2}(2))(67 - 2(16)) = (8\sqrt{2})(67 - 32) = (8\sqrt{2})(35) = 280\sqrt{2}$.
If $k = -2$,$\alpha^3 + \beta^3 = (4\sqrt{2}(-2))(67 - 2(16)) = (-8\sqrt{2})(35) = -280\sqrt{2}$.
Since the question implies a single value and $-280\sqrt{2}$ is an option,we select $D$.

(C) The given quadratic equation is $x^2 + px + \frac{3p}{4} = 0$.
By the properties of roots,we have $\alpha + \beta = -p$ and $\alpha \beta = \frac{3p}{4}$.
Given that $|\alpha - \beta| = \sqrt{10}$,squaring both sides gives $(\alpha - \beta)^2 = 10$.
Using the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$,we substitute the values:
$(-p)^2 - 4 \left( \frac{3p}{4} \right) = 10$.
This simplifies to $p^2 - 3p = 10$,or $p^2 - 3p - 10 = 0$.
Factoring the quadratic equation,we get $(p - 5)(p + 2) = 0$.
Thus,$p = 5$ or $p = -2$.
Therefore,$p \in \{-2, 5\}$.

(B) Given $\alpha^3 + \beta^3 = -p$ and $\alpha \beta = q$.
Let the roots of the required quadratic equation be $x_1 = \frac{\alpha^2}{\beta}$ and $x_2 = \frac{\beta^2}{\alpha}$.
The sum of the roots is $x_1 + x_2 = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha \beta} = \frac{-p}{q}$.
The product of the roots is $x_1 \times x_2 = \frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha} = \alpha \beta = q$.
The required quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - (\frac{-p}{q})x + q = 0$.
$x^2 + \frac{p}{q}x + q = 0$.
Multiplying by $q$,we get $qx^2 + px + q^2 = 0$.

(D) Let the roots of the equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$ be $x_1$ and $x_2$.
From the properties of quadratic equations,we have:
$x_1 + x_2 = \sin \alpha - 2$
$x_1 x_2 = -(1 + \sin \alpha)$
We want to minimize $S = x_1^2 + x_2^2$.
$S = (x_1 + x_2)^2 - 2x_1 x_2$
$S = (\sin \alpha - 2)^2 - 2(-(1 + \sin \alpha))$
$S = \sin^2 \alpha - 4 \sin \alpha + 4 + 2 + 2 \sin \alpha$
$S = \sin^2 \alpha - 2 \sin \alpha + 6$
To minimize $S$,let $u = \sin \alpha$,where $u \in [-1, 1]$.
$S(u) = u^2 - 2u + 6 = (u - 1)^2 + 5$.
The minimum value occurs at $u = 1$.
Since $\sin \alpha = 1$,we have $\alpha = \frac{\pi}{2}$.

(D) Let the roots of the quadratic equation $81x^2 + kx + 256 = 0$ be $\alpha$ and $\alpha^3$.
From the relation between roots and coefficients:
Sum of roots: $\alpha + \alpha^3 = -\frac{k}{81}$ $(1)$
Product of roots: $\alpha \cdot \alpha^3 = \alpha^4 = \frac{256}{81}$ $(2)$
From $(2)$,$\alpha^4 = (\frac{4}{3})^4$,so $\alpha = \pm \frac{4}{3}$.
Case $1$: If $\alpha = \frac{4}{3}$,then $\alpha + \alpha^3 = \frac{4}{3} + (\frac{4}{3})^3 = \frac{4}{3} + \frac{64}{27} = \frac{36+64}{27} = \frac{100}{27}$.
Substituting into $(1)$: $\frac{100}{27} = -\frac{k}{81} \implies k = -\frac{100 \times 81}{27} = -300$.
Case $2$: If $\alpha = -\frac{4}{3}$,then $\alpha + \alpha^3 = -\frac{4}{3} - \frac{64}{27} = -\frac{100}{27}$.
Substituting into $(1)$: $-\frac{100}{27} = -\frac{k}{81} \implies k = 300$.
Since $300$ is not in the options,the value is $-300$.

