QUADRATIC EQUATION Questions in English

(B) Given inequality: $2x^2 + 3x - 9 \le 0$
Factorizing the quadratic expression: $2x^2 + 6x - 3x - 9 \le 0$
$2x(x + 3) - 3(x + 3) \le 0$
$(2x - 3)(x + 3) \le 0$
The roots of the equation $2x^2 + 3x - 9 = 0$ are $x = 3/2$ and $x = -3$.
For a quadratic expression $ax^2 + bx + c \le 0$ where $a > 0$,the solution lies between the roots,i.e.,$\text{root}_1 \le x \le \text{root}_2$.
Thus,the solution is $-3 \le x \le 3/2$.

(C) For a quadratic equation $Ax^2 + Bx + C = 0$ to have roots with opposite signs,the product of the roots must be negative.
The product of the roots is given by $\frac{C}{A}$. Here,$A = 3$ and $C = a^2 - 3a + 2$.
Thus,we require $\frac{a^2 - 3a + 2}{3} < 0$,which implies $a^2 - 3a + 2 < 0$.
Factoring the quadratic expression,we get $(a - 1)(a - 2) < 0$.
This inequality holds when $1 < a < 2$.
Additionally,for the roots to be real,the discriminant $D = B^2 - 4AC$ must be greater than or equal to $0$. Since the product of the roots is negative,the roots are guaranteed to be real and distinct,so the discriminant condition is automatically satisfied for all $a$ in the interval $(1, 2)$.

(C) Given the roots of $x^2 + px + q = 0$ are $\alpha$ and $\beta$,we have $\alpha + \beta = -p$ and $\alpha\beta = q$.
Given the roots of $x^2 - xr + s = 0$ are $\alpha^4$ and $\beta^4$,we have $\alpha^4 + \beta^4 = r$ and $\alpha^4\beta^4 = s$.
Now,consider the discriminant $D$ of the equation $x^2 - 4qx + 2q^2 - r = 0$:
$D = (-4q)^2 - 4(1)(2q^2 - r) = 16q^2 - 8q^2 + 4r = 8q^2 + 4r$.
Substitute $q = \alpha\beta$ and $r = \alpha^4 + \beta^4$:
$D = 8(\alpha\beta)^2 + 4(\alpha^4 + \beta^4) = 4(2\alpha^2\beta^2 + \alpha^4 + \beta^4) = 4(\alpha^2 + \beta^2)^2$.
Since $(\alpha^2 + \beta^2)^2 \ge 0$ for all real $\alpha, \beta$,it follows that $D \ge 0$.
Therefore,the roots of the equation $x^2 - 4qx + 2q^2 - r = 0$ are always real.

(C) Given the expression $f(x) = mx - 1 + \frac{1}{x} \ge 0$ for $x > 0$.
Multiplying by $x$ (since $x > 0$),we get $mx^2 - x + 1 \ge 0$.
For a quadratic expression $ax^2 + bx + c$ to be non-negative for all $x > 0$,we must have $a > 0$ and the discriminant $D \le 0$.
Here,$a = m$,$b = -1$,and $c = 1$.
Condition $1$: $a > 0 \implies m > 0$.
Condition $2$: $D = b^2 - 4ac \le 0 \implies (-1)^2 - 4(m)(1) \le 0$.
$1 - 4m \le 0 \implies 4m \ge 1 \implies m \ge \frac{1}{4}$.
Since $m > 0$ and $m \ge \frac{1}{4}$,the minimum value of $m$ is $\frac{1}{4}$.

(C) For a quadratic equation $Ax^2 + Bx + C = 0$,the roots are equal if the discriminant $D = B^2 - 4AC = 0$.
Here,$A = a(b - c)$,$B = b(c - a)$,and $C = c(a - b)$.
Sum of coefficients: $A + B + C = a(b - c) + b(c - a) + c(a - b) = ab - ac + bc - ba + ca - cb = 0$.
Since the sum of coefficients is $0$,$x = 1$ is one of the roots.
Because the roots are equal,both roots must be $1$.
For a quadratic equation $Ax^2 + Bx + C = 0$,the product of roots is $C/A$.
Therefore,$1 \times 1 = \frac{c(a - b)}{a(b - c)}$.
$a(b - c) = c(a - b) \Rightarrow ab - ac = ac - bc$.
$ab + bc = 2ac$.
Dividing both sides by $abc$,we get $\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$.
This condition implies that $a, b, c$ are in Harmonic Progression $(H.P.)$.

(A) Let the roots of the equation $lx^2 + nx + n = 0$ be $\alpha$ and $\beta$.
Given that the ratio of the roots is $\frac{\alpha}{\beta} = \frac{p}{q}$.
From the properties of quadratic equations,we have $\alpha + \beta = -\frac{n}{l}$ and $\alpha\beta = \frac{n}{l}$.
We need to evaluate $\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}}$.
Substituting $\frac{p}{q} = \frac{\alpha}{\beta}$ and $\frac{q}{p} = \frac{\beta}{\alpha}$,the expression becomes $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$.
This simplifies to $\frac{\sqrt{\alpha}}{\sqrt{\beta}} + \frac{\sqrt{\beta}}{\sqrt{\alpha}} + \sqrt{\alpha\beta} = \frac{\alpha + \beta}{\sqrt{\alpha\beta}} + \sqrt{\alpha\beta}$.
Substituting the values $\alpha + \beta = -\frac{n}{l}$ and $\alpha\beta = \frac{n}{l}$,we get $\frac{-n/l}{\sqrt{n/l}} + \sqrt{n/l} = -\sqrt{\frac{n}{l}} + \sqrt{\frac{n}{l}} = 0$.

(D) Let the correct equation be $x^2 + px + q = 0$ .....$(i)$
For the first candidate,the roots are $2$ and $6$. Since the error is only in $p$,the constant term $q$ is correct.
Product of roots $q = 2 \times 6 = 12$.
For the second candidate,the roots are $2$ and $-9$. Since the error is only in $q$,the coefficient $p$ is correct.
Sum of roots $= 2 + (-9) = -7 = -p$,which implies $p = 7$.
Substituting the correct values of $p$ and $q$ into equation $(i)$,we get:
$x^2 + 7x + 12 = 0$
Factorizing the quadratic equation:
$x^2 + 4x + 3x + 12 = 0$
$x(x + 4) + 3(x + 4) = 0$
$(x + 3)(x + 4) = 0$
Therefore,the roots are $x = -3$ and $x = -4$.

