Solution of quadratic equations and Nature of roots Questions in English

(B) For the quadratic equation $ax^2 + bx + c = 0$ to have unequal real roots,the discriminant $D = b^2 - 4ac$ must be greater than $0$ $(D > 0)$.
Also,for it to be a quadratic equation,$a \neq 0$.
Given the set $S = \{0, 1, 2, 4\}$ and the condition $a \neq b \neq c$,we test combinations of $(a, b, c)$ where $a \in \{1, 2, 4\}$ and $b, c \in \{0, 1, 2, 4\}$.
$1$. If $a=1$: $b^2 - 4c > 0 \implies b^2 > 4c$. Possible $(b, c)$ pairs from $\{0, 2, 4\}$ (since $b, c \neq 1$):
- $(b=2, c=0) \implies 4 > 0$ (Valid)
- $(b=4, c=0) \implies 16 > 0$ (Valid)
- $(b=4, c=2) \implies 16 > 8$ (Valid)
$2$. If $a=2$: $b^2 - 8c > 0 \implies b^2 > 8c$. Possible $(b, c)$ pairs from $\{0, 1, 4\}$ (since $b, c \neq 2$):
- $(b=1, c=0) \implies 1 > 0$ (Valid)
- $(b=4, c=0) \implies 16 > 0$ (Valid)
- $(b=4, c=1) \implies 16 > 8$ (Valid)
$3$. If $a=4$: $b^2 - 16c > 0 \implies b^2 > 16c$. Possible $(b, c)$ pairs from $\{0, 1, 2\}$ (since $b, c \neq 4$):
- $(b=1, c=0) \implies 1 > 0$ (Valid)
- $(b=2, c=0) \implies 4 > 0$ (Valid)
Total valid equations = $3 + 3 + 2 = 8$. However,checking the constraint $a \neq b \neq c$ strictly,we re-evaluate:
Valid sets $(a, b, c)$ are $(1, 2, 0), (1, 4, 0), (1, 4, 2), (2, 1, 0), (2, 4, 0), (2, 4, 1), (4, 1, 0), (4, 2, 0)$. All satisfy $a \neq b, b \neq c, a \neq c$. Total is $8$. Since $8$ is not an option,re-checking the set: The question implies $a, b, c$ are distinct. The count is $8$.

(D) Let $\alpha, \beta, \gamma, \delta$ be the roots of $x^4-x^3-8 x^2+2 x+12=0$. Given $\alpha+\beta=0$.
We can write the polynomial as $(x^2+a)(x^2-x+b) = x^4-x^3+(a+b)x^2-ax+ab$.
Comparing the coefficients with $x^4-x^3-8x^2+2x+12=0$:
$-a=2 \implies a=-2$.
$ab=12 \implies -2b=12 \implies b=-6$.
Thus,$x^4-x^3-8x^2+2x+12 = (x^2-2)(x^2-x-6) = (x^2-2)(x-3)(x+2)$.
The roots are $\pm\sqrt{2}, 3, -2$.
Since $\alpha+\beta=0$,we have $\alpha=\sqrt{2}, \beta=-\sqrt{2}$.
The other roots are $\gamma=3, \delta=-2$ (given $\gamma > \delta$).
Therefore,$3\gamma+2\delta = 3(3)+2(-2) = 9-4 = 5$.

(C) Given the quadratic equation $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$.
Sum of coefficients $= a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ba + ca - cb = 0$.
Since the sum of coefficients is $0$,$x = 1$ is a root of the equation.
Let the roots be $\alpha$ and $\beta$. We know $\beta = 1$.
From the product of roots formula,$\alpha \times \beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{c(a-b)}{a(b-c)}$.
Therefore,$\alpha \times 1 = \frac{c(a-b)}{a(b-c)}$.
The roots are $1$ and $\frac{c(a-b)}{a(b-c)}$.

(D) Since the coefficients of the quadratic equation $x^2+ax+b=0$ are real,the complex roots must occur in conjugate pairs.
Given one root is $x_1 = 3+i$,the other root must be $x_2 = 3-i$.
The sum of the roots of a quadratic equation $x^2+ax+b=0$ is given by $-a$.
Therefore,$(3+i) + (3-i) = -a$.
$6 = -a$.
$a = -6$.

(B) To find the equation whose roots are shifted by $-1$,we replace $x$ with $(x - (-1))$,which is $(x+1)$.
Substituting $(x+1)$ for $x$ in the original equation $f(x) = x^4+5x^3+6x^2+7x+9=0$:
$f(x+1) = (x+1)^4 + 5(x+1)^3 + 6(x+1)^2 + 7(x+1) + 9 = 0$
Expanding each term:
$(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1$
$5(x+1)^3 = 5(x^3 + 3x^2 + 3x + 1) = 5x^3 + 15x^2 + 15x + 5$
$6(x+1)^2 = 6(x^2 + 2x + 1) = 6x^2 + 12x + 6$
$7(x+1) = 7x + 7$
Adding these together:
$x^4 + (4+5)x^3 + (6+15+6)x^2 + (4+15+12+7)x + (1+5+6+7+9) = 0$
$x^4 + 9x^3 + 27x^2 + 38x + 28 = 0$

(B) The given equation is $x^2 - 3ax + a^2 - 2a - 4 = 0$.
For the roots to be rational,the discriminant $D$ must be a perfect square of a rational number.
$D = b^2 - 4ac = (-3a)^2 - 4(1)(a^2 - 2a - 4) = 9a^2 - 4a^2 + 8a + 16 = 5a^2 + 8a + 16$.
Since $5a^2 + 8a + 16$ is a quadratic in '$a$' with discriminant $D_a = 8^2 - 4(5)(16) = 64 - 320 = -256 < 0$,the expression $5a^2 + 8a + 16$ is always positive for all real '$a$'.
However,for the roots to be rational,$D$ must be a perfect square.
Since $5a^2 + 8a + 16$ is not a perfect square for all rational '$a$',the roots are real and distinct but not necessarily rational.

