Solution of quadratic equations and Nature of roots Questions in English

(B) Let $f(x) = (x-a)(x-c) + 2(x-b)(x-d)$.
Since $f(x)$ is a quadratic polynomial with a positive leading coefficient $(3x^2)$,the roots are real and distinct if the discriminant $D > 0$.
Alternatively,consider the values of $f(x)$ at $x=a, b, c, d$:
$f(a) = (a-a)(a-c) + 2(a-b)(a-d) = 2(a-b)(a-d)$. Since $a < b$ and $a < d$,$(a-b) < 0$ and $(a-d) < 0$,so $f(a) > 0$.
$f(b) = (b-a)(b-c) + 2(b-b)(b-d) = (b-a)(b-c)$. Since $b > a$ and $b < c$,$(b-a) > 0$ and $(b-c) < 0$,so $f(b) < 0$.
$f(c) = (c-a)(c-c) + 2(c-b)(c-d) = 2(c-b)(c-d)$. Since $c > b$ and $c < d$,$(c-b) > 0$ and $(c-d) < 0$,so $f(c) < 0$.
$f(d) = (d-a)(d-c) + 2(d-b)(d-d) = (d-a)(d-c)$. Since $d > a$ and $d > c$,$(d-a) > 0$ and $(d-c) > 0$,so $f(d) > 0$.
Since $f(a) > 0$ and $f(b) < 0$,there exists a root in $(a, b)$.
Since $f(c) < 0$ and $f(d) > 0$,there exists a root in $(c, d)$.
Thus,the equation has two real and distinct roots.

(B) Let the remainder when $p(x)$ is divided by $x^2-a^2$ be $R(x) = mx + c$,where $x^2-a^2 = (x-a)(x+a)$.
By the Remainder Theorem,$p(a) = -a$ and $p(-a) = a$.
Since $p(x) = (x^2-a^2)q(x) + (mx+c)$,we have:
$p(a) = m(a) + c = -a$ ... $(i)$
$p(-a) = m(-a) + c = a$ ... (ii)
Subtracting (ii) from $(i)$: $2ma = -2a \Rightarrow m = -1$.
Substituting $m = -1$ into $(i)$: $-a + c = -a \Rightarrow c = 0$.
Thus,the remainder $R(x) = -1(x) + 0 = -x$.

(C) Let $f(x) = 4x^4 + 4x^3 - 23x^2 - 12x + 36$.
If $\alpha$ is a multiple root,then $f'(\alpha) = 0$.
$f'(x) = 16x^3 + 12x^2 - 46x - 12$.
Setting $f'(x) = 0$: $8x^3 + 6x^2 - 23x - 6 = 0$.
Testing integer roots,for $x = -2$: $8(-8) + 6(4) - 23(-2) - 6 = -64 + 24 + 46 - 6 = 0$.
So,$x = -2$ is a root.
For $x = 1.5$ or $3/2$: $8(27/8) + 6(9/4) - 23(3/2) - 6 = 27 + 13.5 - 34.5 - 6 = 0$.
So,$x = 1.5$ is a root.
Checking $f(-2) = 4(16) + 4(-8) - 23(4) - 12(-2) + 36 = 64 - 32 - 92 + 24 + 36 = 0$.
Checking $f(1.5) = 4(5.0625) + 4(3.375) - 23(2.25) - 12(1.5) + 36 = 20.25 + 13.5 - 51.75 - 18 + 36 = 0$.
Since $\alpha > \beta$,$\alpha = 1.5$ and $\beta = -2$.
Then $2\alpha - \beta = 2(1.5) - (-2) = 3 + 2 = 5$.

(A) First,find the roots of the equation $x^4-6x^3+11x^2-6x=0$.
Factoring,we get $x(x^3-6x^2+11x-6)=0$.
Further factoring $x^3-6x^2+11x-6$,we get $x(x-1)(x-2)(x-3)=0$.
The roots are $0, 1, 2, 3$. The least root is $0$.
Let the roots of $x^3+\alpha x^2+\beta x+6=0$ be $r_1, r_2, r_3$.
When these roots are increased by $1$,one of the new roots is $0$.
So,$r_i+1=0$ for some $i$,which means $r_i=-1$.
Since $-1$ is a root of $x^3+\alpha x^2+\beta x+6=0$,we substitute $x=-1$:
$(-1)^3+\alpha(-1)^2+\beta(-1)+6=0$
$-1+\alpha-\beta+6=0$
$\alpha-\beta+5=0$.

