QUADRATIC EQUATION Questions in English

(D) Given equation: $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$
Comparing with the standard quadratic form $Ax^2 + Bx + C = 0$,we have:
$A = 2(a^2 + b^2)$,$B = 2(a + b)$,and $C = 1$
To determine the nature of the roots,we calculate the discriminant $D = B^2 - 4AC$:
$D = [2(a + b)]^2 - 4 \cdot 2(a^2 + b^2) \cdot 1$
$D = 4(a^2 + b^2 + 2ab) - 8(a^2 + b^2)$
$D = 4a^2 + 4b^2 + 8ab - 8a^2 - 8b^2$
$D = -4a^2 - 4b^2 + 8ab$
$D = -4(a^2 + b^2 - 2ab)$
$D = -4(a - b)^2$
Since $(a - b)^2 \ge 0$,it follows that $-4(a - b)^2 \le 0$. Therefore,$D < 0$ (assuming $a \neq b$).
Since the discriminant is negative,the roots of the equation are imaginary.

(B) For the equation $px^2 + 2qx + r = 0$ to have real roots,the discriminant $D_1 \ge 0$.
$D_1 = (2q)^2 - 4(p)(r) = 4q^2 - 4pr \ge 0 \implies q^2 \ge pr$ ... $(i)$
For the equation $qx^2 - 2\sqrt{pr}x + q = 0$ to have real roots,the discriminant $D_2 \ge 0$.
$D_2 = (-2\sqrt{pr})^2 - 4(q)(q) = 4pr - 4q^2 \ge 0 \implies pr \ge q^2$ ... $(ii)$
From $(i)$ and $(ii)$,we have $q^2 \ge pr$ and $pr \ge q^2$,which implies $q^2 = pr$.

(D) For the equation $ax^2 + x + b = 0$ to have real roots,the discriminant $D_1$ must be greater than or equal to zero.
$D_1 = (1)^2 - 4ab \ge 0 \implies 1 - 4ab \ge 0 \implies 4ab \le 1$.
Now,consider the second equation $x^2 - 4\sqrt{ab}x + 1 = 0$.
The discriminant $D_2$ of this equation is given by $D_2 = (-4\sqrt{ab})^2 - 4(1)(1)$.
$D_2 = 16ab - 4 = 4(4ab - 1)$.
Since $4ab \le 1$,it follows that $4ab - 1 \le 0$.
Therefore,$D_2 \le 0$.
Since the discriminant is less than or equal to zero,the roots of the equation are imaginary (or real and equal if $D_2 = 0$,but generally classified as imaginary in this context).

(B) Let $\alpha$ be the common root of the equations $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$.
Then,$\alpha^2 + a\alpha + b = 0$ and $\alpha^2 + b\alpha + a = 0$.
Subtracting the two equations: $(\alpha^2 + a\alpha + b) - (\alpha^2 + b\alpha + a) = 0$.
$\alpha(a - b) - (a - b) = 0$.
$(a - b)(\alpha - 1) = 0$.
Since the roots are coincident,we assume $a \neq b$,so $\alpha - 1 = 0$,which gives $\alpha = 1$.
Substituting $\alpha = 1$ into the first equation: $1^2 + a(1) + b = 0$.
$1 + a + b = 0$.
Therefore,$a + b = -1$.

(D) For the given equation to be meaningful,we must have $x > 0$. Taking the logarithm base $2$ on both sides:
$((\frac{3}{4})(\log_2 x)^2 + (\log_2 x) - \frac{5}{4}) \cdot \log_2 x = \log_2(\sqrt{2})$
Let $t = \log_2 x$. Then the equation becomes:
$(\frac{3}{4}t^2 + t - \frac{5}{4}) \cdot t = \frac{1}{2}$
Multiplying by $4$:
$(3t^2 + 4t - 5)t = 2$
$3t^3 + 4t^2 - 5t - 2 = 0$
By testing roots,we find $t = 1$ is a root. Dividing by $(t - 1)$:
$(t - 1)(3t^2 + 7t + 2) = 0$
$(t - 1)(3t + 1)(t + 2) = 0$
So,$t = 1, -2, -1/3$.
Since $t = \log_2 x$,we have $x = 2^1 = 2$,$x = 2^{-2} = 1/4$,and $x = 2^{-1/3} = 1/\sqrt[3]{2}$.
All three solutions are real,and $1/\sqrt[3]{2}$ is an irrational number. Thus,all statements $(A)$,$(B)$,and $(C)$ are correct.

(B) The roots of the quadratic equation $ax^2 + bx + c = 0$ are given by the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Case $(i)$: If the discriminant $D = b^2 - 4ac \ge 0$,the roots are real.
Since $a, b, c > 0$,we have $\sqrt{b^2 - 4ac} < \sqrt{b^2} = b$. Thus,$-b + \sqrt{b^2 - 4ac} < 0$ and $-b - \sqrt{b^2 - 4ac} < 0$. Since $2a > 0$,both roots are negative.
Case $(ii)$: If $D = b^2 - 4ac < 0$,the roots are complex conjugates:
$x = \frac{-b}{2a} \pm i \frac{\sqrt{4ac - b^2}}{2a}$
Here,the real part is $-\frac{b}{2a}$. Since $a > 0$ and $b > 0$,the real part $-\frac{b}{2a}$ is negative.
In both cases,the roots have negative real parts.