(B) Given that $\alpha_1, \alpha_2$ are the roots of $ax^2 + bx + c = 0$.
By Vieta's formulas,$\alpha_1 + \alpha_2 = -b/a$ and $\alpha_1 \alpha_2 = c/a$.
Similarly,$\beta_1, \beta_2$ are the roots of $px^2 + qx + r = 0$,so $\beta_1 + \beta_2 = -q/p$ and $\beta_1 \beta_2 = r/p$.
The system $\alpha_1 y + \alpha_2 z = 0$ and $\beta_1 y + \beta_2 z = 0$ has a non-zero solution if the determinant of the coefficient matrix is zero: $\alpha_1 \beta_2 - \alpha_2 \beta_1 = 0$,which implies $\alpha_1 / \beta_1 = \alpha_2 / \beta_2 = k$ (say).
Then $\alpha_1 = k \beta_1$ and $\alpha_2 = k \beta_2$.
From the sum and product of roots:
$\alpha_1 + \alpha_2 = k(\beta_1 + \beta_2) \implies -b/a = k(-q/p) \implies b/a = k(q/p) \implies k = bp/aq$.
$\alpha_1 \alpha_2 = k^2(\beta_1 \beta_2) \implies c/a = k^2(r/p)$.
Substituting $k = bp/aq$:
$c/a = (b^2 p^2 / a^2 q^2) \cdot (r/p) = (b^2 pr) / (a^2 q^2)$.
$c/a = (b^2 pr) / (a^2 q^2) \implies c = (b^2 pr) / (a q^2) \implies a q^2 c = b^2 pr$.
Thus,$b^2 pr = q^2 ac$.

(A) Given $\alpha + \beta = p$ and $\alpha\beta = q$.
Let the roots of the required quadratic equation be $A$ and $B$.
$A = (\alpha^2 - \beta^2)(\alpha^3 - \beta^3) = (\alpha - \beta)(\alpha + \beta)(\alpha - \beta)(\alpha^2 + \alpha\beta + \beta^2) = (\alpha - \beta)^2(\alpha + \beta)(\alpha^2 + \alpha\beta + \beta^2)$.
Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = p^2 - 4q$ and $\alpha^2 + \alpha\beta + \beta^2 = (\alpha + \beta)^2 - \alpha\beta = p^2 - q$,we have $A = (p^2 - 4q)p(p^2 - q) = p[p^4 - 5p^2q + 4q^2]$.
$B = \alpha^3\beta^2 + \alpha^2\beta^3 = \alpha^2\beta^2(\alpha + \beta) = q^2p$.
Sum of roots $S = A + B = p[p^4 - 5p^2q + 4q^2] + pq^2 = p[p^4 - 5p^2q + 5q^2]$.
Product of roots $P = A \cdot B = p[p^4 - 5p^2q + 4q^2] \cdot pq^2 = p^2q^2(p^4 - 5p^2q + 4q^2)$.
The required quadratic equation is $x^2 - Sx + P = 0$.

(D) Let the correct quadratic equation be $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$.
The first student copied the constant term incorrectly,but the coefficient of $x^2$ $(a=1)$ and the coefficient of $x$ $(b)$ were correct. The sum of roots is $-b/a$. Since the roots were $3$ and $2$,the sum is $3 + 2 = 5$. Thus,$-b/1 = 5$,which means $b = -5$.
The second student copied the constant term $(c = -6)$ and the coefficient of $x^2$ $(a = 1)$ correctly. The product of roots is $c/a = -6/1 = -6$.
So,we have the equation $x^2 - 5x - 6 = 0$.
Factoring the equation: $x^2 - 6x + x - 6 = 0$.
$x(x - 6) + 1(x - 6) = 0$.
$(x - 6)(x + 1) = 0$.
Therefore,the correct roots are $x = 6$ and $x = -1$.

(A) Given that $\alpha, \beta$ are roots of $ax^2 + 2bx + c = 0$,so $\alpha + \beta = -\frac{2b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Given that $\gamma, \delta$ are roots of $px^2 + 2qx + r = 0$,so $\gamma + \delta = -\frac{2q}{p}$ and $\gamma\delta = \frac{r}{p}$.
Since $\alpha, \beta, \gamma, \delta$ are in $G.P.$,let the common ratio be $k$. Then $\beta = \alpha k, \gamma = \alpha k^2, \delta = \alpha k^3$.
From the roots,$\frac{\alpha\beta}{\gamma\delta} = \frac{c/a}{r/p} = \frac{cp}{ar}$.
Substituting the $G.P.$ terms: $\frac{\alpha(\alpha k)}{\alpha k^2(\alpha k^3)} = \frac{\alpha^2 k}{\alpha^2 k^5} = \frac{1}{k^4} = \frac{cp}{ar}$.
Also,$\frac{\alpha + \beta}{\gamma + \delta} = \frac{\alpha(1+k)}{\alpha k^2(1+k)} = \frac{1}{k^2}$.
Substituting the sum of roots: $\frac{-2b/a}{-2q/p} = \frac{bp}{aq} = \frac{1}{k^2}$.
Squaring this gives $\frac{b^2p^2}{a^2q^2} = \frac{1}{k^4}$.
Equating the two expressions for $\frac{1}{k^4}$: $\frac{b^2p^2}{a^2q^2} = \frac{cp}{ar}$.
Simplifying: $b^2p^2ar = a^2q^2cp \Rightarrow b^2pr = q^2ac$.

(D) Given the quadratic equation $x^2 + 6x - 2 = 0$ with roots $\alpha$ and $\beta$.
From the properties of roots,$\alpha + \beta = -6$ and $\alpha\beta = -2$.
$1$. Since $\alpha$ is a root,$\alpha^2 + 6\alpha - 2 = 0$. Thus,$\alpha^2 = 2 - 6\alpha$.
Substituting this into option $(A)$: $\alpha^2 + 5\alpha - 8 = (2 - 6\alpha) + 5\alpha - 8 = -\alpha - 6$. Since $\beta = -6 - \alpha$,option $(A)$ is correct.
$2$. From $\alpha\beta = -2$,we have $\beta = -2/\alpha$. Substituting $\alpha^2 + 6\alpha - 2 = 0$ into the numerator of option $(C)$: $\frac{2(\alpha^2 + 6\alpha - 2) - 2}{\alpha} = \frac{2(0) - 2}{\alpha} = -2/\alpha$. Thus,option $(C)$ is correct.
$3$. From $\alpha + \beta = -6$ and $\alpha\beta = -2$,we have $\frac{\alpha + \beta}{\alpha\beta} = \frac{-6}{-2} = 3$. This implies $\frac{1}{\beta} + \frac{1}{\alpha} = 3$,so $\frac{\alpha + \beta}{\alpha\beta} = 3$. Rearranging for $\beta$: $\frac{1}{\beta} = 3 - \frac{1}{\alpha} = \frac{3\alpha - 1}{\alpha}$,which gives $\beta = \frac{\alpha}{3\alpha - 1}$. Thus,option $(B)$ is correct.
Since all options are equivalent to $\beta$,the correct answer is $(D)$.