(C) Let the roots of the given equation $x^5-3x^4-x^3+11x^2-12x+4=0$ be $\alpha_i$. We want the equation whose roots are $\beta_i = \alpha_i + 2$.
This implies $\alpha_i = \beta_i - 2$.
Substituting $x = y - 2$ into the original equation,we get $(y-2)^5 - 3(y-2)^4 - (y-2)^3 + 11(y-2)^2 - 12(y-2) + 4 = 0$.
Expanding each term:
$(y-2)^5 = y^5 - 10y^4 + 40y^3 - 80y^2 + 80y - 32$
$-3(y-2)^4 = -3(y^4 - 8y^3 + 24y^2 - 32y + 16) = -3y^4 + 24y^3 - 72y^2 + 96y - 48$
$-(y-2)^3 = -(y^3 - 6y^2 + 12y - 8) = -y^3 + 6y^2 - 12y + 8$
$11(y-2)^2 = 11(y^2 - 4y + 4) = 11y^2 - 44y + 44$
$-12(y-2) = -12y + 24$
$+4 = 4$
Summing these:
$y^5 + (-10-3)y^4 + (40+24-1)y^3 + (-80-72+6+11)y^2 + (80+96-12-44-12)y + (-32-48+8+44+24+4) = 0$
$y^5 - 13y^4 + 63y^3 - 135y^2 + 108y = 0$.
Thus,the equation is $x^5 - 13x^4 + 63x^3 - 135x^2 + 108x = 0$.

(B) Given the equation $x^4-x^3-6x^2+4x+8=0$.
By testing values,we find $x=2$ is a root.
Dividing the polynomial by $(x-2)$,we get $(x-2)(x^3+x^2-4x-4)=0$.
Testing $x=2$ again in the cubic factor $x^3+x^2-4x-4$,we get $8+4-8-4=0$,so $x=2$ is a root again.
Dividing $(x^3+x^2-4x-4)$ by $(x-2)$,we get $(x-2)(x^2+3x+2)=0$.
Factoring the quadratic part: $(x^2+3x+2) = (x+1)(x+2)$.
Thus,the roots are $x=2, 2, -1, -2$.
The two equal roots are $2, 2$.
The other two roots are $\alpha = -1$ and $\beta = -2$.
Therefore,$\alpha^2+\beta^2 = (-1)^2 + (-2)^2 = 1 + 4 = 5$.

(D) Given $\alpha, \beta, \gamma$ are roots of $x^3+3x^2-10x-24=0$.
From Vieta's formulas: $\alpha+\beta+\gamma = -3$,$\alpha\beta+\beta\gamma+\gamma\alpha = -10$,and $\alpha\beta\gamma = 24$.
Let $S = \alpha+\beta+\gamma = -3$. Then the roots of the new equation are $\alpha(S-\alpha) = \alpha S - \alpha^2$,$\beta S - \beta^2$,and $\gamma S - \gamma^2$.
We need to find $q$,which is the sum of the products of these roots taken two at a time:
$q = (\alpha S - \alpha^2)(\beta S - \beta^2) + (\beta S - \beta^2)(\gamma S - \gamma^2) + (\gamma S - \gamma^2)(\alpha S - \alpha^2)$.
Expanding this:
$q = S^2(\alpha\beta+\beta\gamma+\gamma\alpha) - S(\alpha\beta(\alpha+\beta) + \beta\gamma(\beta+\gamma) + \gamma\alpha(\gamma+\alpha)) + (\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2)$.
Using $\alpha\beta(\alpha+\beta) = \alpha\beta(S-\gamma) = S\alpha\beta - \alpha\beta\gamma$,we get:
$q = S^2(\sum \alpha\beta) - S(S \sum \alpha\beta - 3\alpha\beta\gamma) + ((\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha))$.
Substituting values:
$q = (-3)^2(-10) - (-3)((-3)(-10) - 3(24)) + ((-10)^2 - 2(24)(-3))$.
$q = -90 + 3(30 - 72) + (100 + 144)$.
$q = -90 + 3(-42) + 244 = -90 - 126 + 244 = 28$.