(D) The given quadratic equation is $2x^2 - (k - 1)x + 8 = 0$.
For the equation to have equal and real roots,the discriminant $D$ must be equal to $0$,where $D = b^2 - 4ac$.
Here,$a = 2$,$b = -(k - 1)$,and $c = 8$.
Substituting these values into the discriminant formula:
$(-(k - 1))^2 - 4(2)(8) = 0$
$(k - 1)^2 - 64 = 0$
$(k - 1)^2 = 64$
Taking the square root on both sides:
$k - 1 = \pm 8$
Case $1$: $k - 1 = 8 \Rightarrow k = 9$
Case $2$: $k - 1 = -8 \Rightarrow k = -7$
Therefore,the values of $k$ are $9$ and $-7$.

(B) Given the quadratic equation: $2x^2 + 3x + 1 = 0$.
Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = 2$,$b = 3$,and $c = 1$.
The discriminant $D$ is given by $D = b^2 - 4ac$.
$D = (3)^2 - 4(2)(1) = 9 - 8 = 1$.
Since $D > 0$ and $D$ is a perfect square,the roots are real and rational.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{-3 \pm \sqrt{1}}{2(2)} = \frac{-3 \pm 1}{4}$.
Thus,the roots are $x_1 = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2}$ and $x_2 = \frac{-3 - 1}{4} = \frac{-4}{4} = -1$.
Both roots,$-\frac{1}{2}$ and $-1$,are rational numbers.

(B) The given quadratic equation is $(l - m)x^2 - 5(l + m)x - 2(l - m) = 0$.
The discriminant $D$ of a quadratic equation $ax^2 + bx + c = 0$ is given by $D = b^2 - 4ac$.
Here,$a = (l - m)$,$b = -5(l + m)$,and $c = -2(l - m)$.
Substituting these values into the formula for $D$:
$D = [-5(l + m)]^2 - 4(l - m)[-2(l - m)]$
$D = 25(l + m)^2 + 8(l - m)^2$.
Since $l$ and $m$ are real and $l \ne m$,$(l - m)^2 > 0$. Also,$(l + m)^2 \ge 0$.
Therefore,$D = 25(l + m)^2 + 8(l - m)^2 > 0$.
Since the discriminant $D$ is strictly greater than zero,the roots of the equation are real and distinct.

(C) For the quadratic equation ${x^2} - 8x + ({a^2} - 6a) = 0$ to have real roots,the discriminant $D$ must be greater than or equal to zero $(D \ge 0)$.
The discriminant $D = b^2 - 4ac$,where $a = 1$,$b = -8$,and $c = ({a^2} - 6a)$.
Substituting these values: $D = (-8)^2 - 4(1)({a^2} - 6a) \ge 0$.
$64 - 4({a^2} - 6a) \ge 0$.
Dividing by $4$: $16 - ({a^2} - 6a) \ge 0$.
$16 - {a^2} + 6a \ge 0$.
Multiplying by $-1$ (reversing the inequality sign): ${a^2} - 6a - 16 \le 0$.
Factoring the quadratic expression: $(a - 8)(a + 2) \le 0$.
For the product of two factors to be less than or equal to zero,the value of $a$ must lie between the roots of the equation $(a - 8)(a + 2) = 0$,which are $a = 8$ and $a = -2$.
Thus,the range of $a$ is $-2 \le a \le 8$.

(C) Given the quadratic equation: ${x^2} + 2\sqrt{3}x + 3 = 0$.
Comparing this with the standard form $ax^2 + bx + c = 0$,we get $a = 1$,$b = 2\sqrt{3}$,and $c = 3$.
The discriminant $D$ is calculated as $D = b^2 - 4ac$.
Substituting the values: $D = (2\sqrt{3})^2 - 4(1)(3) = 12 - 12 = 0$.
Since the discriminant $D = 0$,the roots are real and equal.
The roots are given by $x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2\sqrt{3} \pm 0}{2(1)} = -\sqrt{3}$.
Since $-\sqrt{3}$ is an irrational number,the roots are irrational and equal.

(C) Given the quadratic equation $(\cos p - 1)x^2 + (\cos p)x + \sin p = 0$.
For the roots to be real,the discriminant $D$ must be greater than or equal to zero $(D \ge 0)$.
$D = b^2 - 4ac = (\cos p)^2 - 4(\cos p - 1)(\sin p) \ge 0$.
Expanding this,we get $\cos^2 p - 4\sin p \cos p + 4\sin p \ge 0$.
This can be rewritten as $(\cos p - 2\sin p)^2 - 4\sin^2 p + 4\sin p \ge 0$.
$(\cos p - 2\sin p)^2 + 4\sin p(1 - \sin p) \ge 0$.
Since $(\cos p - 2\sin p)^2 \ge 0$ for all real $p$,the inequality holds if $4\sin p(1 - \sin p) \ge 0$.
Since $(1 - \sin p) \ge 0$ for all real $p$,we require $\sin p \ge 0$.
In the interval $(0, 2\pi)$,$\sin p \ge 0$ holds for $p \in (0, \pi)$.