(B) Let the roots of the equation be $\alpha$ and $\beta$. Given $\lambda = \frac{\alpha}{\beta}$.
Given $\lambda + \frac{1}{\lambda} = 1$,we have $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 1$.
This simplifies to $\frac{\alpha^2 + \beta^2}{\alpha\beta} = 1$,or $\alpha^2 + \beta^2 = \alpha\beta$.
Adding $2\alpha\beta$ to both sides,we get $(\alpha + \beta)^2 = 3\alpha\beta$.
From the quadratic equation $3m^2x^2 + m(m - 4)x + 2 = 0$,the sum of roots $\alpha + \beta = -\frac{m(m - 4)}{3m^2} = -\frac{m - 4}{3m}$ and the product of roots $\alpha\beta = \frac{2}{3m^2}$.
Substituting these into the equation $(\alpha + \beta)^2 = 3\alpha\beta$:
$\left(-\frac{m - 4}{3m}\right)^2 = 3 \left(\frac{2}{3m^2}\right)$.
$\frac{(m - 4)^2}{9m^2} = \frac{2}{m^2}$.
Since $m \neq 0$,we multiply by $9m^2$:
$(m - 4)^2 = 18$.
$m^2 - 8m + 16 = 18 \Rightarrow m^2 - 8m - 2 = 0$.
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$m = \frac{8 \pm \sqrt{64 - 4(1)(-2)}}{2} = \frac{8 \pm \sqrt{72}}{2} = \frac{8 \pm 6\sqrt{2}}{2} = 4 \pm 3\sqrt{2}$.
The least value is $4 - 3\sqrt{2}$.

(D) Given the equation $x^{2}-x-1=0$,by Vieta's formulas,$\alpha+\beta=1$ and $\alpha\beta=-1$.
Since $\alpha$ and $\beta$ are roots,$\alpha^{2}-\alpha-1=0$ and $\beta^{2}-\beta-1=0$.
Multiplying by $\alpha^{k-2}$ and $\beta^{k-2}$ respectively,we get $\alpha^{k}-\alpha^{k-1}-\alpha^{k-2}=0$ and $\beta^{k}-\beta^{k-1}-\beta^{k-2}=0$.
Adding these,we obtain the recurrence relation $p_{k}=p_{k-1}+p_{k-2}$.
Calculating the values:
$p_{1}=\alpha+\beta=1$
$p_{2}=(\alpha+\beta)^{2}-2\alpha\beta=1^{2}-2(-1)=3$
$p_{3}=p_{2}+p_{1}=3+1=4$
$p_{4}=p_{3}+p_{2}=4+3=7$
$p_{5}=p_{4}+p_{3}=7+4=11$
Checking the options:
$A: p_{1}+p_{2}+p_{3}+p_{4}+p_{5} = 1+3+4+7+11 = 26$ (True)
$B: p_{5}=11$ (True)
$C: p_{5}-p_{4} = 11-7 = 4 = p_{3}$ (True)
$D: p_{2} \cdot p_{3} = 3 \cdot 4 = 12 \neq p_{5}$ (False)
Thus,the statement that is not true is $D$.

(A) Let the roots of the quadratic equation be $a$ and $b$.
According to the given condition,
$A.M. = \frac{a+b}{2} = 8 \Rightarrow a+b = 16$ $(1)$
$G.M. = \sqrt{ab} = 5 \Rightarrow ab = 25$ $(2)$
The quadratic equation is given by,
$x^{2} - x(\text{Sum of roots}) + (\text{Product of roots}) = 0$
$x^{2} - x(a+b) + (ab) = 0$
Substituting values from $(1)$ and $(2)$,
$x^{2} - 16x + 25 = 0$
Thus,the required quadratic equation is $x^{2} - 16x + 25 = 0$.

(A) Since $\alpha$ and $\beta$ are roots of $5x^{2} + 6x - 2 = 0$,they satisfy the equation.
$5\alpha^{2} + 6\alpha - 2 = 0 \implies 5\alpha^{n+2} + 6\alpha^{n+1} - 2\alpha^{n} = 0$ (multiplying by $\alpha^{n}$).
Similarly,$5\beta^{n+2} + 6\beta^{n+1} - 2\beta^{n} = 0$.
Adding these two equations,we get $5(\alpha^{n+2} + \beta^{n+2}) + 6(\alpha^{n+1} + \beta^{n+1}) - 2(\alpha^{n} + \beta^{n}) = 0$.
This simplifies to $5S_{n+2} + 6S_{n+1} - 2S_{n} = 0$.
For $n = 4$,we have $5S_{6} + 6S_{5} - 2S_{4} = 0$,which implies $5S_{6} + 6S_{5} = 2S_{4}$.