(C) Given that $x^2+x-6$ is a factor of $P(x) = 2x^3+x^2+ax+b$.
Factorizing the divisor: $x^2+x-6 = (x+3)(x-2)$.
Since $(x+3)$ and $(x-2)$ are factors,$P(-3) = 0$ and $P(2) = 0$.
For $x = -3$: $2(-3)^3 + (-3)^2 + a(-3) + b = 0$ $\Rightarrow -54 + 9 - 3a + b = 0$ $\Rightarrow -3a + b = 45$ ... $(i)$.
For $x = 2$: $2(2)^3 + (2)^2 + a(2) + b = 0$ $\Rightarrow 16 + 4 + 2a + b = 0$ $\Rightarrow 2a + b = -20$ ... $(ii)$.
Subtracting $(ii)$ from $(i)$: $(-3a - 2a) + (b - b) = 45 - (-20)$ $\Rightarrow -5a = 65$ $\Rightarrow a = -13$.
Substituting $a = -13$ into $(ii)$: $2(-13) + b = -20$ $\Rightarrow -26 + b = -20$ $\Rightarrow b = 6$.
Finally,$6a + 13b = 6(-13) + 13(6) = -78 + 78 = 0$.

(D) Given the equation: $2^{6x} - 3(2^{3x+2}) + 32 = 0$
Since $2^{3x+2} = 2^{3x} \times 2^2 = 4 \times 2^{3x}$,the equation becomes:
$(2^{3x})^2 - 3(4 \times 2^{3x}) + 32 = 0$
$(2^{3x})^2 - 12(2^{3x}) + 32 = 0$
Let $y = 2^{3x}$. Then the equation is $y^2 - 12y + 32 = 0$.
Factoring the quadratic: $(y - 4)(y - 8) = 0$.
So,$y = 4$ or $y = 8$.
Case $1$: $2^{3x} = 4 = 2^2 \implies 3x = 2 \implies x = \frac{2}{3}$.
Case $2$: $2^{3x} = 8 = 2^3 \implies 3x = 3 \implies x = 1$.
Given $\beta < 1$,we have $\beta = \frac{2}{3}$ and $\alpha = 1$.
Therefore,$2\alpha + 3\beta = 2(1) + 3(\frac{2}{3}) = 2 + 2 = 4$.

(A) Let the roots of the equation $x^3-6x^2+3x+10=0$ be $\alpha, \beta, \gamma$.
Given that one root is the average of the other two,let $\beta = \frac{\alpha+\gamma}{2} \Rightarrow \alpha+\gamma = 2\beta$.
From the sum of roots,$\alpha+\beta+\gamma = 6$.
Substituting $\alpha+\gamma = 2\beta$,we get $2\beta+\beta = 6$ $\Rightarrow 3\beta = 6$ $\Rightarrow \beta = 2$.
Since $\beta=2$ is a root,it must satisfy the equation: $(2)^3 - 6(2)^2 + 3(2) + 10 = 8 - 24 + 6 + 10 = 0$.
Now,divide the polynomial by $(x-2)$: $(x-2)(x^2-4x-5) = 0$.
Factoring the quadratic: $(x-2)(x-5)(x+1) = 0$.
The roots are $2, 5, -1$.
The sum of the fourth powers of the roots is $2^4 + 5^4 + (-1)^4 = 16 + 625 + 1 = 642$.

(A) Given the equation: $\frac{k}{kx+3}+\frac{3}{3x-k}=\frac{12x+5}{(kx+3)(3x-k)}$
Multiplying both sides by $(kx+3)(3x-k)$,we get:
$k(3x-k) + 3(kx+3) = 12x+5$
$3kx - k^2 + 3kx + 9 = 12x + 5$
$6kx - k^2 + 9 = 12x + 5$
Comparing the coefficients of $x$ on both sides:
$6k = 12 \Rightarrow k = 2$
Substituting $k=2$ into the quadratic equation $kx^2-7x+3=0$:
$2x^2-7x+3=0$
$2x^2-6x-x+3=0$
$2x(x-3)-1(x-3)=0$
$(2x-1)(x-3)=0$
Thus,the roots are $x = \frac{1}{2}$ and $x = 3$.
Since both $\frac{1}{2}$ and $3$ are rational numbers,the correct option is $A$.