(C) For a cubic equation of the form ${x^3} + px + q = 0$,the discriminant is given by $\Delta = - (4p^3 + 27q^2)$.
Comparing ${x^3} + 3Hx + G = 0$ with the standard form,we have $p = 3H$ and $q = G$.
The discriminant is $\Delta = - (4(3H)^3 + 27G^2) = - (108H^3 + 27G^2) = - 27(4H^3 + G^2)$.
Given that ${G^2} + 4{H^3} > 0$,it follows that $\Delta = - 27(G^2 + 4H^3) < 0$.
For a cubic equation with real coefficients,if the discriminant $\Delta < 0$,the equation has one real root and two complex conjugate (imaginary) roots.
Therefore,the correct option is $(c)$.

(A) Given the quadratic equation $x^2 + px + p^3 = 0$ with roots $\alpha$ and $\beta$.
From the properties of roots,we have $\alpha + \beta = -p$ and $\alpha \beta = p^3$.
Since the point $(\alpha, \beta)$ lies on the parabola $y^2 = x$,we have $\beta^2 = \alpha$.
Substituting $\alpha = \beta^2$ into the product of roots equation: $\beta^2 \cdot \beta = p^3 \implies \beta^3 = p^3 \implies \beta = p$.
Now,substitute $\beta = p$ and $\alpha = p^2$ into the sum of roots equation $\alpha + \beta = -p$:
$p^2 + p = -p \implies p^2 + 2p = 0$.
Factoring gives $p(p + 2) = 0$. Since $p \neq 0$,we have $p = -2$.
Substituting $p = -2$ back into the roots: $\beta = p = -2$ and $\alpha = p^2 = (-2)^2 = 4$.
Thus,the roots are $4$ and $-2$.

(C) Let $r > 1$ be the common ratio of the $G.P.$ $\alpha, \beta, \gamma, \delta$.
Then $\beta = r\alpha, \gamma = r^2\alpha,$ and $\delta = r^3\alpha$.
From the properties of roots of quadratic equations:
$\alpha + \beta = \alpha(1 + r) = 3$ ..... $(i)$
$\alpha \beta = \alpha^2 r = a$ ..... $(ii)$
$\gamma + \delta = \alpha r^2(1 + r) = 12$ ..... $(iii)$
$\gamma \delta = \alpha^2 r^5 = b$ ..... $(iv)$
Dividing equation $(iii)$ by $(i)$,we get $\frac{\alpha r^2(1 + r)}{\alpha(1 + r)} = \frac{12}{3}$,which simplifies to $r^2 = 4$. Since the $G.P.$ is increasing,$r = 2$.
Substituting $r = 2$ into equation $(i)$,we get $\alpha(1 + 2) = 3$,so $\alpha = 1$.
Now,calculate $a$ and $b$:
$a = \alpha^2 r = (1)^2(2) = 2$.
$b = \alpha^2 r^5 = (1)^2(2^5) = 32$.
Thus,$a = 2$ and $b = 32$.

(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
From the given equation $x^2 - (a - 2)x + (a - 3) = 0$,we have:
Sum of roots: $\alpha + \beta = a - 2$
Product of roots: $\alpha \beta = a - 3$
We need to minimize the sum of the cubes of the roots,$S = \alpha^3 + \beta^3$.
Using the identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$:
$S = (a - 2)^3 - 3(a - 3)(a - 2)$
$S = (a - 2) [(a - 2)^2 - 3(a - 3)]$
$S = (a - 2) [a^2 - 4a + 4 - 3a + 9]$
$S = (a - 2) (a^2 - 7a + 13)$
$S = a^3 - 7a^2 + 13a - 2a^2 + 14a - 26$
$S = a^3 - 9a^2 + 27a - 26$
$S = (a - 3)^3 + 1$
Since $a \ge 3$,the expression $(a - 3)^3 + 1$ is minimized when $(a - 3)^3 = 0$,which occurs at $a = 3$.

(A) Given the equation is ${x^3} - 5{x^2} + 9x - 5 = 0$.
Since the coefficients of the polynomial are real,complex roots must occur in conjugate pairs.
Given one root is $z_1 = 2 + i$,therefore,another root must be $z_2 = 2 - i$.
Let the third root be $\alpha$.
According to the relation between roots and coefficients for a cubic equation $ax^3 + bx^2 + cx + d = 0$,the sum of the roots is given by $-b/a$.
Here,the sum of the roots $= (2 + i) + (2 - i) + \alpha = -(-5)/1 = 5$.
$4 + \alpha = 5 \Rightarrow \alpha = 1$.
Thus,the other roots are $1$ and $2 - i$.

(C) Given the quadratic equation $x^2 + px + (1 - p) = 0$ $(i)$.
Since $(1 - p)$ is a root,it must satisfy the equation:
$(1 - p)^2 + p(1 - p) + (1 - p) = 0$
Factor out $(1 - p)$:
$(1 - p) [(1 - p) + p + 1] = 0$
$(1 - p) [2] = 0$
This implies $1 - p = 0$,so $p = 1$.
Substitute $p = 1$ into the original equation $(i)$:
$x^2 + (1)x + (1 - 1) = 0$
$x^2 + x = 0$
$x(x + 1) = 0$
Thus,the roots are $x = 0$ and $x = -1$.

(A) For $y$ to be a real value,the expression inside the square root must be non-negative,i.e.,$\frac{(x + 1)(x - 3)}{(x - 2)} \ge 0$.
Also,the denominator cannot be zero,so $x \neq 2$.
Let $f(x) = \frac{(x + 1)(x - 3)}{(x - 2)}$. We use the wavy curve method (sign scheme) to find the intervals where $f(x) \ge 0$.
The critical points are $x = -1, 2, 3$.
Testing the intervals:
$1$) For $x \in (-1, 2)$,let $x = 0$: $f(0) = \frac{(1)(-3)}{(-2)} = \frac{3}{2} > 0$.
$2$) For $x \in (2, 3)$,let $x = 2.5$: $f(2.5) = \frac{(3.5)(-0.5)}{(0.5)} = -3.5 < 0$.
$3$) For $x \ge 3$,let $x = 4$: $f(4) = \frac{(5)(1)}{(2)} = 2.5 > 0$.
$4$) For $x \le -1$,let $x = -2$: $f(-2) = \frac{(-1)(-5)}{(-4)} = -1.25 < 0$.
Including the points where the numerator is zero $(x = -1, 3)$,the inequality holds for $-1 \le x < 2$ or $x \ge 3$.