(D) For the equation $x^{2}-x+2\lambda=0$,we have $\alpha+\beta=1$ and $\alpha\beta=2\lambda$.
For the equation $3x^{2}-10x+27\lambda=0$,we have $\alpha+\gamma=\frac{10}{3}$ and $\alpha\gamma=\frac{27\lambda}{3}=9\lambda$.
Subtracting the sum of roots: $(\alpha+\gamma)-(\alpha+\beta)=\frac{10}{3}-1 \Rightarrow \gamma-\beta=\frac{7}{3}$.
Dividing the product of roots: $\frac{\alpha\gamma}{\alpha\beta}=\frac{9\lambda}{2\lambda}$ $\Rightarrow \frac{\gamma}{\beta}=\frac{9}{2}$ $\Rightarrow \gamma=\frac{9}{2}\beta$.
Substituting $\gamma$ into $\gamma-\beta=\frac{7}{3}$: $\frac{9}{2}\beta-\beta=\frac{7}{3}$ $\Rightarrow \frac{7}{2}\beta=\frac{7}{3}$ $\Rightarrow \beta=\frac{2}{3}$.
Then $\gamma=\frac{9}{2} \times \frac{2}{3}=3$.
Since $\alpha+\beta=1$,$\alpha=1-\frac{2}{3}=\frac{1}{3}$.
Using $\alpha\beta=2\lambda$: $\frac{1}{3} \times \frac{2}{3}=2\lambda$ $\Rightarrow \frac{2}{9}=2\lambda$ $\Rightarrow \lambda=\frac{1}{9}$.
Finally,$\frac{\beta\gamma}{\lambda}=\frac{(2/3) \times 3}{1/9}=\frac{2}{1/9}=18$.

(A) Given the quadratic equation $7x^{2}-3x-2=0$.
From the properties of roots,we have $\alpha+\beta = \frac{3}{7}$ and $\alpha\beta = \frac{-2}{7}$.
We need to evaluate $S = \frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$.
$S = \frac{\alpha(1-\beta^{2})+\beta(1-\alpha^{2})}{(1-\alpha^{2})(1-\beta^{2})} = \frac{\alpha-\alpha\beta^{2}+\beta-\alpha^{2}\beta}{1-(\alpha^{2}+\beta^{2})+\alpha^{2}\beta^{2}}$.
$S = \frac{(\alpha+\beta)-\alpha\beta(\alpha+\beta)}{1-((\alpha+\beta)^{2}-2\alpha\beta)+(\alpha\beta)^{2}}$.
Substituting the values: $\alpha+\beta = \frac{3}{7}$ and $\alpha\beta = \frac{-2}{7}$.
Numerator: $\frac{3}{7} - (\frac{-2}{7})(\frac{3}{7}) = \frac{3}{7} + \frac{6}{49} = \frac{21+6}{49} = \frac{27}{49}$.
Denominator: $1 - ((\frac{3}{7})^{2} - 2(\frac{-2}{7})) + (\frac{-2}{7})^{2} = 1 - (\frac{9}{49} + \frac{4}{7}) + \frac{4}{49} = 1 - (\frac{9+28}{49}) + \frac{4}{49} = 1 - \frac{37}{49} + \frac{4}{49} = \frac{49-37+4}{49} = \frac{16}{49}$.
Thus,$S = \frac{27/49}{16/49} = \frac{27}{16}$.

(C) The given equation is $2x(2x+1)=1$,which simplifies to $4x^{2}+2x-1=0$.
Since $\alpha$ and $\beta$ are the roots,we have $\alpha+\beta = -\frac{b}{a} = -\frac{2}{4} = -\frac{1}{2}$ and $\alpha\beta = \frac{c}{a} = -\frac{1}{4}$.
From $\alpha+\beta = -\frac{1}{2}$,we get $\beta = -\frac{1}{2} - \alpha$.
Since $\alpha$ is a root,it satisfies $4\alpha^{2}+2\alpha-1=0$,which implies $1 = 4\alpha^{2}+2\alpha$.
Substituting $1 = 4\alpha^{2}+2\alpha$ into the expression for $\beta$:
$\beta = -\frac{4\alpha^{2}+2\alpha}{2} - \alpha$
$\beta = -2\alpha^{2} - \alpha - \alpha$
$\beta = -2\alpha^{2} - 2\alpha$
$\beta = -2\alpha(\alpha+1)$.