(A) Given $(x+a)(x+1991)+1 = (x+b)(x+c)$.
Expanding the left side: $x^2 + (a+1991)x + 1991a + 1 = x^2 + (b+c)x + bc$.
Comparing coefficients,we have $b+c = a+1991$ and $bc = 1991a+1$.
Since $b$ and $c$ are roots of the quadratic equation $x^2 - (b+c)x + bc = 0$,the discriminant $D$ must be a perfect square,say $m^2$.
$D = (b+c)^2 - 4bc = (a+1991)^2 - 4(1991a+1) = m^2$.
$(a+1991)^2 - 4(1991a) - 4 = m^2$.
$a^2 + 2(1991)a + 1991^2 - 4(1991)a - 4 = m^2$.
$a^2 - 2(1991)a + 1991^2 - 4 = m^2$.
$(a-1991)^2 - 4 = m^2$.
$(a-1991)^2 - m^2 = 4$.
$(a-1991-m)(a-1991+m) = 4$.
Let $X = a-1991-m$ and $Y = a-1991+m$. Then $XY = 4$.
Since $Y-X = 2m$,$X$ and $Y$ must have the same parity. Since their product is $4$ (even),both must be even.
The possible pairs $(X, Y)$ are $(2, 2)$ and $(-2, -2)$.
Case $1$: $a-1991-m = 2$ and $a-1991+m = 2$. Adding gives $2(a-1991) = 4$ $\Rightarrow a-1991 = 2$ $\Rightarrow a = 1993$.
Case $2$: $a-1991-m = -2$ and $a-1991+m = -2$. Adding gives $2(a-1991) = -4$ $\Rightarrow a-1991 = -2$ $\Rightarrow a = 1989$.
Thus,$a \in \{1989, 1993\}$.

(D) The given quadratic equation is $x^2 - 2(1 + 3m)x + 7(3 + 2m) = 0$.
For the roots to be distinct,the discriminant $D$ must be greater than $0$.
$D = b^2 - 4ac > 0$
$[-2(1 + 3m)]^2 - 4(1)(7(3 + 2m)) > 0$
$4(1 + 9m^2 + 6m) - 28(3 + 2m) > 0$
$4 + 36m^2 + 24m - 84 - 56m > 0$
$36m^2 - 32m - 80 > 0$
Dividing by $4$:
$9m^2 - 8m - 20 > 0$
Factoring the quadratic expression:
$(9m + 10)(m - 2) > 0$
The roots of the equation $9m^2 - 8m - 20 = 0$ are $m = 2$ and $m = -\frac{10}{9}$.
Thus,the inequality holds for $m \in (-\infty, -\frac{10}{9}) \cup (2, \infty)$.
Since $S$ is a subset of the set of real numbers $\mathbb{R}$ and the interval $(-\infty, -\frac{10}{9}) \cup (2, \infty)$ contains infinitely many real numbers,the number of elements in $S$ is infinite.

(B) For the quadratic equation $x^2+(2m+1)x+m=0$ to have equal roots,the discriminant $D$ must be equal to $0$.
The discriminant is given by $D = b^2 - 4ac$.
Here,$a=1$,$b=(2m+1)$,and $c=m$.
Substituting these values into $D=0$:
$(2m+1)^2 - 4(1)(m) = 0$
$4m^2 + 4m + 1 - 4m = 0$
$4m^2 + 1 = 0$
$4m^2 = -1$
$m^2 = -\frac{1}{4}$
Since the square of any real number $m$ must be non-negative $(m^2 \ge 0)$,there are no real values of $m$ that satisfy this equation.
Therefore,the number of real values of $m$ is $0$.

(B) Given the quadratic equation $7x^2 - 13x + a = 0$.
For the roots to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square.
Here,$D = (-13)^2 - 4(7)(a) = 169 - 28a$.
For $D$ to be a perfect square,$169 - 28a = k^2$ for some non-negative integer $k$.
Testing the given options:
If $a = 5$,$D = 169 - 28(5) = 169 - 140 = 29$ (not a perfect square).
If $a = 6$,$D = 169 - 28(6) = 169 - 168 = 1 = 1^2$ (a perfect square).
Since $a=6$ satisfies the condition and is the smallest value among the options,the smallest possible value of $a$ is $6$.

(C) We know that a cubic equation with rational coefficients having an irrational root $\alpha+\sqrt{\beta}$ (where $\alpha, \beta \in \mathbb{Q}$ and $\beta$ is not a perfect square) must also have the conjugate root $\alpha-\sqrt{\beta}$.
Hence,the roots are $1, 2+\sqrt{5},$ and $2-\sqrt{5}$.
The cubic equation with roots $\alpha, \beta, \gamma$ is given by $x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-(\alpha\beta\gamma)=0$.
Let $\alpha=1, \beta=2+\sqrt{5}, \gamma=2-\sqrt{5}$.
Sum of roots: $\alpha+\beta+\gamma = 1+(2+\sqrt{5})+(2-\sqrt{5}) = 5$.
Sum of product of roots taken two at a time: $\alpha\beta+\beta\gamma+\gamma\alpha = 1(2+\sqrt{5})+(2+\sqrt{5})(2-\sqrt{5})+1(2-\sqrt{5}) = 2+\sqrt{5}+4-5+2-\sqrt{5} = 3$.
Product of roots: $\alpha\beta\gamma = 1(2+\sqrt{5})(2-\sqrt{5}) = 1(4-5) = -1$.
Substituting these into the general form: $x^3-(5)x^2+(3)x-(-1) = 0$,which simplifies to $x^3-5x^2+3x+1=0$.