(D) The domain of definition of the function $y = \sqrt{x(x - 3)}$ is $x(x - 3) \ge 0$,which implies $x \le 0$ or $x \ge 3$ ... $(i)$.
Given the equation $(3|x| - 3)^2 = |x| + 7$,let $t = |x|$. Then $(3t - 3)^2 = t + 7$.
Expanding the left side: $9t^2 - 18t + 9 = t + 7$.
Rearranging the terms: $9t^2 - 19t + 2 = 0$.
Factoring the quadratic: $(9t - 1)(t - 2) = 0$.
Thus,$t = 1/9$ or $t = 2$.
Since $t = |x|$,we have $|x| = 1/9$ or $|x| = 2$.
This gives $x = \pm 1/9$ or $x = \pm 2$.
Checking these values against the domain condition $(i)$ ($x \le 0$ or $x \ge 3$):
For $x = 1/9$,$1/9 \not\le 0$ and $1/9 \not\ge 3$ (Reject).
For $x = -1/9$,$-1/9 \le 0$ (Accept).
For $x = 2$,$2 \not\le 0$ and $2 \not\ge 3$ (Reject).
For $x = -2$,$-2 \le 0$ (Accept).
Therefore,the required solutions are $-1/9$ and $-2$.

(C) Given that ${x^2} + 2ax + {a^2}$ is a factor of ${x^3} - 3px + 2q$.
Note that ${x^2} + 2ax + {a^2} = {(x + a)^2}$.
Let ${x^3} - 3px + 2q = {(x + a)^2}(x - 2a) = (x^2 + 2ax + a^2)(x - 2a)$.
Expanding the right side: $(x^2 + 2ax + a^2)(x - 2a) = x^3 - 2ax^2 + 2ax^2 - 4a^2x + a^2x - 2a^3 = x^3 - 3a^2x - 2a^3$.
Comparing this with ${x^3} - 3px + 2q$,we get:
$-3p = -3a^2 \Rightarrow p = a^2$.
$2q = -2a^3 \Rightarrow q = -a^3$.
Now,substitute $a^2 = p$ into $q = -a^3$:
$q^2 = (-a^3)^2 = a^6 = (a^2)^3 = p^3$.
Thus,the condition is ${p^3} = {q^2}$.

(C) $1$. The parabola opens downwards,which implies that the coefficient of $x^2$ is negative,so $a < 0$.
$2$. The curve intersects the $y$-axis at $x = 0$. Substituting $x = 0$ into the equation $y = ax^2 + bx + c$,we get $y = c$. From the figure,the intersection point on the $y$-axis is above the origin,so $c > 0$.
$3$. The roots of the quadratic equation are $x_1$ and $x_2$. From the figure,$x_1 > 0$ and $x_2 < 0$. Since the vertex of the parabola is to the left of the $y$-axis,the sum of the roots is negative: $x_1 + x_2 < 0$. Since the sum of roots is $-b/a$,we have $-b/a < 0$,which implies $b/a > 0$. Since $a < 0$,it follows that $b < 0$.
$4$. Since the curve intersects the $x$-axis at two distinct points,the discriminant $D = b^2 - 4ac > 0$.
Comparing these findings with the options,$c > 0$ is a correct statement.

(A) Let $f(x) = x^2 - 3x + k$. For the roots $\alpha, \beta$ to be real,distinct,and lie in the interval $(0, 1)$,the following conditions must be satisfied:
$1$. Discriminant $D > 0$: $D = (-3)^2 - 4(1)(k) = 9 - 4k > 0 \implies k < 2.25$.
$2$. $f(0) > 0$: $0^2 - 3(0) + k > 0 \implies k > 0$.
$3$. $f(1) > 0$: $1^2 - 3(1) + k > 0 \implies 1 - 3 + k > 0 \implies k > 2$.
$4$. The vertex of the parabola $x = -b/(2a)$ must lie in $(0, 1)$: $x = 3/2 = 1.5$. Since $1.5$ is not in the interval $(0, 1)$,it is impossible for both roots to lie in $(0, 1)$.
Alternatively,for roots to lie in $(0, 1)$,the sum of roots $\alpha + \beta$ must satisfy $0 < \alpha + \beta < 2$. Here,$\alpha + \beta = 3$,which is not less than $2$. Therefore,no such $k$ exists.

(A) Given the cubic equation $x^3 + px + q = 0$.
Since $\alpha, \beta, \gamma$ are the roots,by Vieta's formulas:
$\alpha + \beta + \gamma = 0$ (coefficient of $x^2$ is $0$).
$\alpha\beta + \beta\gamma + \gamma\alpha = p$.
$\alpha\beta\gamma = -q$.
We know 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,$\alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma$.
Substituting $\alpha\beta\gamma = -q$,we get $\alpha^3 + \beta^3 + \gamma^3 = 3(-q) = -3q$.

(A) Let $x = \frac{p}{q}$ be a rational root in simplest form,where $p, q \in \mathbb{Z}$,$q > 0$,and $\gcd(p, q) = 1$.
According to the Rational Root Theorem,for a polynomial with integer coefficients $a_n x^n + \dots + a_0 = 0$,any rational root $\frac{p}{q}$ must satisfy $p \mid a_0$ and $q \mid a_n$.
Here,$a_n = 1$ and $a_0 = 1$.
Thus,$p \mid 1$ and $q \mid 1$,which implies $p, q \in \{-1, 1\}$.
Therefore,the only possible rational roots are $x = 1$ or $x = -1$.
Let $f(x) = x^{2016} - x^{2015} + x^{1008} + x^{1003} + 1$.
For $x = 1$: $f(1) = 1^{2016} - 1^{2015} + 1^{1008} + 1^{1003} + 1 = 1 - 1 + 1 + 1 + 1 = 3 \neq 0$.
For $x = -1$: $f(-1) = (-1)^{2016} - (-1)^{2015} + (-1)^{1008} + (-1)^{1003} + 1 = 1 - (-1) + 1 - 1 + 1 = 1 + 1 + 1 - 1 + 1 = 3 \neq 0$.
Since neither $1$ nor $-1$ is a root,the equation has no rational roots.
Thus,the number of rational roots is $0$.

(B) Given the equation: $\frac{P^2}{x} + \frac{Q^2}{x - 1} = 1$.
Multiplying by $x(x - 1)$,we get: $P^2(x - 1) + Q^2x = x(x - 1)$.
Expanding the terms: $P^2x - P^2 + Q^2x = x^2 - x$.
Rearranging into a standard quadratic form $ax^2 + bx + c = 0$: $x^2 - (P^2 + Q^2 + 1)x + P^2 = 0$.
The discriminant $D$ of this quadratic equation is given by $D = b^2 - 4ac$.
Here,$a = 1$,$b = -(P^2 + Q^2 + 1)$,and $c = P^2$.
$D = (P^2 + Q^2 + 1)^2 - 4(1)(P^2) = (P^2 + Q^2 + 1)^2 - 4P^2$.
Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $D = (P^2 + Q^2 + 1 - 2P)(P^2 + Q^2 + 1 + 2P) = ((P - 1)^2 + Q^2)((P + 1)^2 + Q^2)$.
Since $P$ and $Q$ are non-zero real numbers,$D > 0$. Therefore,the equation has two distinct real roots.