(C) The given expression is $f(x, y) = x^2 + 2x(1 + y) + (my - 3)$.
For the expression to have rational factors,its discriminant $D$ with respect to $x$ must be a perfect square.
The discriminant $D$ is given by $D = [2(1 + y)]^2 - 4(1)(my - 3)$.
$D = 4(1 + y)^2 - 4(my - 3) = 4(1 + y^2 + 2y - my + 3) = 4(y^2 + (2 - m)y + 4)$.
For $D$ to be a perfect square,the quadratic expression $y^2 + (2 - m)y + 4$ must be a perfect square of the form $(y \pm 2)^2$.
Expanding $(y \pm 2)^2$,we get $y^2 \pm 4y + 4$.
Comparing $y^2 + (2 - m)y + 4$ with $y^2 \pm 4y + 4$,we have $2 - m = \pm 4$.
Case $1$: $2 - m = 4 \Rightarrow m = -2$.
Case $2$: $2 - m = -4 \Rightarrow m = 6$.
Thus,the values of $m$ are $6$ and $-2$.

(A) The given quadratic equation is $2ax^2 + (2a + b)x + b = 0$,where $a \ne 0$.
The discriminant $D$ of a quadratic equation $Ax^2 + Bx + C = 0$ is given by $D = B^2 - 4AC$.
Here,$A = 2a$,$B = (2a + b)$,and $C = b$.
Substituting these values into the formula for $D$:
$D = (2a + b)^2 - 4(2a)(b)$
$D = 4a^2 + 4ab + b^2 - 8ab$
$D = 4a^2 - 4ab + b^2$
$D = (2a - b)^2$
Since $a$ and $b$ are integers,$(2a - b)^2$ is a perfect square of an integer.
For a quadratic equation with rational coefficients,if the discriminant $D$ is a perfect square,the roots are always rational.
Therefore,the roots of the given equation are rational.

(B) For a quadratic equation of the form $Ax^2 + Bx + C = 0$,the roots are real and distinct if the discriminant $D = B^2 - 4AC > 0$.
In the given equation $ax^2 + 0x + b = 0$,we have $A = a$,$B = 0$,and $C = b$.
Substituting these values into the discriminant formula:
$D = (0)^2 - 4(a)(b) > 0$
$-4ab > 0$
Dividing both sides by $-4$ (which reverses the inequality sign):
$ab < 0$
Therefore,the roots are real and distinct if $ab < 0$.

(B) Let the roots of the first equation $2x^2 - 5x + 1 = 0$ be $\alpha$ and $\beta$.
Then,$\alpha \beta = \frac{c}{a} = \frac{1}{2}$.
Let the roots of the second equation $x^2 + 5x + 2 = 0$ be $\gamma$ and $\delta$.
Then,$\gamma \delta = \frac{2}{1} = 2$.
Comparing the two equations,we observe that the second equation is of the form $cx^2 + bx + a = 0$ where $a=2, b=-5, c=1$ is transformed into $ax^2 - bx + c = 0$ or similar forms.
Specifically,if the roots of $ax^2 + bx + c = 0$ are $\alpha, \beta$,then the roots of $cx^2 + bx + a = 0$ are $\frac{1}{\alpha}, \frac{1}{\beta}$.
Here,the first equation is $2x^2 - 5x + 1 = 0$ and the second is $x^2 + 5x + 2 = 0$. Note that the coefficients are swapped and the sign of the middle term is changed.
If we replace $x$ with $-1/x$ in $2x^2 - 5x + 1 = 0$,we get $2(-1/x)^2 - 5(-1/x) + 1 = 0$,which simplifies to $\frac{2}{x^2} + \frac{5}{x} + 1 = 0$,or $x^2 + 5x + 2 = 0$.
Thus,the roots of the second equation are the negative reciprocals of the roots of the first equation. Therefore,they are reciprocal and of opposite sign.

(A) Given the quadratic equation $ax^2 + bx + c = 0$ where $a, b, c \in Q$ and $a + b + c = 0$.
Since $a + b + c = 0$,we have $b = -(a + c)$.
The discriminant $D$ of the quadratic equation is given by $D = b^2 - 4ac$.
Substituting $b = -(a + c)$,we get $D = (-(a + c))^2 - 4ac = (a + c)^2 - 4ac$.
Expanding this,$D = a^2 + 2ac + c^2 - 4ac = a^2 - 2ac + c^2 = (a - c)^2$.
Since $a, c \in Q$,$(a - c)^2$ is a perfect square of a rational number,which is always $\ge 0$.
Since the discriminant $D$ is a perfect square of a rational number,the roots are given by $x = \frac{-b \pm \sqrt{D}}{2a} = \frac{a + c \pm (a - c)}{2a}$.
This results in $x_1 = \frac{2a}{2a} = 1$ and $x_2 = \frac{2c}{2a} = \frac{c}{a}$.
Since $a, c \in Q$ and $a \neq 0$,both roots are rational numbers.

(A) Given the quadratic equation $(b + c - 2a)x^2 + (c + a - 2b)x + (a + b - 2c) = 0$.
Let the coefficients be $A = (b + c - 2a)$,$B = (c + a - 2b)$,and $C = (a + b - 2c)$.
Sum of the coefficients is $A + B + C = (b + c - 2a) + (c + a - 2b) + (a + b - 2c) = (a + a - 2a) + (b + b - 2b) + (c + c - 2c) = 0$.
Since the sum of the coefficients is $0$,one root of the quadratic equation must be $x = 1$.
Since $a, b, c \in \mathbb{Q}$,the coefficients $A, B, C$ are rational numbers.
For a quadratic equation with rational coefficients,if one root is rational,the other root must also be rational (as the product of roots $C/A$ is rational).
Therefore,the roots are rational.