(C) Given the equation: $x^2-5|x|-14=0$.
Since $x^2 = |x|^2$,we can rewrite the equation as:
$|x|^2-5|x|-14=0$.
Let $t = |x|$,where $t \geq 0$. The equation becomes $t^2-5t-14=0$.
Factoring the quadratic: $(t-7)(t+2)=0$.
This gives $t=7$ or $t=-2$.
Since $t = |x| \geq 0$,we reject $t=-2$.
Thus,$|x|=7$,which implies $x=7$ or $x=-7$.
Both roots are real. Therefore,there are exactly two real roots.

(C) Given the quadratic equation $\frac{1}{4}x^2 + bx + c = 0$.
Multiplying the entire equation by $4$,we get $x^2 + 4bx + 4c = 0$.
Using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ for the equation $x^2 + 4bx + 4c = 0$,where $A=1, B=4b, C=4c$:
$x = \frac{-4b \pm \sqrt{(4b)^2 - 4(1)(4c)}}{2(1)}$
$x = \frac{-4b \pm \sqrt{16b^2 - 16c}}{2}$
$x = \frac{-4b \pm 4\sqrt{b^2 - c}}{2}$
$x = -2b \pm 2\sqrt{b^2 - c}$.
For $x$ to be an integer,$b$ must be an integer and $b^2 - c$ must be a perfect square of an integer.

(B) For the quadratic equation $x^2+mx+n=0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = m^2 - 4n \geq 0$,which implies $m^2 \geq 4n$.
Given $0 \leq m, n \leq 10$,we count the pairs $(m, n)$ satisfying $n \leq \frac{m^2}{4}$:
- If $m=10$,$n \leq 25$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values).
- If $m=9$,$n \leq 20.25$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values).
- If $m=8$,$n \leq 16$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values).
- If $m=7$,$n \leq 12.25$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values).
- If $m=6$,$n \leq 9$,so $n \in \{0, 1, \dots, 9\}$ ($10$ values).
- If $m=5$,$n \leq 6.25$,so $n \in \{0, 1, \dots, 6\}$ ($7$ values).
- If $m=4$,$n \leq 4$,so $n \in \{0, 1, 2, 3, 4\}$ ($5$ values).
- If $m=3$,$n \leq 2.25$,so $n \in \{0, 1, 2\}$ ($3$ values).
- If $m=2$,$n \leq 1$,so $n \in \{0, 1\}$ ($2$ values).
- If $m=1$,$n \leq 0.25$,so $n \in \{0\}$ ($1$ value).
- If $m=0$,$n \leq 0$,so $n \in \{0\}$ ($1$ value).
Total number of pairs $= 11+11+11+11+10+7+5+3+2+1+1 = 73$.

(D) Given,$|x^2-x-6|=x+2$.
This implies two cases:
Case $1$: $x^2-x-6 = x+2$
$\Rightarrow x^2-2x-8 = 0$
$\Rightarrow (x-4)(x+2) = 0$
$\Rightarrow x = 4, -2$.
Case $2$: $x^2-x-6 = -(x+2)$
$\Rightarrow x^2-x-6 = -x-2$
$\Rightarrow x^2-4 = 0$
$\Rightarrow x^2 = 4$
$\Rightarrow x = 2, -2$.
Combining the results from both cases,the set of roots is $\{-2, 2, 4\}$.

(A) Given that $a, b, c$ are in Arithmetic Progression $(AP)$,we have $2b = a + c$.
Substituting this into the quadratic equation $ax^2 - 2bx + c = 0$,we get:
$ax^2 - (a + c)x + c = 0$
$ax^2 - ax - cx + c = 0$
$ax(x - 1) - c(x - 1) = 0$
$(x - 1)(ax - c) = 0$
Thus,the roots are $x = 1$ and $x = \frac{c}{a}$.
Hence,option $A$ is correct.

(C) Given equation: $(\cos p-1) x^2+(\cos p) x+\sin p=0$.
Since the roots are real,the discriminant $\Delta \geq 0$.
$\Delta = b^2 - 4ac = (\cos p)^2 - 4(\cos p - 1)(\sin p) \geq 0$.
$\cos^2 p - 4\sin p \cos p + 4\sin p \geq 0$.
For the quadratic equation to exist,the coefficient of $x^2$ must be non-zero: $\cos p - 1 \neq 0 \Rightarrow \cos p \neq 1$.
Since $\cos^2 p \geq 0$ and $(\cos p - 1) < 0$ for all $p \neq 2n\pi$,the condition $\Delta \geq 0$ is satisfied when $\sin p > 0$.
Thus,$p \in (0, \pi)$.
Hence,option $C$ is correct.

(A) Given equation: $x^2-5|x|+6=0$
Since $x^2 = |x|^2$,we can rewrite the equation as:
$|x|^2-5|x|+6=0$
Let $|x| = t$,then the equation becomes $t^2-5t+6=0$.
Factoring the quadratic: $(t-3)(t-2)=0$.
So,$|x|=3$ or $|x|=2$.
If $|x|=3$,then $x = 3$ or $x = -3$.
If $|x|=2$,then $x = 2$ or $x = -2$.
Thus,the solutions are $x \in \{-3, -2, 2, 3\}$.
Therefore,there are $4$ solutions.