(C) To find the number of solutions for the equation $|x^2 - 2|x|| = 2^x$,we can analyze the graphs of $f(x) = |x^2 - 2|x||$ and $g(x) = 2^x$.
$1$. Analyze $f(x) = |x^2 - 2|x||$:
- Since $f(x)$ is an even function $(f(x) = f(-x))$,the graph is symmetric about the $y$-axis.
- For $x \ge 0$,$f(x) = |x^2 - 2x|$.
- The roots of $x^2 - 2x = 0$ are $x = 0$ and $x = 2$.
- Between $x = 0$ and $x = 2$,$x^2 - 2x$ is negative,so $|x^2 - 2x| = -(x^2 - 2x) = 2x - x^2$.
- For $x > 2$,$x^2 - 2x$ is positive,so $|x^2 - 2x| = x^2 - 2x$.
$2$. Analyze $g(x) = 2^x$:
- This is an exponential growth function that is always positive and strictly increasing.
$3$. Intersection points:
- By plotting both functions,we observe the intersection points.
- For $x < 0$,the exponential function $2^x$ approaches $0$ from above,while the parabola $|x^2 + 2x|$ intersects it at two points.
- For $x > 0$,the function $f(x) = |x^2 - 2x|$ intersects $g(x) = 2^x$ at exactly one point (since $2^x$ grows much faster than $x^2 - 2x$ for large $x$).
- Thus,there are a total of $3$ intersection points.
Therefore,the number of solutions is $3$.

(C) To find the number of solutions for the equation $2^x = x^2$, we can analyze the intersection points of the functions $f(x) = 2^x$ and $g(x) = x^2$.
$1$. For $x < 0$: As $x$ becomes more negative, $2^x$ approaches $0$ from above, while $x^2$ increases. There is one intersection point in the interval $(-1, 0)$ because at $x = -1$, $2^{-1} = 0.5$ and $(-1)^2 = 1$, and at $x = -0.5$, $2^{-0.5} \approx 0.707$ and $(-0.5)^2 = 0.25$. Thus, there is one solution here.
$2$. For $x > 0$: We check specific integer values:
At $x = 2$, $2^2 = 4$ and $2^2 = 4$. So, $x = 2$ is a solution.
At $x = 4$, $2^4 = 16$ and $4^2 = 16$. So, $x = 4$ is a solution.
$3$. Between $x = 2$ and $x = 4$, the function $x^2$ is greater than $2^x$ (e.g., at $x = 3$, $2^3 = 8$ and $3^2 = 9$).
For $x > 4$, the exponential function $2^x$ grows much faster than the quadratic function $x^2$, so there are no more solutions for $x > 4$.
Thus, the solutions are $x = -0.766$ (approx), $x = 2$, and $x = 4$. There are a total of $3$ solutions.

(C) Let $f(x) = \frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2}$ and $g(x) = 3x^3$.
We need to find the number of intersection points of the graphs of $f(x)$ and $g(x)$.
The function $f(x)$ has vertical asymptotes at $x = 0, x = 1,$ and $x = 2$.
$1$. For $x < 0$,$f(x)$ is negative and $g(x) = 3x^3$ is negative. As $x \to -\infty$,$f(x) \to 0$ and $g(x) \to -\infty$. As $x \to 0^-$,$f(x) \to -\infty$ and $g(x) \to 0$. By the Intermediate Value Theorem,there is one intersection point in $(-\infty, 0)$.
$2$. For $0 < x < 1$,$f(x)$ goes from $+\infty$ to $-\infty$. $g(x)$ is positive. There is one intersection point in $(0, 1)$.
$3$. For $1 < x < 2$,$f(x)$ goes from $+\infty$ to $-\infty$. $g(x)$ is positive. There is one intersection point in $(1, 2)$.
$4$. For $x > 2$,$f(x)$ goes from $+\infty$ to $0$ as $x \to \infty$. $g(x)$ goes from $24$ to $+\infty$. There is one intersection point in $(2, \infty)$.
Thus,there are $4$ real roots.

(B) Let the roots of the quadratic equation $x^2 - (p + 3)x + (5p - 2) = 0$ be $\alpha$ and $\beta$.
From the properties of quadratic equations,we have the sum of roots $\alpha + \beta = p + 3$ and the product of roots $\alpha \beta = 5p - 2$.
We want to minimize the sum of the squares of the roots,which is $S = \alpha^2 + \beta^2$.
Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$,we substitute the values:
$S = (p + 3)^2 - 2(5p - 2)$
$S = p^2 + 6p + 9 - 10p + 4$
$S = p^2 - 4p + 13$
To find the minimum value,we complete the square:
$S = (p^2 - 4p + 4) + 9$
$S = (p - 2)^2 + 9$
Since $(p - 2)^2 \ge 0$,the minimum value of $S$ occurs when $p - 2 = 0$,which means $p = 2$.

(A) Given equation: $\log_{\sqrt{3}}(x^3 - 1) = \log_{\sqrt{3}}(x - 1) + 2$.
For the logarithmic functions to be defined,we must have $x^3 - 1 > 0$ and $x - 1 > 0$,which implies $x > 1$.
Using the property $\log_b(A) - \log_b(B) = \log_b(A/B)$,we get $\log_{\sqrt{3}}((x^3 - 1)/(x - 1)) = 2$.
Since $x^3 - 1 = (x - 1)(x^2 + x + 1)$,the equation becomes $\log_{\sqrt{3}}(x^2 + x + 1) = 2$.
Converting to exponential form: $x^2 + x + 1 = (\sqrt{3})^2 = 3$.
So,$x^2 + x - 2 = 0$.
Factoring the quadratic: $(x + 2)(x - 1) = 0$.
This gives $x = -2$ or $x = 1$.
Since the domain requires $x > 1$,both $x = -2$ and $x = 1$ are rejected.
Thus,there are $0$ solutions.

(D) The given inequality is $kx^2 - 2x + k \geq 0$.
If $k = 0$,the inequality becomes $-2x \geq 0$,which implies $x \leq 0$. Since this holds for at least one real $x$,$k = 0$ is a solution.
If $k \neq 0$,we can rewrite the inequality as $k(x^2 + 1) \geq 2x$,which gives $k \geq \frac{2x}{x^2 + 1}$.
Let $f(x) = \frac{2x}{x^2 + 1}$. To find the range of $f(x)$,we analyze its derivative $f'(x) = \frac{2(x^2 + 1) - 2x(2x)}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2}$.
Setting $f'(x) = 0$ gives $x^2 = 1$,so $x = 1$ or $x = -1$.
The maximum value is $f(1) = \frac{2}{2} = 1$ and the minimum value is $f(-1) = \frac{-2}{2} = -1$.
Thus,the range of $f(x)$ is $[-1, 1]$.
For $k \geq f(x)$ to hold for at least one $x$,$k$ must be greater than or equal to the minimum value of $f(x)$.
Therefore,$k \geq -1$. Combining this with the case $k=0$ and the condition for the quadratic to exist,the set of values is $k \in [-1, \infty) \setminus \{0\}$ is incorrect; rather,checking the boundary,the condition $k \geq -1$ covers all valid $k$ values.