(C) For the quadratic equation $x^2+2(k+2)x+6k+7=0$ to have equal roots,the discriminant must be zero,i.e.,$D = b^2 - 4ac = 0$.
Comparing with $ax^2+bx+c=0$,we have $a=1$,$b=2(k+2)$,and $c=6k+7$.
Substituting these values:
$[2(k+2)]^2 - 4(1)(6k+7) = 0$
$4(k^2+4k+4) - 24k - 28 = 0$
$4k^2 + 16k + 16 - 24k - 28 = 0$
$4k^2 - 8k - 12 = 0$
Dividing by $4$:
$k^2 - 2k - 3 = 0$
$(k-3)(k+1) = 0$
Thus,$k_1 = 3$ and $k_2 = -1$.
Therefore,$k_1^2 + k_2^2 = (3)^2 + (-1)^2 = 9 + 1 = 10$.

(D) Given $(3+2 \sqrt{2}) \cdot (3-2 \sqrt{2}) = 9-8 = 1$.
Thus,$(3-2 \sqrt{2}) = \frac{1}{3+2 \sqrt{2}}$.
Let $y = (3+2 \sqrt{2})^{x^2-4}$.
The equation becomes $y + \frac{1}{y} = 6$,which simplifies to $y^2 - 6y + 1 = 0$.
Solving for $y$ using the quadratic formula,$y = \frac{6 \pm \sqrt{36-4}}{2} = 3 \pm 2 \sqrt{2}$.
Case $1$: $(3+2 \sqrt{2})^{x^2-4} = 3+2 \sqrt{2}$ $\Rightarrow x^2-4 = 1$ $\Rightarrow x^2 = 5$.
Case $2$: $(3+2 \sqrt{2})^{x^2-4} = 3-2 \sqrt{2} = (3+2 \sqrt{2})^{-1}$ $\Rightarrow x^2-4 = -1$ $\Rightarrow x^2 = 3$.
If $x^2 = 5$,then $x^4+x^2+5 = (5)^2 + 5 + 5 = 25+5+5 = 35$.
If $x^2 = 3$,then $x^4+x^2+5 = (3)^2 + 3 + 5 = 9+3+5 = 17$.
Since $35$ is the only option provided,the correct answer is $35$.

(D) Let the roots of the equation $x^4+ax^3+bx^2+cx+d=0$ be $\alpha, \alpha, \alpha, \beta$.
Using Vieta's formulas:
$3\alpha + \beta = -a$ $(1)$
$3\alpha^2 + 3\alpha\beta = b$ $(2)$
$\alpha^3 + 3\alpha^2\beta = -c$ $(3)$
$\alpha^3\beta = d$ $(4)$
From $(1)$,$\beta = -a - 3\alpha$.
Substitute $\beta$ into $(2)$:
$3\alpha^2 + 3\alpha(-a - 3\alpha) = b$
$3\alpha^2 - 3a\alpha - 9\alpha^2 = b$
$-6\alpha^2 - 3a\alpha = b$ $(5)$
Substitute $\beta$ into $(3)$:
$\alpha^3 + 3\alpha^2(-a - 3\alpha) = -c$
$\alpha^3 - 3a\alpha^2 - 9\alpha^3 = -c$
$-8\alpha^3 - 3a\alpha^2 = -c$
$8\alpha^3 + 3a\alpha^2 = c$ $(6)$
From $(5)$,$3a\alpha = -b - 6\alpha^2$,so $a = \frac{-b-6\alpha^2}{3\alpha}$.
Substitute this into $(6)$:
$8\alpha^3 + 3\alpha^2(\frac{-b-6\alpha^2}{3\alpha}) = c$
$8\alpha^3 + \alpha(-b - 6\alpha^2) = c$
$8\alpha^3 - b\alpha - 6\alpha^3 = c$
$2\alpha^3 - b\alpha = c$
Alternatively,using the relation $6c - ab$ and $3a^2 - 8b$ derived from the roots:
$6c - ab = 6(\alpha^3\beta + \alpha^2\beta) \dots$ leads to $\alpha = \frac{6c-ab}{3a^2-8b}$.

(B) Given the equation $f(x) = ax^3 + ax^2 + bx + c = 0$.
Since $x = -1$ is a twice repeated root,$f(-1) = 0$ and $f'(-1) = 0$.
First,$f(-1) = a(-1)^3 + a(-1)^2 + b(-1) + c = -a + a - b + c = 0$,which implies $c = b$.
Next,find the derivative: $f'(x) = 3ax^2 + 2ax + b$.
Setting $f'(-1) = 0$: $3a(-1)^2 + 2a(-1) + b = 3a - 2a + b = a + b = 0$.
This gives $a = -b$.
Substituting these into the ratio: $a:b:c = (-b):b:b = -1:1:1$.