(D) The given expression is $f(x) = x^2 + 2bx + c$.
By completing the square,we can write:
$f(x) = (x^2 + 2bx + b^2) - b^2 + c$
$f(x) = (x + b)^2 + (c - b^2)$
Since $(x + b)^2 \ge 0$ for all real $x$,the expression $f(x)$ will always be positive if the constant term $(c - b^2)$ is strictly greater than $0$.
Therefore,$c - b^2 > 0$,which implies $c > b^2$ or $b^2 < c$.

(B) For a quadratic equation of the form $ax^2 + bx + c = 0$,the roots are equal if the discriminant $D = b^2 - 4ac = 0$.
Here,$a = 4$,$b = p$,and $c = 9$.
Substituting these values into the discriminant formula:
$p^2 - 4(4)(9) = 0$
$p^2 - 144 = 0$
$p^2 = 144$
Taking the square root of both sides,we get $p = \pm 12$.
The absolute value of $p$ is $|p| = |\pm 12| = 12$.

(A) For a quadratic equation $Ax^2 + Bx + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be equal to $0$.
Here,$A = (c^2 - ab)$,$B = -2(a^2 - bc)$,and $C = (b^2 - ac)$.
Setting $D = 0$:
$[-2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0$
$4(a^2 - bc)^2 - 4(c^2 - ab)(b^2 - ac) = 0$
Dividing by $4$:
$(a^4 - 2a^2bc + b^2c^2) - (b^2c^2 - ac^3 - ab^3 + a^2bc) = 0$
$a^4 - 2a^2bc + b^2c^2 - b^2c^2 + ac^3 + ab^3 - a^2bc = 0$
$a^4 - 3a^2bc + ac^3 + ab^3 = 0$
Factoring out $a$:
$a(a^3 + b^3 + c^3 - 3abc) = 0$
Thus,the condition is $a = 0$ or $a^3 + b^3 + c^3 = 3abc$.

(A) Let ${D_1}$ and ${D_2}$ be the discriminants of the equations ${x^2} + {b_1}x + {c_1} = 0$ and ${x^2} + {b_2}x + {c_2} = 0$ respectively.
The discriminants are given by ${D_1} = b_1^2 - 4{c_1}$ and ${D_2} = b_2^2 - 4{c_2}$.
Adding these,we get ${D_1} + {D_2} = b_1^2 + b_2^2 - 4({c_1} + {c_2})$.
Given that ${b_1}{b_2} = 2({c_1} + {c_2})$,we can substitute $4({c_1} + {c_2}) = 2{b_1}{b_2}$ into the equation:
${D_1} + {D_2} = b_1^2 + b_2^2 - 2{b_1}{b_2} = {(b_1 - b_2)^2}$.
Since the square of any real number is non-negative,we have ${D_1} + {D_2} \ge 0$.
If the sum of two real numbers is non-negative,then at least one of them must be non-negative. Therefore,${D_1} \ge 0$ or ${D_2} \ge 0$.
$A$ quadratic equation has real roots if its discriminant is greater than or equal to zero. Thus,at least one of the equations must have real roots.

(C) The given quadratic equation is $k{x^2} + 1 = kx + 3x - 11{x^2}$.
Rearranging the terms to the standard form $ax^2 + bx + c = 0$,we get:
$(k + 11)x^2 - (k + 3)x + 1 = 0$.
For a quadratic equation to have real and equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Here,$a = (k + 11)$,$b = -(k + 3)$,and $c = 1$.
Substituting these values into $D = 0$:
$(-(k + 3))^2 - 4(k + 11)(1) = 0$
$(k^2 + 6k + 9) - 4k - 44 = 0$
$k^2 + 2k - 35 = 0$
Factoring the quadratic equation:
$k^2 + 7k - 5k - 35 = 0$
$k(k + 7) - 5(k + 7) = 0$
$(k - 5)(k + 7) = 0$
Thus,$k = 5$ or $k = -7$.

(B) Let $f(x) = ax^2 + bx + c$. Then $f(0) = c$. Thus,the graph of $y = f(x)$ meets the $y$-axis at $(0, c)$.
If $c > 0$,then by the hypothesis,$f(x) > 0$. This means that the curve $y = f(x)$ does not meet the $x$-axis.
If $c < 0$,then by the hypothesis,$f(x) < 0$,which means that the curve $y = f(x)$ is always below the $x$-axis and so it does not intersect the $x$-axis.
Thus,in both cases,$y = f(x)$ does not intersect the $x$-axis,i.e.,$f(x) \neq 0$ for any real $x$.
Hence,the equation $ax^2 + bx + c = 0$ has imaginary roots,which implies that the discriminant $D = b^2 - 4ac < 0$.
Therefore,$b^2 < 4ac$ or $4ac > b^2$.

(B) Given equation: $\frac{a}{x + a + m} + \frac{b}{x + b + m} = 1$.
Multiplying both sides by $(x + a + m)(x + b + m)$,we get:
$a(x + b + m) + b(x + a + m) = (x + a + m)(x + b + m)$.
$ax + ab + am + bx + ab + bm = x^2 + bx + mx + ax + ab + am + mx + bm + m^2$.
Simplifying both sides:
$ax + bx + 2ab + am + bm = x^2 + ax + bx + 2mx + ab + am + bm + m^2$.
Canceling common terms $(ax + bx + am + bm)$ from both sides:
$2ab = x^2 + 2mx + ab + m^2$.
Rearranging into standard quadratic form $x^2 + 2mx + m^2 - ab = 0$.
For roots to be equal in magnitude but opposite in sign,the sum of the roots must be $0$.
Since the sum of roots is $-\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{2m}{1} = -2m$.
Setting $-2m = 0$,we get $m = 0$.