(C) The given equation is $[x^2] = 2x - 1$.
Since $[x^2]$ is an integer, $2x - 1$ must be an integer, which implies $2x$ is an integer.
By definition of the greatest integer function, $x^2 - 1 < [x^2] \le x^2$.
Substituting $[x^2] = 2x - 1$, we get $x^2 - 1 < 2x - 1 \le x^2$.
From $x^2 - 1 < 2x - 1$, we get $x^2 - 2x < 0$, so $x(x - 2) < 0$, which means $0 < x < 2$.
From $2x - 1 \le x^2$, we get $x^2 - 2x + 1 \ge 0$, which means $(x - 1)^2 \ge 0$, which is true for all real $x$.
Case $1$: $0 < x < 1$. Then $0 < x^2 < 1$, so $[x^2] = 0$. The equation becomes $0 - 2x + 1 = 0$, so $x = 1/2$. Since $1/2$ is in $(0, 1)$, it is a solution.
Case $2$: $1 \le x < \sqrt{2}$. Then $1 \le x^2 < 2$, so $[x^2] = 1$. The equation becomes $1 - 2x + 1 = 0$, so $2x = 2$, $x = 1$. Since $1$ is in $[1, \sqrt{2})$, it is a solution.
Case $3$: $\sqrt{2} \le x < \sqrt{3}$. Then $2 \le x^2 < 3$, so $[x^2] = 2$. The equation becomes $2 - 2x + 1 = 0$, so $2x = 3$, $x = 3/2$. Since $3/2 = 1.5$ and $\sqrt{2} \approx 1.414$, $1.5$ is in $[\sqrt{2}, \sqrt{3})$, so it is a solution.
Case $4$: $\sqrt{3} \le x < 2$. Then $3 \le x^2 < 4$, so $[x^2] = 3$. The equation becomes $3 - 2x + 1 = 0$, so $2x = 4$, $x = 2$. But $x < 2$, so this is not a solution.
The solutions are $1/2, 1, 3/2$. The sum is $1/2 + 1 + 3/2 = 3$.

(C) Let the number be $x$. According to the problem,the difference between the number and its reciprocal is $1$,so $|x - \frac{1}{x}| = 1$.
This gives two equations: $x - \frac{1}{x} = 1$ and $x - \frac{1}{x} = -1$.
For $x - \frac{1}{x} = 1$,we have $x^2 - x - 1 = 0$. Using the quadratic formula,$x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Since $x$ must be positive,$a = \frac{1 + \sqrt{5}}{2}$.
For $x - \frac{1}{x} = -1$,we have $x^2 + x - 1 = 0$. Using the quadratic formula,$x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Since $x$ must be positive,$b = \frac{\sqrt{5} - 1}{2}$.
The sum $a + b = \frac{1 + \sqrt{5}}{2} + \frac{\sqrt{5} - 1}{2} = \frac{1 + \sqrt{5} + \sqrt{5} - 1}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}$.

(C) Let $f(x) = ||x - 2| - |3 - x||$.
By the triangle inequality property of absolute values,$||x - 2| - |3 - x|| \leq |(x - 2) - (3 - x)| = |2x - 5|$.
More simply,the expression $||x - 2| - |3 - x||$ represents the absolute difference between the distances of a point $x$ from $2$ and $3$.
For $x < 2$,$f(x) = |(2 - x) - (3 - x)| = |-1| = 1$.
For $2 \leq x \leq 3$,$f(x) = |(x - 2) - (3 - x)| = |2x - 5|$. As $x$ goes from $2$ to $3$,$2x - 5$ goes from $-1$ to $1$,so $|2x - 5|$ goes from $1$ to $0$ and back to $1$.
For $x > 3$,$f(x) = |(x - 2) - (x - 3)| = |1| = 1$.
Thus,the range of $f(x)$ is $[0, 1]$.
For the equation $f(x) = 2 - a$ to have a solution,we must have $0 \leq 2 - a \leq 1$.
Solving for $a$: $-2 \leq -a \leq -1$,which implies $1 \leq a \leq 2$.
The integral values of $a$ are $1$ and $2$.
The sum of these values is $1 + 2 = 3$.

(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 $S_2 = 5$ and $3S_5 = 11S_3$.
Using the identity $\alpha^5 + \beta^5 = (\alpha^2 + \beta^2)(\alpha^3 + \beta^3) - \alpha^2 \beta^2(\alpha + \beta)$,this is not direct. Instead,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 $p^2 - 2q = 5$.
Also,$S_n = p S_{n-1} - q S_{n-2}$.
For $n=3$: $S_3 = p S_2 - q S_1 = 5p - qp = p(5-q)$.
For $n=5$: $S_5 = p S_4 - q S_3$.
$S_4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2 \beta^2 = 25 - 2q^2$.
So,$S_5 = p(25 - 2q^2) - q(p(5-q)) = p(25 - 2q^2 - 5q + q^2) = p(25 - 5q - q^2)$.
Substitute into $3S_5 = 11S_3$:
$3p(25 - 5q - q^2) = 11p(5 - q)$.
Assuming $p \neq 0$,$3(25 - 5q - q^2) = 11(5 - q) \Rightarrow 75 - 15q - 3q^2 = 55 - 11q$.
$3q^2 + 4q - 20 = 0 \Rightarrow (3q + 10)(q - 2) = 0$.
Since $\alpha, \beta$ are real,$p^2 = 5 + 2q \ge 0$. If $q=2$,$p^2 = 9$ (real). If $q = -10/3$,$p^2 = 5 - 20/3 = -5/3$ (not real).
Thus,$\alpha \beta = 2$.

(A) Given the equation $x^3 - x - 1 = 0$ with roots $\alpha, \beta, \gamma$.
From the properties of roots,the sum of roots is $\alpha + \beta + \gamma = 0$.
Therefore,we can write: $\beta + \gamma = -\alpha$,$\gamma + \alpha = -\beta$,and $\alpha + \beta = -\gamma$.
The roots of the required equation are $\frac{1}{-\alpha}, \frac{1}{-\beta}, \frac{1}{-\gamma}$,which simplifies to $-\frac{1}{\alpha}, -\frac{1}{\beta}, -\frac{1}{\gamma}$.
To find the equation with roots $-\frac{1}{x}$,we substitute $x$ with $-x$ in the original equation to get $(-x)^3 - (-x) - 1 = 0$,which is $-x^3 + x - 1 = 0$,or $x^3 - x + 1 = 0$.
Wait,let's re-evaluate: If the roots are $y = -\frac{1}{x}$,then $x = -\frac{1}{y}$.
Substituting $x = -\frac{1}{y}$ into $x^3 - x - 1 = 0$:
$(-\frac{1}{y})^3 - (-\frac{1}{y}) - 1 = 0$
$-\frac{1}{y^3} + \frac{1}{y} - 1 = 0$
Multiplying by $-y^3$: $1 - y^2 + y^3 = 0$,which is $y^3 - y^2 + 1 = 0$.