(B) The given quadratic equation is $2x^2 + 5x + k = 0$.
For the roots of a quadratic equation with rational coefficients to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square of a rational number.
Here,$a = 2$,$b = 5$,and $c = k$.
$D = (5)^2 - 4(2)(k) = 25 - 8k$.
For $D$ to be a perfect square,we test the given options:
If $k = \frac{25}{8}$,then $D = 25 - 8(\frac{25}{8}) = 25 - 25 = 0$.
Since $0$ is a perfect square $(0^2 = 0)$,the roots are rational.
Therefore,$k = \frac{25}{8}$ is the correct value.
Thus,option $B$ is correct.

(A) Given the equation $f(x)=x^4+2x^3-16x^2-22x+7=0$.
Since the coefficients are rational,if $2+\sqrt{3}$ is a root,its conjugate $2-\sqrt{3}$ must also be a root.
Let the four roots be $2+\sqrt{3}, 2-\sqrt{3}, \alpha, \beta$.
From the sum of roots: $(2+\sqrt{3})+(2-\sqrt{3})+\alpha+\beta = -2 \implies 4+\alpha+\beta = -2 \implies \alpha+\beta = -6$.
From the product of roots: $(2+\sqrt{3})(2-\sqrt{3}) \alpha \beta = 7 \implies (4-3) \alpha \beta = 7 \implies \alpha \beta = 7$.
Forming a quadratic equation for $\alpha$ and $\beta$: $t^2 - (\alpha+\beta)t + \alpha \beta = 0 \implies t^2+6t+7=0$.
Solving for $t$: $t = \frac{-6 \pm \sqrt{36-28}}{2} = -3 \pm \sqrt{2}$.
Thus,the roots are $2+\sqrt{3}, 2-\sqrt{3}, -3+\sqrt{2}, -3-\sqrt{2}$.
Comparing with the options,$3-\sqrt{2}$ is not a root.

(A) Given the quadratic equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ has real roots,the discriminant $D \geq 0$.
$D = [2(a+b+c)]^2 - 4(1)(3\lambda(ab+bc+ca)) \geq 0$
$4(a+b+c)^2 - 12\lambda(ab+bc+ca) \geq 0$
$(a+b+c)^2 \geq 3\lambda(ab+bc+ca)$
$\lambda \leq \frac{(a+b+c)^2}{3(ab+bc+ca)}$
For a scalene triangle,we know that $(a-b)^2 + (b-c)^2 + (c-a)^2 > 0$.
Expanding this,$2(a^2+b^2+c^2) - 2(ab+bc+ca) > 0$,so $a^2+b^2+c^2 > ab+bc+ca$.
Adding $2(ab+bc+ca)$ to both sides,we get $(a+b+c)^2 > 3(ab+bc+ca)$.
Thus,$\frac{(a+b+c)^2}{3(ab+bc+ca)} > 1$.
Also,from the triangle inequality,$(a+b+c)^2 < 4(ab+bc+ca)$,which implies $\frac{(a+b+c)^2}{3(ab+bc+ca)} < \frac{4}{3}$.
Since $\lambda \leq \frac{(a+b+c)^2}{3(ab+bc+ca)}$ and the expression is strictly less than $\frac{4}{3}$,the range for $\lambda$ is $\left(-\infty, \frac{4}{3}\right)$.

(D) Given that the polynomial equation of degree $4$ has real coefficients. Since complex roots occur in conjugate pairs,if $1+2i$ is a root,then $1-2i$ must also be a root. The roots are $2+\sqrt{3}$,$2-\sqrt{3}$,$1+2i$,and $1-2i$.
The quadratic factor corresponding to roots $2 \pm \sqrt{3}$ is:
$(x-(2+\sqrt{3}))(x-(2-\sqrt{3})) = (x-2)^2 - 3 = x^2-4x+4-3 = x^2-4x+1$.
The quadratic factor corresponding to roots $1 \pm 2i$ is:
$(x-(1+2i))(x-(1-2i)) = (x-1)^2 - (2i)^2 = x^2-2x+1+4 = x^2-2x+5$.
The required polynomial is the product of these two factors:
$(x^2-4x+1)(x^2-2x+5) = x^2(x^2-2x+5) - 4x(x^2-2x+5) + 1(x^2-2x+5)$
$= x^4-2x^3+5x^2-4x^3+8x^2-20x+x^2-2x+5$
$= x^4-6x^3+14x^2-22x+5 = 0$.
Thus,the correct option is $D$.

(A) Given $\sin \alpha = p$. We know that $\sin \alpha = \frac{2 \tan \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} = p$.
Let $t = \tan \frac{\alpha}{2}$. Then $\frac{2t}{1 + t^2} = p$,which implies $2t = p + pt^2$,or $pt^2 - 2t + p = 0$.
Dividing by $p$,we get $t^2 - \frac{2}{p} t + 1 = 0$.
The roots of this equation are $t_1 = \tan \frac{\alpha}{2}$ and $t_2 = \frac{1}{\tan \frac{\alpha}{2}} = \cot \frac{\alpha}{2}$.
The sum of roots is $\tan \frac{\alpha}{2} + \cot \frac{\alpha}{2} = \frac{2}{p}$ and the product of roots is $\tan \frac{\alpha}{2} \times \cot \frac{\alpha}{2} = 1$.
The required quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - \frac{2}{p} x + 1 = 0$,which simplifies to $px^2 - 2x + p = 0$.
Thus,option $(A)$ is correct.