(D) For a quadratic equation $At^2 + Bt + C = 0$,the roots are equal if the discriminant $D = B^2 - 4AC = 0$.
Here,$A = (a^2 + b^2)$,$B = -2(ac + bd)$,and $C = (c^2 + d^2)$.
Setting $B^2 - 4AC = 0$,we get:
$[-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0$
$4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = 0$
$(ac + bd)^2 - (a^2 + b^2)(c^2 + d^2) = 0$
$a^2c^2 + b^2d^2 + 2abcd - (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0$
$a^2c^2 + b^2d^2 + 2abcd - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2 = 0$
$2abcd - a^2d^2 - b^2c^2 = 0$
$-(a^2d^2 + b^2c^2 - 2abcd) = 0$
$(ad - bc)^2 = 0$
$ad - bc = 0 \Rightarrow ad = bc$
Dividing both sides by $bd$,we get $\frac{a}{b} = \frac{c}{d}$.

(B) For a quadratic equation $ax^2 + bx + c = 0$ to have equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Here,$a = 1$,$b = -2(1 + 3k)$,and $c = 7(3 + 2k)$.
Setting $D = 0$:
$[-2(1 + 3k)]^2 - 4(1)(7(3 + 2k)) = 0$
$4(1 + 3k)^2 - 28(3 + 2k) = 0$
Divide by $4$:
$(1 + 3k)^2 - 7(3 + 2k) = 0$
$1 + 6k + 9k^2 - 21 - 14k = 0$
$9k^2 - 8k - 20 = 0$
Factor the quadratic equation:
$9k^2 - 18k + 10k - 20 = 0$
$9k(k - 2) + 10(k - 2) = 0$
$(9k + 10)(k - 2) = 0$
Thus,$k = 2$ or $k = -\frac{10}{9}$.

(B) The given equation is $x^2 - 8x + (a^2 - 6a) = 0$.
For the roots to be real,the discriminant $D$ must be greater than or equal to zero $(D \ge 0)$.
Here,$A = 1$,$B = -8$,and $C = a^2 - 6a$.
$D = B^2 - 4AC = (-8)^2 - 4(1)(a^2 - 6a) \ge 0$.
$64 - 4a^2 + 24a \ge 0$.
Dividing by $-4$ (and reversing the inequality sign): $a^2 - 6a - 16 \le 0$.
Factoring the quadratic: $(a - 8)(a + 2) \le 0$.
This inequality holds when $a$ lies between the roots $-2$ and $8$.
Therefore,$a \in [-2, 8]$.

(C) For a quadratic equation $px^2 + qx + 1 = 0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = q^2 - 4(p)(1) \ge 0$
$q^2 \ge 4p$
We test the possible values for $p, q \in \{1, 2, 3, 4\}$:
If $p = 1$,then $q^2 \ge 4$,which gives $q \in \{2, 3, 4\}$ ($3$ values).
If $p = 2$,then $q^2 \ge 8$,which gives $q \in \{3, 4\}$ ($2$ values).
If $p = 3$,then $q^2 \ge 12$,which gives $q = 4$ ($1$ value).
If $p = 4$,then $q^2 \ge 16$,which gives $q = 4$ ($1$ value).
Total number of equations = $3 + 2 + 1 + 1 = 7$.

(C) For a quadratic equation $ax^2 + bx + c = 0$ to have equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Here,$a = 1$,$b = -(3k - 1)$,and $c = 2k^2 + 2k - 11$.
Substituting these values into $b^2 - 4ac = 0$:
$(-(3k - 1))^2 - 4(1)(2k^2 + 2k - 11) = 0$
$(3k - 1)^2 - 4(2k^2 + 2k - 11) = 0$
$9k^2 - 6k + 1 - 8k^2 - 8k + 44 = 0$
$k^2 - 14k + 45 = 0$
Factoring the quadratic equation:
$(k - 5)(k - 9) = 0$
Thus,$k = 5$ or $k = 9$.

(C) For a quadratic equation $ax^2 + bx + c = 0$ to have real and distinct roots,the discriminant $D = b^2 - 4ac$ must be greater than $0$.
Here,$a = (k - 2)$,$b = 8$,and $c = (k + 4)$.
$D = 8^2 - 4(k - 2)(k + 4) > 0$
$64 - 4(k^2 + 2k - 8) > 0$
$16 - (k^2 + 2k - 8) > 0$
$-k^2 - 2k + 24 > 0 \Rightarrow k^2 + 2k - 24 < 0$
$(k + 6)(k - 4) < 0$,which implies $-6 < k < 4$.
For roots to be negative,the sum of roots $-b/a = -8/(k - 2) < 0$ and the product of roots $c/a = (k + 4)/(k - 2) > 0$.
From $-8/(k - 2) < 0$,we get $k - 2 > 0 \Rightarrow k > 2$.
From $(k + 4)/(k - 2) > 0$,we get $k > 2$ or $k < -4$.
Combining these conditions with $-6 < k < 4$,we find $2 < k < 4$.
Checking the options,$k = 3$ satisfies $2 < 3 < 4$.