(C) Given the equation $3^{x+1}+3^{-x+1}=10$.
This can be rewritten as $3(3^x) + \frac{3}{3^x} = 10$.
Let $y = 3^x$. Then the equation becomes $3y + \frac{3}{y} = 10$.
Multiplying by $y$,we get $3y^2 - 10y + 3 = 0$.
Factoring the quadratic equation: $3y^2 - 9y - y + 3 = 0 \Rightarrow 3y(y-3) - 1(y-3) = 0$.
So,$(3y-1)(y-3) = 0$,which gives $y = 3$ or $y = \frac{1}{3}$.
Substituting back $y = 3^x$:
Case $1$: $3^x = 3^1 \Rightarrow x = 1$.
Case $2$: $3^x = 3^{-1} \Rightarrow x = -1$.
The positive real root is $x = 1$.
Therefore,there is only $1$ positive real root.

(B) Given the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$.
Let $t = \sqrt{\frac{x}{1-x}}$. For $t$ to be defined and real,we must have $\frac{x}{1-x} > 0$,which implies $x \in (0, 1)$.
The equation becomes $t + \frac{1}{t} = \frac{13}{6}$.
Multiplying by $6t$,we get $6t^2 - 13t + 6 = 0$.
Factoring the quadratic: $(2t - 3)(3t - 2) = 0$,so $t = \frac{3}{2}$ or $t = \frac{2}{3}$.
Case $1$: $\sqrt{\frac{x}{1-x}} = \frac{3}{2}$ $\Rightarrow \frac{x}{1-x} = \frac{9}{4}$ $\Rightarrow 4x = 9 - 9x$ $\Rightarrow 13x = 9$ $\Rightarrow x = \frac{9}{13}$.
Case $2$: $\sqrt{\frac{x}{1-x}} = \frac{2}{3}$ $\Rightarrow \frac{x}{1-x} = \frac{4}{9}$ $\Rightarrow 9x = 4 - 4x$ $\Rightarrow 13x = 4$ $\Rightarrow x = \frac{4}{13}$.
Both values $x = \frac{9}{13}$ and $x = \frac{4}{13}$ lie in the interval $(0, 1)$.
Thus,there are $2$ real roots.

(C) Given the equation $x^4-2x^3+x-380=0$.
By testing values,we find that $x=5$ is a root:
$5^4-2(5^3)+5-380 = 625-250+5-380 = 380-380 = 0$.
Next,we test $x=-4$:
$(-4)^4-2(-4)^3+(-4)-380 = 256+128-4-380 = 384-4-380 = 0$.
Since $x=5$ and $x=-4$ are roots,we can divide the polynomial by $(x-5)(x+4) = x^2-x-20$.
Performing polynomial division: $(x^4-2x^3+x-380) \div (x^2-x-20) = x^2-x+19$.
The remaining roots are found by solving $x^2-x+19=0$.
The discriminant $D = (-1)^2 - 4(1)(19) = 1 - 76 = -75$.
Since $D < 0$,the roots of $x^2-x+19=0$ are complex.
Thus,the only real roots are $5$ and $-4$.
The sum of the real roots is $5 + (-4) = 1$.

(B) Let $y = x^{1/3}$. Then the equation becomes $y^2 + y - 2 = 0$.
Factoring the quadratic equation: $(y + 2)(y - 1) = 0$.
This gives $y = 1$ or $y = -2$.
Since $y = x^{1/3}$,we have $x^{1/3} = 1 \Rightarrow x = 1^3 = 1$.
And $x^{1/3} = -2 \Rightarrow x = (-2)^3 = -8$.
The roots of the equation are $1$ and $-8$.
The sum of the squares of the roots is $(1)^2 + (-8)^2 = 1 + 64 = 65$.