(A) Given equation: $\sqrt{\frac{x}{2x+1}} + \sqrt{\frac{2x+1}{x}} = 2$
Let $y = \sqrt{\frac{x}{2x+1}}$. Then the equation becomes $y + \frac{1}{y} = 2$.
Multiplying by $y$,we get $y^2 - 2y + 1 = 0$,which is $(y-1)^2 = 0$.
Thus,$y = 1$.
Substituting back: $\sqrt{\frac{x}{2x+1}} = 1 \implies \frac{x}{2x+1} = 1 \implies x = 2x + 1 \implies x = -1$.
Since $\alpha$ satisfies this equation,$\alpha = -1$.
Now,substitute $\alpha = -1$ into the quadratic equation $\alpha^2 x^2 + 4\alpha x + 3 = 0$:
$(-1)^2 x^2 + 4(-1)x + 3 = 0
\implies x^2 - 4x + 3 = 0
\implies (x-1)(x-3) = 0$.
Therefore,the roots are $x = 1, 3$.

(B) Given equation is $(p^2+p-3)(p^2+p-2)-12=0$.
Let $y = p^2+p-2$.
Then the equation becomes $(y-1)y - 12 = 0$.
$y^2 - y - 12 = 0$.
$(y-4)(y+3) = 0$.
So,$y = 4$ or $y = -3$.
Case $1$: $p^2+p-2 = 4 \Rightarrow p^2+p-6 = 0$.
$(p+3)(p-2) = 0$,so $p = -3, 2$ (These are real roots).
Case $2$: $p^2+p-2 = -3 \Rightarrow p^2+p+1 = 0$.
For this quadratic,the discriminant $D = b^2 - 4ac = 1^2 - 4(1)(1) = -3 < 0$.
Thus,the roots are non-real.
The sum of the roots of $p^2+p+1=0$ is given by $-\frac{b}{a} = -\frac{1}{1} = -1$.

(C) For the roots of the quadratic equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ to be real,the discriminant $D$ must be greater than or equal to $0$.
$D = [2(a+b+c)]^2 - 4(1)(3\lambda(ab+bc+ca)) \geq 0$
$4(a+b+c)^2 - 12\lambda(ab+bc+ca) \geq 0$
$(a+b+c)^2 \geq 3\lambda(ab+bc+ca)$
$a^2+b^2+c^2+2(ab+bc+ca) \geq 3\lambda(ab+bc+ca)$
Dividing by $(ab+bc+ca)$,we get:
$\frac{a^2+b^2+c^2}{ab+bc+ca} + 2 \geq 3\lambda$
Since $a, b, c$ are sides of a triangle,$a+b > c$,$b+c > a$,and $c+a > b$.
It is a known inequality that $a^2+b^2+c^2 < 2(ab+bc+ca)$.
Thus,$\frac{a^2+b^2+c^2}{ab+bc+ca} < 2$.
Substituting this into the inequality:
$3\lambda \leq \frac{a^2+b^2+c^2}{ab+bc+ca} + 2 < 2 + 2 = 4$
$3\lambda < 4 \Rightarrow \lambda < \frac{4}{3}$.

(A) Let $t = x + \frac{1}{x}$. Then $t^2 = x^2 + \frac{1}{x^2} + 2$,so $x^2 + \frac{1}{x^2} = t^2 - 2$.
Substituting this into the equation: $2(t^2 - 2) - 7t + 9 = 0$.
$2t^2 - 4 - 7t + 9 = 0 \implies 2t^2 - 7t + 5 = 0$.
Factoring the quadratic: $(2t - 5)(t - 1) = 0$.
So,$t = 1$ or $t = \frac{5}{2}$.
Case $1$: $x + \frac{1}{x} = 1 \implies x^2 - x + 1 = 0$. The discriminant $D = (-1)^2 - 4(1)(1) = -3 < 0$. No real solutions.
Case $2$: $x + \frac{1}{x} = \frac{5}{2} \implies 2x^2 - 5x + 2 = 0$.
Factoring: $(2x - 1)(x - 2) = 0$,so $x = \frac{1}{2}$ or $x = 2$.
Since the question asks for the number of integral solutions,we check $x = 2$ (which is an integer) and $x = \frac{1}{2}$ (which is not).
Thus,there is only $1$ integral solution,which is $x = 2$.

(A) Given equation: $|x|^2 - 5|x| - 24 = 0$.
Let $|x| = t$,where $t \ge 0$.
The equation becomes $t^2 - 5t - 24 = 0$.
Factoring the quadratic: $(t - 8)(t + 3) = 0$.
This gives $t = 8$ or $t = -3$.
Since $|x| \ge 0$,we reject $t = -3$.
Thus,$|x| = 8$,which implies $x = 8$ or $x = -8$.
The roots of the equation are $8$ and $-8$.
Product of the roots: $8 \times (-8) = -64$.
Sum of the roots: $8 + (-8) = 0$.
Therefore,the product and sum are $-64$ and $0$ respectively.