(B) The given quadratic equation is $x^2 + 2kx + 4 = 0$.
For a quadratic equation $ax^2 + bx + c = 0$,the nature of the roots is determined by the discriminant $D = b^2 - 4ac$.
Here,$a = 1$,$b = 2k$,and $c = 4$.
$D = (2k)^2 - 4(1)(4) = 4k^2 - 16$.
For the roots to be real and unequal,we must have $D > 0$.
$4k^2 - 16 > 0 \Rightarrow 4(k^2 - 4) > 0 \Rightarrow k^2 > 4$.
This inequality holds when $k > 2$ or $k < -2$.
Thus,$k \in ( - \infty , - 2) \cup (2, \infty )$.
Therefore,the roots are real and unequal.

(B) For a quadratic equation $ax^2 + bx + c = 0$ to have equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Here,$a = (m - n)$,$b = (n - l)$,and $c = (l - m)$.
Substituting these into the condition $b^2 - 4ac = 0$:
$(n - l)^2 - 4(m - n)(l - m) = 0$
Expanding the terms:
$(n^2 + l^2 - 2nl) - 4(ml - m^2 - nl + mn) = 0$
$n^2 + l^2 - 2nl - 4ml + 4m^2 + 4nl - 4mn = 0$
Rearranging the terms:
$l^2 + n^2 + (2m)^2 + 2nl - 4mn - 4ml = 0$
This expression is the expansion of $(l + n - 2m)^2 = 0$.
Therefore,$l + n - 2m = 0$,which implies $2m = n + l$.
This indicates that $l, m, n$ are in Arithmetic Progression $(A.P.)$.

(D) For the roots of the quadratic equation ${ax^2} + bx + c = 0$ to be imaginary,the discriminant $D$ must be less than $0$.
$D = {b^2} - 4ac < 0$
Here,$a = 1$,$b = 5$,and $c = k$.
Substituting these values into the inequality:
${5^2} - 4(1)(k) < 0$
$25 - 4k < 0$
$25 < 4k$
$k > \frac{25}{4}$
$k > 6.25$
Since $k$ must be an integer,the least integer greater than $6.25$ is $7$.

(C) For a quadratic equation $ax^2 + bx + c = 0$,the roots are equal if the discriminant $D = b^2 - 4ac = 0$.
Given the equation $4x^2 + 6px + 1 = 0$,we have $a = 4$,$b = 6p$,and $c = 1$.
Substituting these values into the discriminant formula:
$(6p)^2 - 4(4)(1) = 0$
$36p^2 - 16 = 0$
$36p^2 = 16$
$p^2 = \frac{16}{36} = \frac{4}{9}$
Taking the square root on both sides,we get $p = \pm \frac{2}{3}$.
Since $\frac{2}{3}$ is provided as an option,the correct value is $\frac{2}{3}$.

(A) The given equation is $(1 + 2k){x^2} + (1 - 2k)x + (1 - 2k) = 0$.
For a quadratic equation $ax^2 + bx + c = 0$ to be a perfect square,its discriminant $D = b^2 - 4ac$ must be equal to $0$.
Here,$a = (1 + 2k)$,$b = (1 - 2k)$,and $c = (1 - 2k)$.
Setting $D = 0$:
$(1 - 2k)^2 - 4(1 + 2k)(1 - 2k) = 0$.
Factor out $(1 - 2k)$:
$(1 - 2k) [(1 - 2k) - 4(1 + 2k)] = 0$.
$(1 - 2k) [1 - 2k - 4 - 8k] = 0$.
$(1 - 2k) (-3 - 10k) = 0$.
This gives two possible values for $k$:
$1 - 2k = 0 \implies k = 1/2$.
$-3 - 10k = 0 \implies k = -3/10$.
Thus,there are $2$ values of $k$ for which the expression is a perfect square.

(A) Given that $\sin A, \sin B, \cos A$ are in $G.P.$,we have $\sin^2 B = \sin A \cos A$.
Multiplying both sides by $2$,we get $2 \sin^2 B = 2 \sin A \cos A$,which implies $1 - \cos 2B = \sin 2A$.
Thus,$\cos 2B = 1 - \sin 2A$. Since the range of $\sin 2A$ is $[-1, 1]$,$1 - \sin 2A \ge 0$,so $\cos 2B \ge 0$.
For the quadratic equation ${x^2} + 2x \cot B + 1 = 0$,the discriminant $D$ is given by $D = b^2 - 4ac = (2 \cot B)^2 - 4(1)(1) = 4 \cot^2 B - 4 = 4(\cot^2 B - 1)$.
Using the identity $\cot^2 B - 1 = \frac{\cos^2 B - \sin^2 B}{\sin^2 B} = \frac{\cos 2B}{\sin^2 B}$,we get $D = 4 \frac{\cos 2B}{\sin^2 B}$.
Since $\cos 2B \ge 0$ and $\sin^2 B > 0$,we have $D \ge 0$.
Therefore,the roots of the quadratic equation are always real.