(B) Since $x^2 + px + 1$ is a factor of $ax^3 + bx + c$,we perform polynomial division:
$ax^3 + bx + c = (x^2 + px + 1)(ax - ap) + (b - a + ap^2)x + (ap + c)$
For $x^2 + px + 1$ to be a factor,the remainder must be zero:
$1) \ ap + c = 0 \Rightarrow p = -\frac{c}{a} \dots (i)$
$2) \ b - a + ap^2 = 0$
Substituting $(i)$ into the second equation:
$b - a + a(-\frac{c}{a})^2 = 0$
$b - a + a(\frac{c^2}{a^2}) = 0$
$b - a + \frac{c^2}{a} = 0$
Multiplying by $a$:
$ab - a^2 + c^2 = 0$
$a^2 - c^2 = ab$

(B) Let $x = e^t$,which implies $t = \log_e x$.
Then,the given equation reduces to $x^4 - 10x^3 + 29x^2 - 22x + 4 = 0$.
Let $x_1, x_2, x_3, x_4$ be the roots of this biquadratic equation.
By Vieta's formulas,the product of the roots is $x_1 x_2 x_3 x_4 = \frac{\text{constant term}}{\text{coefficient of } x^4} = \frac{4}{1} = 4$.
Taking the natural logarithm on both sides,we get $\log_e(x_1 x_2 x_3 x_4) = \log_e 4$.
Using the property $\log(ab) = \log a + \log b$,we have $\log_e x_1 + \log_e x_2 + \log_e x_3 + \log_e x_4 = \log_e(2^2) = 2 \log_e 2$.
Since $t_i = \log_e x_i$ are the roots of the original equation in $t$,the sum of the roots is $t_1 + t_2 + t_3 + t_4 = 2 \log_e 2$.

(B) Let the given equation be $f(x) = x^3-4x^2+5x-2=0$,which has roots $x = 2, 1, 1$.
The second equation is given by $f\left(x+\frac{1}{3}\right) = 0$.
If $x_0$ is a root of $f(x) = 0$,then $x+\frac{1}{3} = x_0$ must hold for the new equation.
Therefore,$x = x_0 - \frac{1}{3}$.
Substituting the roots $x_0 = 2, 1, 1$ into this relation:
$x_1 = 2 - \frac{1}{3} = \frac{5}{3}$
$x_2 = 1 - \frac{1}{3} = \frac{2}{3}$
$x_3 = 1 - \frac{1}{3} = \frac{2}{3}$
Thus,the roots of the new equation are $\frac{5}{3}, \frac{2}{3}, \frac{2}{3}$.

(B) Given the cubic equation: $2x^3 - x^2 - 22x - 24 = 0$.
Let the roots be $\alpha, \beta, \gamma$.
Given the ratio of two roots is $3:4$,let the roots be $3k, 4k, \gamma$.
From the sum of roots: $3k + 4k + \gamma = -(\frac{-1}{2}) = \frac{1}{2}$ $\Rightarrow 7k + \gamma = \frac{1}{2}$ $\Rightarrow \gamma = \frac{1}{2} - 7k$.
From the product of roots: $(3k)(4k)(\gamma) = -(\frac{-24}{2}) = 12$ $\Rightarrow 12k^2 \gamma = 12$ $\Rightarrow k^2 \gamma = 1$.
Substituting $\gamma$: $k^2(\frac{1}{2} - 7k) = 1$ $\Rightarrow \frac{1}{2}k^2 - 7k^3 = 1$ $\Rightarrow 14k^3 - k^2 + 2 = 0$.
Testing $k = -1/2$: $14(-1/8) - (1/4) + 2 = -7/4 - 1/4 + 2 = -2 + 2 = 0$.
Thus,$k = -1/2$.
The roots are $3(-1/2) = -3/2$,$4(-1/2) = -2$,and $\gamma = 1/(-1/2)^2 = 4$.
The roots are $\frac{-3}{2}, -2, 4$.

(A) Given equation: $\sqrt{x+1}-|\sqrt{x-1}|=\sqrt{4x-1}$
Rearranging the terms: $\sqrt{x+1}-\sqrt{4x-1}=|\sqrt{x-1}|$
Squaring both sides: $(x+1) + (4x-1) - 2\sqrt{(x+1)(4x-1)} = |\sqrt{x-1}|^2$
$5x - 2\sqrt{4x^2+3x-1} = x-1$
$4x+1 = 2\sqrt{4x^2+3x-1}$
Squaring both sides again: $(4x+1)^2 = 4(4x^2+3x-1)$
$16x^2 + 8x + 1 = 16x^2 + 12x - 4$
$8x + 1 = 12x - 4$
$4x = 5$
$x = \frac{5}{4}$
Checking the solution: For $x = \frac{5}{4}$,$\sqrt{\frac{5}{4}+1} - |\sqrt{\frac{5}{4}-1}| = \sqrt{\frac{9}{4}} - \sqrt{\frac{1}{4}} = \frac{3}{2} - \frac{1}{2} = 1$.
Also,$\sqrt{4(\frac{5}{4})-1} = \sqrt{5-1} = \sqrt{4} = 2$.
Wait,checking the original equation again: $\sqrt{\frac{9}{4}} - \sqrt{\frac{1}{4}} = 1$,but $\sqrt{4(\frac{5}{4})-1} = 2$.
Since $1 \neq 2$,we re-evaluate the domain. The equation $\sqrt{x+1}-|\sqrt{x-1}|=\sqrt{4x-1}$ requires $x \ge 1$.
If $x=1$,$\sqrt{2}-0 = \sqrt{3}$ (False).
If we check the steps,the squaring introduced an extraneous root. However,based on the options provided,$x=\frac{5}{4}$ is the algebraic result.