(A) Given that the roots of the quadratic equation $(b-c)x^2 + (c-a)x + (a-b) = 0$ are equal,the discriminant $D$ must be $0$.
$D = (c-a)^2 - 4(b-c)(a-b) = 0$
Expanding the terms:
$(c^2 + a^2 - 2ac) - 4(ab - b^2 - ac + bc) = 0$
$c^2 + a^2 - 2ac - 4ab + 4b^2 + 4ac - 4bc = 0$
$c^2 + a^2 + 2ac + 4b^2 - 4ab - 4bc = 0$
$(c+a)^2 - 4b(a+c) + (2b)^2 = 0$
This is in the form $X^2 - 2XY + Y^2 = 0$,where $X = (c+a)$ and $Y = 2b$.
$(c+a - 2b)^2 = 0$
$c+a - 2b = 0$
$2b = a+c$
Since $2b = a+c$,the terms $a, b, c$ are in arithmetic progression $(AP)$.

(C) Given equation: $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$. \\ Let $t = x-a$. Then the equation becomes: \\ $t(t-1) + (t-1)(t-2) + t(t-2) = 0$ \\ $t^2 - t + t^2 - 3t + 2 + t^2 - 2t = 0$ \\ $3t^2 - 6t + 2 = 0$ \\ The discriminant $D = b^2 - 4ac = (-6)^2 - 4(3)(2) = 36 - 24 = 12$. \\ Since $D > 0$,the roots for $t$ are real and distinct. \\ Consequently,$x = a + t$ will also have real and distinct roots.

(B) Given,$f(x) = x^2 + ax + b$ has imaginary roots.
Therefore,the discriminant $D < 0$,which implies $a^2 - 4b < 0$.
Now,we calculate the derivatives:
$f'(x) = 2x + a$
$f''(x) = 2$
Substituting these into the equation $f(x) + f'(x) + f''(x) = 0$:
$(x^2 + ax + b) + (2x + a) + 2 = 0$
$x^2 + (a + 2)x + (b + a + 2) = 0$
The discriminant $D'$ of this new quadratic equation is:
$D' = (a + 2)^2 - 4(1)(b + a + 2)$
$D' = a^2 + 4a + 4 - 4b - 4a - 8$
$D' = a^2 - 4b - 4$
Since $a^2 - 4b < 0$,it follows that $a^2 - 4b - 4 < -4$.
Thus,$D' < 0$.
Since the discriminant is negative,the roots of the equation $f(x) + f'(x) + f''(x) = 0$ are imaginary.

(A) Given equation is $x^4+7x^3+18x^2+20x+8=0$.
We can factorize the polynomial by testing small integer roots.
For $x = -2$: $(-2)^4 + 7(-2)^3 + 18(-2)^2 + 20(-2) + 8 = 16 - 56 + 72 - 40 + 8 = 0$.
So,$(x+2)$ is a factor.
Dividing $x^4+7x^3+18x^2+20x+8$ by $(x+2)$ gives $x^3+5x^2+8x+4$.
Testing $x = -2$ again for $x^3+5x^2+8x+4$: $(-2)^3 + 5(-2)^2 + 8(-2) + 4 = -8 + 20 - 16 + 4 = 0$.
So,$(x+2)$ is a factor again.
Dividing $x^3+5x^2+8x+4$ by $(x+2)$ gives $x^2+3x+2 = (x+1)(x+2)$.
Thus,the equation is $(x+2)^3(x+1) = 0$.
The repeated root is $-2$.

(A) Given equation: $2x^3 + 5x^2 - 4x - 12 = 0$
Factorizing the cubic equation:
$2x^3 + 4x^2 + x^2 + 2x - 6x - 12 = 0$
$2x^2(x + 2) + x(x + 2) - 6(x + 2) = 0$
$(2x^2 + x - 6)(x + 2) = 0$
$(2x - 3)(x + 2)(x + 2) = 0$
The roots are $x = -2, -2, \frac{3}{2}$.
The distinct roots are $-2$ and $\frac{3}{2}$.
The quadratic equation with these roots is:
$(x - (-2))(x - \frac{3}{2}) = 0$
$(x + 2)(x - \frac{3}{2}) = 0$
$x^2 - \frac{3}{2}x + 2x - 3 = 0$
$x^2 + \frac{1}{2}x - 3 = 0$
Multiplying by $2$ to clear the fraction:
$2x^2 + x - 6 = 0$
The constant term is $-6$.