(D) Given the quadratic equation $x^{2}-64x+256=0$.
From the properties of roots,$\alpha+\beta = 64$ and $\alpha\beta = 256$.
We need to evaluate the expression $E = \left(\frac{\alpha^{3}}{\beta^{5}}\right)^{\frac{1}{8}}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{\frac{1}{8}}$.
$E = \frac{\alpha^{3/8}}{\beta^{5/8}} + \frac{\beta^{3/8}}{\alpha^{5/8}}$.
Taking the common denominator,$E = \frac{\alpha^{3/8} \cdot \alpha^{5/8} + \beta^{3/8} \cdot \beta^{5/8}}{(\alpha\beta)^{5/8}}$.
$E = \frac{\alpha^{(3/8+5/8)} + \beta^{(3/8+5/8)}}{(\alpha\beta)^{5/8}} = \frac{\alpha+\beta}{(\alpha\beta)^{5/8}}$.
Since $\alpha\beta = 256 = 2^{8}$,then $(\alpha\beta)^{5/8} = (2^{8})^{5/8} = 2^{5} = 32$.
Substituting the values,$E = \frac{64}{32} = 2$.

(D) Given $p + q = 2$ and $p^{4} + q^{4} = 272$.
We know that $p^{2} + q^{2} = (p + q)^{2} - 2pq = 2^{2} - 2pq = 4 - 2pq$.
Also,$p^{4} + q^{4} = (p^{2} + q^{2})^{2} - 2p^{2}q^{2} = 272$.
Substituting the expression for $p^{2} + q^{2}$:
$(4 - 2pq)^{2} - 2(pq)^{2} = 272$.
$16 - 16pq + 4(pq)^{2} - 2(pq)^{2} = 272$.
$2(pq)^{2} - 16pq - 256 = 0$.
$(pq)^{2} - 8pq - 128 = 0$.
Let $t = pq$. Then $t^{2} - 8t - 128 = 0$.
Using the quadratic formula: $t = \frac{8 \pm \sqrt{64 - 4(1)(-128)}}{2} = \frac{8 \pm \sqrt{64 + 512}}{2} = \frac{8 \pm \sqrt{576}}{2} = \frac{8 \pm 24}{2}$.
$t = 16$ or $t = -8$.
Since $p$ and $q$ are positive,$pq$ must be positive,so $pq = 16$.
The quadratic equation with roots $p$ and $q$ is $x^{2} - (p+q)x + pq = 0$.
Substituting the values,we get $x^{2} - 2x + 16 = 0$.

(D) The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
Given vertices are $(a, c), (2, b), (a, b)$ and centroid is $\left(\frac{10}{3}, \frac{7}{3}\right)$.
Equating the $x$-coordinates: $\frac{a+2+a}{3} = \frac{10}{3} \implies 2a + 2 = 10 \implies 2a = 8 \implies a = 4$.
Equating the $y$-coordinates: $\frac{c+b+b}{3} = \frac{7}{3} \implies c + 2b = 7$.
Since $a, b, c$ are in arithmetic progression,$2b = a + c$. Substituting $a=4$,we get $2b = 4 + c \implies c = 2b - 4$.
Substitute $c$ into the equation $c + 2b = 7$: $(2b - 4) + 2b = 7 \implies 4b = 11 \implies b = \frac{11}{4}$.
The quadratic equation is $4x^{2} + \frac{11}{4}x + 1 = 0$.
For roots $\alpha, \beta$,we have $\alpha + \beta = -\frac{b}{a} = -\frac{11/4}{4} = -\frac{11}{16}$ and $\alpha\beta = \frac{c}{a} = \frac{1}{4}$.
We need to find $\alpha^{2} + \beta^{2} - \alpha\beta = (\alpha + \beta)^{2} - 3\alpha\beta$.
Substituting the values: $\left(-\frac{11}{16}\right)^{2} - 3\left(\frac{1}{4}\right) = \frac{121}{256} - \frac{3}{4} = \frac{121 - 192}{256} = -\frac{71}{256}$.

(B) Since $\alpha, \beta \in \mathbb{R}$,the complex roots must occur in conjugate pairs. Therefore,the other root is $1+2i$.
Using the relationship between roots and coefficients:
Sum of roots $= -\alpha = (1-2i) + (1+2i) = 2 \implies \alpha = -2$.
Product of roots $= \beta = (1-2i)(1+2i) = 1^{2} - (2i)^{2} = 1 + 4 = 5$.
Therefore,$\alpha - \beta = -2 - 5 = -7$.

(A) Since $\alpha$ and $\beta$ are the roots of $x^{2}-6x-2=0$,we have $\alpha^{2}-6\alpha-2=0$ and $\beta^{2}-6\beta-2=0$.
Multiplying the first equation by $\alpha^{8}$,we get $\alpha^{10}-6\alpha^{9}-2\alpha^{8}=0$.
Similarly,multiplying the second equation by $\beta^{8}$,we get $\beta^{10}-6\beta^{9}-2\beta^{8}=0$.
Subtracting the two equations,we obtain $(\alpha^{10}-\beta^{10})-6(\alpha^{9}-\beta^{9})-2(\alpha^{8}-\beta^{8})=0$.
Using the definition $a_{n}=\alpha^{n}-\beta^{n}$,this simplifies to $a_{10}-6a_{9}-2a_{8}=0$.
Rearranging the terms,we get $a_{10}-2a_{8}=6a_{9}$.
Therefore,$\frac{a_{10}-2a_{8}}{3a_{9}} = \frac{6a_{9}}{3a_{9}} = 2$.