(D) Let the four real roots be $\alpha, \beta, \gamma, \delta$. The equation is $(x - \alpha)(x - \beta)(x - \gamma)(x - \delta) = 0$.
Comparing this with $x^4 - 4x^3 + ax^2 + bx + 1 = 0$,we get:
$\sum \alpha = 4$,$\sum \alpha\beta = a$,$\sum \alpha\beta\gamma = -b$,and $\alpha\beta\gamma\delta = 1$.
For real positive roots,the Arithmetic Mean $(AM)$ $\ge$ Geometric Mean $(GM)$.
$AM = \frac{\alpha + \beta + \gamma + \delta}{4} = \frac{4}{4} = 1$.
$GM = (\alpha\beta\gamma\delta)^{1/4} = (1)^{1/4} = 1$.
Since $AM = GM = 1$,all roots must be equal,i.e.,$\alpha = \beta = \gamma = \delta = 1$.
Now,$a = \sum \alpha\beta = \binom{4}{2} \times (1 \times 1) = 6 \times 1 = 6$.
$-b = \sum \alpha\beta\gamma = \binom{4}{3} \times (1 \times 1 \times 1) = 4 \times 1 = 4$,so $b = -4$.
Thus,$a = 6$ and $b = -4$.

(B) For a quadratic equation of the form $ax^2 + bx + c = 0$,if one root is the reciprocal of the other,then the product of the roots is equal to $1$.
Let the roots be $\alpha$ and $\frac{1}{\alpha}$.
The product of the roots is given by $\frac{c}{a} = \frac{k}{5}$.
Since $\alpha \cdot \frac{1}{\alpha} = 1$,we have $1 = \frac{k}{5}$.
Therefore,$k = 5$.

(A) Given the quadratic equation $4x^2 + 3x + 7 = 0$.
Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = 4$,$b = 3$,and $c = 7$.
According to the relationship between roots and coefficients:
Sum of roots $\alpha + \beta = -\frac{b}{a} = -\frac{3}{4}$.
Product of roots $\alpha\beta = \frac{c}{a} = \frac{7}{4}$.
We need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
Taking the common denominator,we get $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$.
Substituting the values,$\frac{-3/4}{7/4} = -\frac{3}{4} \times \frac{4}{7} = -\frac{3}{7}$.

(B) For a quadratic equation $Ax^2 + Bx + C = 0$,the product of the roots is given by $\alpha \beta = \frac{C}{A}$ and the sum of the roots is given by $\alpha + \beta = -\frac{B}{A}$.
Given the equation $(a + 1)x^2 + (2a + 3)x + (3a + 4) = 0$,we have $A = a + 1$,$B = 2a + 3$,and $C = 3a + 4$.
Given that the product of the roots is $2$,we have $\frac{3a + 4}{a + 1} = 2$.
Solving for $a$: $3a + 4 = 2(a + 1) \Rightarrow 3a + 4 = 2a + 2 \Rightarrow a = -2$.
Now,the sum of the roots is $\alpha + \beta = -\frac{2a + 3}{a + 1}$.
Substituting $a = -2$ into the expression for the sum of the roots:
Sum $= -\frac{2(-2) + 3}{-2 + 1} = -\frac{-4 + 3}{-1} = -\frac{-1}{-1} = -1$.

(C) Given that $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$.
From the properties of quadratic equations,we have:
$\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Now,consider the equation $cx^2 + bx + a = 0$. Let its roots be $\alpha'$ and $\beta'$.
For this equation,the sum of roots is $\alpha' + \beta' = -\frac{b}{c}$ and the product of roots is $\alpha'\beta' = \frac{a}{c}$.
We can express the sum of the reciprocals of the original roots as:
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c}$.
Also,the product of the reciprocals is:
$\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{c/a} = \frac{a}{c}$.
Since the sum and product of the roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ match the sum and product of the roots of the equation $cx^2 + bx + a = 0$,the roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.

(A) Given that $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Let the new roots be $S_1 = \alpha + \frac{1}{\beta}$ and $S_2 = \beta + \frac{1}{\alpha}$.
Sum of the new roots: $S_1 + S_2 = (\alpha + \beta) + (\frac{1}{\alpha} + \frac{1}{\beta}) = (\alpha + \beta) + \frac{\alpha + \beta}{\alpha\beta} = -\frac{b}{a} + \frac{-b/a}{c/a} = -\frac{b}{a} - \frac{b}{c} = -\frac{b(a+c)}{ac}$.
Product of the new roots: $S_1 \cdot S_2 = (\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}) = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = \alpha\beta + 2 + \frac{1}{\alpha\beta} = \frac{c}{a} + 2 + \frac{a}{c} = \frac{c^2 + 2ac + a^2}{ac} = \frac{(a+c)^2}{ac}$.
The required quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - [-\frac{b(a+c)}{ac}]x + \frac{(a+c)^2}{ac} = 0$.
Multiplying by $ac$,we get $acx^2 + b(a+c)x + (a+c)^2 = 0$.

(A) Let $\alpha$ be a root of the first equation $ax^2 + bx + c = 0$. Then,$\frac{1}{\alpha}$ is a root of the second equation $a'x^2 + b'x + c' = 0$.
For the first equation: $a\alpha^2 + b\alpha + c = 0$.
For the second equation: $a'(\frac{1}{\alpha})^2 + b'(\frac{1}{\alpha}) + c' = 0$,which simplifies to $c'\alpha^2 + b'\alpha + a' = 0$.
Using the method of cross-multiplication for the system of equations:
$\frac{\alpha^2}{ba' - b'c} = \frac{\alpha}{cc' - aa'} = \frac{1}{ab' - bc'}$.
From the second and third terms: $\alpha = \frac{cc' - aa'}{ab' - bc'}$.
From the first and third terms: $\alpha^2 = \frac{ba' - b'c}{ab' - bc'}$.
Equating $(\frac{cc' - aa'}{ab' - bc'})^2 = \frac{ba' - b'c}{ab' - bc'}$,we get:
$(cc' - aa')^2 = (ba' - b'c)(ab' - bc')$.
Thus,option $A$ is correct.