(D) Since $(x-2)$ is a common factor of the expressions $x^2+ax+b$ and $x^2+cx+d$,the value of these expressions at $x=2$ must be $0$.
For $x^2+ax+b$: $(2)^2+a(2)+b=0 \Rightarrow 4+2a+b=0 \dots(i)$
For $x^2+cx+d$: $(2)^2+c(2)+d=0 \Rightarrow 4+2c+d=0 \dots(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(4+2a+b) - (4+2c+d) = 0 - 0$
$2a+b-2c-d = 0$
$b-d = 2c-2a$
$b-d = 2(c-a)$
Dividing both sides by $(c-a)$ (assuming $c \neq a$):
$\frac{b-d}{c-a} = 2$

(B) For the quadratic equation $x^2+3(a+3)x-9a=0$ to have equal roots,the discriminant $D$ must be $0$.
$D = [3(a+3)]^2 - 4(1)(-9a) = 0$
$9(a^2+6a+9) + 36a = 0$
$9a^2 + 54a + 81 + 36a = 0$
$9a^2 + 90a + 81 = 0$
$a^2 + 10a + 9 = 0$
$(a+9)(a+1) = 0 \Rightarrow a = -9, -1$.
For $a = -9$,the equation is $x^2 - 18x + 81 = 0$ $\Rightarrow (x-9)^2 = 0$ $\Rightarrow x = 9$. Let $\alpha = 9$.
For $a = -1$,the equation is $x^2 + 6x + 9 = 0$ $\Rightarrow (x+3)^2 = 0$ $\Rightarrow x = -3$. Let $\beta = -3$.
We want the minimum value of the expression $f(x) = x^2 + \alpha x - \beta = x^2 + 9x - (-3) = x^2 + 9x + 3$.
Completing the square: $f(x) = (x + \frac{9}{2})^2 + 3 - \frac{81}{4} = (x + \frac{9}{2})^2 - \frac{69}{4}$.
Since $(x + \frac{9}{2})^2 \geq 0$,the minimum value is $-\frac{69}{4}$.

(C) The given equation is $|x^2+2x-8| = -(x-2) = 2-x$.
Case $1$: $x^2+2x-8 \ge 0$.
$(x+4)(x-2) \ge 0 \implies x \in (-\infty, -4] \cup [2, \infty)$.
Then $x^2+2x-8 = 2-x \implies x^2+3x-10 = 0 \implies (x+5)(x-2) = 0$.
So $x = -5$ and $x = 2$. Both satisfy the condition.
Case $2$: $x^2+2x-8 < 0$.
$(x+4)(x-2) < 0 \implies x \in (-4, 2)$.
Then $-(x^2+2x-8) = 2-x \implies -x^2-2x+8 = 2-x \implies x^2+x-6 = 0 \implies (x+3)(x-2) = 0$.
So $x = -3$ and $x = 2$.
Since $x=2$ is not in $(-4, 2)$,we only take $x = -3$.
The distinct real solutions are $\{-5, 2, -3\}$.
Thus,there are $3$ distinct real solutions.

(B) To solve the inequality $x^2 - 4x - 21 \leq 0$,we first find the roots of the quadratic equation $x^2 - 4x - 21 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 1, b = -4, c = -21$:
$x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-21)}}{2(1)}$
$x = \frac{4 \pm \sqrt{16 + 84}}{2} = \frac{4 \pm \sqrt{100}}{2} = \frac{4 \pm 10}{2}$
So,$x_1 = \frac{14}{2} = 7$ and $x_2 = \frac{-6}{2} = -3$.
The inequality can be written as $(x - 7)(x + 3) \leq 0$.
For the product to be less than or equal to zero,$x$ must lie between the roots inclusive.
Thus,$x \in [-3, 7]$.

(A) Let $y = \sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$
$\Rightarrow y = \sqrt{42+y}$
On squaring both sides,we get:
$y^2 = 42 + y$
$\Rightarrow y^2 - y - 42 = 0$
Factorizing the quadratic equation:
$(y - 7)(y + 6) = 0$
$\Rightarrow y = 7$ or $y = -6$
Since the square root function must yield a non-negative value,$y = -6$ is rejected.
Therefore,the required solution is $y = 7$.

(C) Given the quadratic equation $3x^2 + 4kx + 3 = 0$.
For the roots to be non-real,the discriminant $D$ must be less than $0$.
$D = b^2 - 4ac < 0$
$(4k)^2 - 4(3)(3) < 0$
$16k^2 - 36 < 0$
$16k^2 < 36$
$k^2 < 36/16$
$k^2 < 9/4$
Taking the square root on both sides,we get $|k| < 3/2$.
Therefore,$-3/2 < k < 3/2$.
Thus,$k$ lies in the interval $(-3/2, 3/2)$.

(A) Given the quadratic equation: $16x^2-10x+1=0$
Factorizing the equation: $16x^2-8x-2x+1=0$
$\Rightarrow 8x(2x-1)-1(2x-1)=0$
$\Rightarrow (8x-1)(2x-1)=0$
The roots are $x_1 = \frac{1}{8}$ and $x_2 = \frac{1}{2}$
The sum of the fourth powers of the roots is:
$(\frac{1}{8})^4 + (\frac{1}{2})^4 = \frac{1}{4096} + \frac{1}{16}$
$= \frac{1 + 256}{4096} = \frac{257}{4096}$

(B) Given equation: $|x-2|^2+|x-2|-2=0$
Let $y = |x-2|$. Since $|x-2| \geq 0$,we must have $y \geq 0$.
The equation becomes $y^2 + y - 2 = 0$.
Factoring the quadratic: $(y+2)(y-1) = 0$.
This gives $y = -2$ or $y = 1$.
Since $y \geq 0$,we discard $y = -2$.
Thus,$|x-2| = 1$.
This implies $x-2 = 1$ or $x-2 = -1$.
Solving for $x$: $x = 3$ or $x = 1$.
The sum of the real roots is $3 + 1 = 4$.