(B) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^{2} - (\alpha+\beta)x + \alpha\beta = 0$.
Substituting the values,we get $x^{2} - x - 1 = 0$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$\alpha^{2} - \alpha - 1 = 0 \Rightarrow \alpha^{n+1} = \alpha^{n} + \alpha^{n-1}$
$\beta^{2} - \beta - 1 = 0 \Rightarrow \beta^{n+1} = \beta^{n} + \beta^{n-1}$
Adding these two equations,we get:
$(\alpha^{n+1} + \beta^{n+1}) = (\alpha^{n} + \beta^{n}) + (\alpha^{n-1} + \beta^{n-1})$
This implies the recurrence relation $p_{n+1} = p_{n} + p_{n-1}$.
Given $p_{n+1} = 29$ and $p_{n-1} = 11$,we have:
$29 = p_{n} + 11$
$p_{n} = 29 - 11 = 18$.
Therefore,$p_{n}^{2} = 18^{2} = 324$.

(D) Given the equation $x^{2}+5 \sqrt{2} x+10=0$. Since $\alpha$ and $\beta$ are roots,they satisfy the equation: $\alpha^{2} + 5 \sqrt{2} \alpha + 10 = 0 \implies \alpha^{2} = -5 \sqrt{2} \alpha - 10$ and $\beta^{2} = -5 \sqrt{2} \beta - 10$.
Also,$P_{n} = \alpha^{n} - \beta^{n}$.
Consider the expression $E = \frac{P_{17} P_{20} + 5 \sqrt{2} P_{17} P_{19}}{P_{18} P_{19} + 5 \sqrt{2} P_{18}^{2}}$.
Factor out $P_{17}$ in the numerator and $P_{18}$ in the denominator:
$E = \frac{P_{17}(P_{20} + 5 \sqrt{2} P_{19})}{P_{18}(P_{19} + 5 \sqrt{2} P_{18})}$.
Substitute $P_{n} = \alpha^{n} - \beta^{n}$:
$P_{20} + 5 \sqrt{2} P_{19} = (\alpha^{20} - \beta^{20}) + 5 \sqrt{2} (\alpha^{19} - \beta^{19}) = \alpha^{19}(\alpha + 5 \sqrt{2}) - \beta^{19}(\beta + 5 \sqrt{2})$.
From the quadratic equation,$\alpha + 5 \sqrt{2} = -\frac{10}{\alpha}$ and $\beta + 5 \sqrt{2} = -\frac{10}{\beta}$.
Substituting these:
$P_{20} + 5 \sqrt{2} P_{19} = \alpha^{19}(-\frac{10}{\alpha}) - \beta^{19}(-\frac{10}{\beta}) = -10 \alpha^{18} + 10 \beta^{18} = -10 P_{18}$.
Similarly,$P_{19} + 5 \sqrt{2} P_{18} = \alpha^{18}(\alpha + 5 \sqrt{2}) - \beta^{18}(\beta + 5 \sqrt{2}) = -10 \alpha^{17} + 10 \beta^{17} = -10 P_{17}$.
Thus,$E = \frac{P_{17}(-10 P_{18})}{P_{18}(-10 P_{17})} = 1$.

(B) Given $a+b+c=1$,$ab+bc+ca=2$,and $abc=3$.
First,find $a^{2}+b^{2}+c^{2}$ using the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2} + 2(ab+bc+ca)$:
$1^{2} = a^{2}+b^{2}+c^{2} + 2(2)$
$a^{2}+b^{2}+c^{2} = 1 - 4 = -3$.
Next,find $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$ using the identity $(ab+bc+ca)^{2} = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 2abc(a+b+c)$:
$2^{2} = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 2(3)(1)$
$4 = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 6$
$a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} = 4 - 6 = -2$.
Finally,use the identity $a^{4}+b^{4}+c^{4} = (a^{2}+b^{2}+c^{2})^{2} - 2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$:
$a^{4}+b^{4}+c^{4} = (-3)^{2} - 2(-2)$
$a^{4}+b^{4}+c^{4} = 9 + 4 = 13$.