(B) Given that $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4 + x^2 + 1 = 0$.
Let $y = x^2$,which implies $x = \pm \sqrt{y}$.
Substituting $x^2 = y$ into the original equation $x^4 + x^2 + 1 = 0$,we get $(x^2)^2 + x^2 + 1 = 0$.
Replacing $x^2$ with $y$,we obtain $y^2 + y + 1 = 0$.
Since the roots of the new equation are $\alpha^2, \beta^2, \gamma^2, \delta^2$,and each root of the original equation satisfies $x^4 + x^2 + 1 = 0$,the transformation $y = x^2$ maps the four roots to two values of $y$ that satisfy $y^2 + y + 1 = 0$.
Since each root is squared,the resulting equation must account for the multiplicity of the roots. The equation $y^2 + y + 1 = 0$ has two roots,say $y_1$ and $y_2$. Since the original equation is of degree $4$,the new equation must also be of degree $4$.
Thus,the required equation is $(y^2 + y + 1)^2 = 0$,which in terms of $x$ is $(x^2 + x + 1)^2 = 0$.

(D) For $x^2 - (a - 1)x + 3 = 0$ to have both roots positive:
$1$. Discriminant $D_1 \ge 0 \Rightarrow (a - 1)^2 - 12 \ge 0 \Rightarrow (a - 1)^2 \ge 12 \Rightarrow a - 1 \ge 2\sqrt{3}$ or $a - 1 \le -2\sqrt{3}$. Since $2\sqrt{3} \approx 3.46$,$a \ge 4.46$ or $a \le -2.46$.
$2$. Sum of roots $> 0 \Rightarrow a - 1 > 0 \Rightarrow a > 1$.
$3$. Product of roots $> 0 \Rightarrow 3 > 0$ (always true).
Combining these,$a \ge 4.46$.
For $x^2 + 3x + 6 - a = 0$ to have both roots negative:
$1$. Discriminant $D_2 \ge 0 \Rightarrow 3^2 - 4(6 - a) \ge 0 \Rightarrow 9 - 24 + 4a \ge 0 \Rightarrow 4a \ge 15 \Rightarrow a \ge 3.75$.
$2$. Sum of roots $< 0 \Rightarrow -3 < 0$ (always true).
$3$. Product of roots $> 0 \Rightarrow 6 - a > 0 \Rightarrow a < 6$.
Combining these,$3.75 \le a < 6$.
Intersection of both conditions: $a \ge 4.46$ and $3.75 \le a < 6 \Rightarrow 4.46 \le a < 6$.
The integral values of $a$ are $5$. Thus,there is only $1$ such value.

(C) Let $f(x) = x^2 - 3x + a$.
For the roots $\alpha$ and $\beta$ to satisfy the condition $\alpha < 1 < \beta$,the value of the quadratic function at $x = 1$ must be negative,i.e.,$f(1) < 0$.
Substituting $x = 1$ into the equation:
$f(1) = (1)^2 - 3(1) + a < 0$
$1 - 3 + a < 0$
$-2 + a < 0$
$a < 2$.
Since $a < 2$ automatically satisfies the condition for real roots $(D > 0)$,where $D = (-3)^2 - 4(1)(a) = 9 - 4a > 0 \Rightarrow a < 9/4$,the intersection of $a < 2$ and $a < 9/4$ is $a < 2$.
Thus,$a \in (-\infty, 2)$.

(A) $1$. The parabola opens downwards,which implies that the coefficient of $x^2$ is negative. Therefore,$a < 0$.
$2$. The $y$-intercept of the graph is where $x = 0$. Substituting $x = 0$ into the equation $y = ax^2 - bx + c$,we get $y = c$. From the graph,the $y$-intercept is below the $x$-axis,so $c < 0$.
$3$. The $x$-coordinate of the vertex of the parabola $y = ax^2 + Bx + C$ is given by $-B / (2A)$. In our equation $y = ax^2 - bx + c$,the coefficient of $x$ is $-b$. Thus,the $x$-coordinate of the vertex is $-(-b) / (2a) = b / (2a)$.
$4$. From the graph,the vertex lies to the right of the $y$-axis,meaning the $x$-coordinate of the vertex is positive. Therefore,$b / (2a) > 0$.
$5$. Since we already know $a < 0$,for the fraction $b / (2a)$ to be positive,$b$ must also be negative. Thus,$b < 0$.
$6$. Combining these,we have $a < 0, b < 0, c < 0$.

(D) Let the roots of the equations be:
$x^2 + ax + bc = 0 \rightarrow \alpha, \beta$ $(1)$
$x^2 + bx + ac = 0 \rightarrow \alpha, \gamma$ $(2)$
Subtracting $(2)$ from $(1)$:
$(a - b)x + (bc - ac) = 0$
$(a - b)x - c(a - b) = 0$
Since $a \ne b$,we divide by $(a - b)$ to get $x = c$.
Thus,the common root is $\alpha = c$.
Substituting $x = c$ into $(1)$:
$c^2 + ac + bc = 0 \Rightarrow c(c + a + b) = 0$.
Since $c \ne 0$,we have $a + b + c = 0$.
From the product of roots in $(1)$,$\alpha \cdot \beta = bc \Rightarrow c \cdot \beta = bc \Rightarrow \beta = b$.
From the product of roots in $(2)$,$\alpha \cdot \gamma = ac \Rightarrow c \cdot \gamma = ac \Rightarrow \gamma = a$.
The equation with roots $\beta$ and $\gamma$ is $x^2 - (\beta + \gamma)x + \beta \gamma = 0$.
$x^2 - (b + a)x + ab = 0$.
Since $a + b = -c$,the equation becomes $x^2 - (-c)x + ab = 0$,which is $x^2 + cx + ab = 0$.
Both statements are true,and Statement $-2$ explains Statement $-1$.

(B) The incorrect equation is $x^2 + 17x + q = 0$.
Given that the roots of this incorrect equation are $-2$ and $-15$,we can find $q$ using the product of roots formula:
Product of roots $= (-2) \times (-15) = 30$.
Since the product of roots for $x^2 + px + q = 0$ is $q$,we have $q = 30$.
The original equation had $13$ as the coefficient of $x$,so the original equation is $x^2 + 13x + 30 = 0$.
To find the roots,we factorize: $x^2 + 10x + 3x + 30 = 0$.
$x(x + 10) + 3(x + 10) = 0$.
$(x + 3)(x + 10) = 0$.
Thus,the roots are $x = -3$ and $x = -10$.