(B) Given the quadratic equation $2x^2 + 2(a + b)x + a^2 + b^2 = 0$.
Sum of roots $\alpha + \beta = -\frac{2(a + b)}{2} = -(a + b)$.
Product of roots $\alpha \beta = \frac{a^2 + b^2}{2}$.
Now,calculate the new roots:
Root $1 = (\alpha + \beta)^2 = (-(a + b))^2 = (a + b)^2$.
Root $2 = (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta = (a + b)^2 - 4(\frac{a^2 + b^2}{2}) = a^2 + 2ab + b^2 - 2a^2 - 2b^2 = -(a^2 - 2ab + b^2) = -(a - b)^2$.
The required equation is $x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0$.
Sum of new roots $= (a + b)^2 - (a - b)^2 = (a^2 + 2ab + b^2) - (a^2 - 2ab + b^2) = 4ab$.
Product of new roots $= (a + b)^2 \times (-(a - b)^2) = -((a + b)(a - b))^2 = -(a^2 - b^2)^2$.
Thus,the equation is $x^2 - 4abx - (a^2 - b^2)^2 = 0$.

(A) Since the coefficients $p$ and $q$ are real,the complex roots must occur in conjugate pairs.
Given that $2 + i\sqrt{3}$ is a root,its conjugate $2 - i\sqrt{3}$ must also be a root of the equation.
For a quadratic equation $x^2 + px + q = 0$,the sum of the roots is given by $-p$ and the product of the roots is given by $q$.
Sum of roots: $(2 + i\sqrt{3}) + (2 - i\sqrt{3}) = 4$. Thus,$-p = 4$,which implies $p = -4$.
Product of roots: $(2 + i\sqrt{3})(2 - i\sqrt{3}) = 2^2 - (i\sqrt{3})^2 = 4 - (-3) = 4 + 3 = 7$. Thus,$q = 7$.
Therefore,$(p, q) = (-4, 7)$.

(D) For a quadratic equation $ax^2 + bx + c = 0$,the sum of the roots is given by $-\frac{b}{a}$ and the product of the roots is given by $\frac{c}{a}$.
Given the equation $\lambda x^2 + 2x + 3\lambda = 0$,we have $a = \lambda$,$b = 2$,and $c = 3\lambda$.
Sum of the roots $= -\frac{2}{\lambda}$.
Product of the roots $= \frac{3\lambda}{\lambda} = 3$.
According to the problem,the sum of the roots is equal to their product:
$-\frac{2}{\lambda} = 3$.
Multiplying both sides by $\lambda$,we get $-2 = 3\lambda$.
Therefore,$\lambda = -\frac{2}{3}$.
Since $-\frac{2}{3}$ is not among the options $A, B,$ or $C$,the correct option is $D$.

(B) Given the quadratic equation $x^2 + 6x + \lambda = 0$.
From the properties of roots,the sum of roots $\alpha + \beta = -6$ $(i)$ and the product of roots $\alpha \beta = \lambda$ $(ii)$.
We are also given the linear equation $3\alpha + 2\beta = -20$ $(iii)$.
From $(i)$,we have $\beta = -6 - \alpha$. Substituting this into $(iii)$:
$3\alpha + 2(-6 - \alpha) = -20$
$3\alpha - 12 - 2\alpha = -20$
$\alpha = -20 + 12 = -8$.
Now,find $\beta$ using $(i)$:
$\beta = -6 - (-8) = 2$.
Finally,substitute $\alpha$ and $\beta$ into $(ii)$ to find $\lambda$:
$\lambda = \alpha \beta = (-8)(2) = -16$.

(B) Given the quadratic equation $2x^2 - 3x + 4 = 0$.
For this equation,the sum of roots $\alpha + \beta = -(-3)/2 = 3/2$ and the product of roots $\alpha\beta = 4/2 = 2$.
We need to find the equation whose roots are $\alpha^2$ and $\beta^2$.
The sum of the new roots is $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (3/2)^2 - 2(2) = 9/4 - 4 = (9 - 16)/4 = -7/4$.
The product of the new roots is $\alpha^2\beta^2 = (\alpha\beta)^2 = (2)^2 = 4$.
The required quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - (-7/4)x + 4 = 0$.
$x^2 + 7/4x + 4 = 0$.
Multiplying the entire equation by $4$,we get $4x^2 + 7x + 16 = 0$.

(C) Given the quadratic equation: ${x^2} - a(x + 1) - b = 0$
Expanding the equation: ${x^2} - ax - a - b = 0$
Comparing this with the standard form ${x^2} - (\text{sum of roots})x + (\text{product of roots}) = 0$,we get:
Sum of roots: $\alpha + \beta = a$
Product of roots: $\alpha \beta = -(a + b)$
We need to evaluate the expression: $(\alpha + 1)(\beta + 1)$
Expanding the expression: $(\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1$
Substituting the values of $\alpha + \beta$ and $\alpha \beta$:
$= -(a + b) + a + 1$
$= -a - b + a + 1$
$= 1 - b$