(B) For the quadratic equation $ax^2 + bx + 1 = 0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = b^2 - 4ac \geq 0$
Given $c = 1$,the condition becomes $b^2 - 4a \geq 0$,or $b^2 \geq 4a$.
We test the values for $b \in \{1, 2, 3, 4\}$:
$1$. If $b = 1$,$b^2 = 1$. $1 \geq 4a \Rightarrow a \leq 0.25$. No value of $a \in \{1, 2, 3, 4\}$ satisfies this.
$2$. If $b = 2$,$b^2 = 4$. $4 \geq 4a \Rightarrow a \leq 1$. Thus,$a = 1$ ($1$ solution).
$3$. If $b = 3$,$b^2 = 9$. $9 \geq 4a \Rightarrow a \leq 2.25$. Thus,$a \in \{1, 2\}$ ($2$ solutions).
$4$. If $b = 4$,$b^2 = 16$. $16 \geq 4a \Rightarrow a \leq 4$. Thus,$a \in \{1, 2, 3, 4\}$ ($4$ solutions).
Total number of equations = $0 + 1 + 2 + 4 = 7$.

(A) Let $y = 2\alpha + 1$,which implies $\alpha = \frac{y - 1}{2}$.
Since $\alpha$ is a root of $x^3 + 2x - 5 = 0$,we substitute $\alpha$ into the equation:
$\left(\frac{y - 1}{2}\right)^3 + 2\left(\frac{y - 1}{2}\right) - 5 = 0$
$\frac{y^3 - 3y^2 + 3y - 1}{8} + (y - 1) - 5 = 0$
Multiply the entire equation by $8$:
$y^3 - 3y^2 + 3y - 1 + 8(y - 6) = 0$
$y^3 - 3y^2 + 3y - 1 + 8y - 48 = 0$
$y^3 - 3y^2 + 11y - 49 = 0$
Comparing this with $x^3 + bx^2 + cx + d = 0$,we get $b = -3, c = 11, d = -49$.
Thus,$|b + c + d| = |-3 + 11 - 49| = |-41| = 41$.

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

(C) Let $P(z) = z^4 + z^3 + z^2 + z + 1 = 0$. The roots are $z_1, z_2, z_3, z_4$.
We can write $P(z) = (z - z_1)(z - z_2)(z - z_3)(z - z_4)$.
We want to find the value of $\prod_{i=1}^{4} (z_i + 2)$.
Note that $\prod_{i=1}^{4} (z_i + 2) = (-1)^4 \prod_{i=1}^{4} (-2 - z_i) = P(-2)$.
Substitute $z = -2$ into the polynomial $P(z)$:
$P(-2) = (-2)^4 + (-2)^3 + (-2)^2 + (-2) + 1$
$P(-2) = 16 - 8 + 4 - 2 + 1$
$P(-2) = 11$.
Therefore,the product is $11$.

(B) Given that the cubic equation has real coefficients and one root is $\alpha = 3 + 4i$.
Since complex roots occur in conjugate pairs for equations with real coefficients,the second root must be $\beta = 3 - 4i$.
Let the cubic equation be $x^3 + ax^2 + bx + c = 0$. Since the coefficient of $x^2$ is zero,we have $a = 0$.
According to Vieta's formulas,the sum of the roots is $\alpha + \beta + \gamma = -a = 0$.
Substituting the known roots: $(3 + 4i) + (3 - 4i) + \gamma = 0$.
$6 + \gamma = 0$,which gives $\gamma = -6$.
The product of the roots is $\alpha \beta \gamma = (3 + 4i)(3 - 4i)(-6)$.
Calculating the product of the conjugate pair: $(3 + 4i)(3 - 4i) = 3^2 - (4i)^2 = 9 + 16 = 25$.
Therefore,$\alpha \beta \gamma = 25 \times (-6) = -150$.

(B) Let $f(x) = x^2 + (a - 1)x + 2a$. For exactly one root to lie in $(0, 3)$,we consider the following cases:
Case $1$: $f(0) \cdot f(3) < 0$
$f(0) = 2a$
$f(3) = 9 + 3(a - 1) + 2a = 9 + 3a - 3 + 2a = 5a + 6$
So,$2a(5a + 6) < 0 \Rightarrow a(5a + 6) < 0 \Rightarrow a \in (-1.2, 0)$.
Case $2$: One root is at the boundary.
If $x = 0$ is a root,$f(0) = 0 \Rightarrow 2a = 0 \Rightarrow a = 0$. The equation becomes $x^2 - x = 0$,roots are $0, 1$. Since $1 \in (0, 3)$,$a = 0$ is a solution.
If $x = 3$ is a root,$f(3) = 0 \Rightarrow 5a + 6 = 0 \Rightarrow a = -1.2$. The equation becomes $x^2 - 2.2x - 2.4 = 0$. Roots are $3$ and $-0.8$. Since only one root $3$ is not in $(0, 3)$ (it is on the boundary),we check if the other root is in $(0, 3)$. $-0.8$ is not in $(0, 3)$. Thus $a = -1.2$ is not included.
Combining these,$a \in (-1.2, 0]$.
Wait,re-evaluating the original equation $x^2 + (a-1)x + 2a = 0$:
$f(0) = 2a$,$f(3) = 9 + 3a - 3 + 2a = 5a + 6$.
$f(0)f(3) < 0 \Rightarrow 2a(5a+6) < 0 \Rightarrow a \in (-1.2, 0)$.
Checking $a=0$: $x^2-x=0 \Rightarrow x=0, 1$. Root $1 \in (0,3)$. Correct.
Checking $a=-1.2$: $x^2-2.2x-2.4=0 \Rightarrow (x-3)(x+0.8)=0$. Roots $3, -0.8$. Neither is in $(0,3)$.
Thus,the set is $(-1.2, 0]$.
Given the options provided in the prompt,there is a discrepancy. Based on the standard interpretation of such problems,the correct set is $(-1.2, 0]$.

(C) Let $f(x) = 3ax^2 + 4bx + c$.
Since the equation $f(x) = 0$ has no real roots and $c > 0$ (which is $f(0) > 0$),the parabola $y = f(x)$ must lie entirely above the $x$-axis.
This implies that $f(x) > 0$ for all real values of $x$.
Specifically,for $x = -1$,we have $f(-1) > 0$.
Substituting $x = -1$ into the equation,we get $f(-1) = 3a(-1)^2 + 4b(-1) + c = 3a - 4b + c$.
Since $f(-1) > 0$,it follows that $3a - 4b + c > 0$.
Rearranging the terms,we get $3a + c > 4b$.