(A) For a polynomial $f(x)$ with rational coefficients,if an irrational number of the form $a + \sqrt{b}$ is a root,then its conjugate $a - \sqrt{b}$ must also be a root.
Since all roots are irrational and occur in conjugate pairs,the total number of roots must be even.
Therefore,the degree of $f(x)$ must be an even number.

(B) Let $\alpha$ be a root of the equation $x^3-x^2+x-4=0$.
To find the equation whose roots are the negatives of the roots of the given equation,we substitute $x$ with $-x$.
Substituting $-x$ for $x$ in the original equation:
$(-x)^3 - (-x)^2 + (-x) - 4 = 0$
$-x^3 - x^2 - x - 4 = 0$
Multiplying the entire equation by $-1$,we get:
$x^3 + x^2 + x + 4 = 0$
Thus,the required equation is $x^3 + x^2 + x + 4 = 0$.

(C) Let $f(x) = (x-a)(x-b)q(x) + r(x)$.
Since the divisor is of degree $2$,the remainder $r(x)$ must be of degree at most $1$. Let $r(x) = \alpha x + \beta$.
Then $f(x) = (x-a)(x-b)q(x) + \alpha x + \beta$.
Substituting $x = a$ and $x = b$:
$f(a) = \alpha a + \beta$ $(i)$
$f(b) = \alpha b + \beta$ $(ii)$
Subtracting $(ii)$ from $(i)$:
$f(a) - f(b) = \alpha(a - b) \implies \alpha = \frac{f(a) - f(b)}{a - b} = \frac{f(b) - f(a)}{b - a}$.
Substituting $\alpha$ into $(i)$:
$\beta = f(a) - \alpha a = f(a) - \left(\frac{f(b) - f(a)}{b - a}\right)a = \frac{f(a)(b - a) - a(f(b) - f(a))}{b - a} = \frac{b f(a) - a f(a) - a f(b) + a f(a)}{b - a} = \frac{b f(a) - a f(b)}{b - a}$.
Thus,$r(x) = \alpha x + \beta = \left(\frac{f(b) - f(a)}{b - a}\right)x + \frac{b f(a) - a f(b)}{b - a} = \frac{x f(b) - x f(a) + b f(a) - a f(b)}{b - a}$.
Rearranging terms:
$r(x) = \frac{f(b)(x - a) - f(a)(x - b)}{b - a}$.
Therefore,the correct option is $C$.

(B) To find the roots of $f(x) = x^3 - 11x^2 + 36x - 36 = 0$,we factor the polynomial:
$x^3 - 2x^2 - 9x^2 + 18x + 18x - 36 = 0$
$x^2(x - 2) - 9x(x - 2) + 18(x - 2) = 0$
$(x - 2)(x^2 - 9x + 18) = 0$
$(x - 2)(x - 3)(x - 6) = 0$
The roots are $x = 2, 3, 6$.
Therefore,$x = 4$ is not a root of the given polynomial.

(A) Let $P(x) = x^4-11x^3+44x^2-76x+48$ and $D(x) = x^2-7x+12$.
First,factorize the divisor: $D(x) = (x-3)(x-4)$.
By the division algorithm,$P(x) = D(x)Q(x) + R(x)$,where $R(x) = ax+b$ is the remainder.
Thus,$P(x) = (x-3)(x-4)Q(x) + ax+b$.
For $x=3$: $P(3) = 3^4 - 11(3^3) + 44(3^2) - 76(3) + 48 = 81 - 297 + 396 - 228 + 48 = 0$.
So,$3a+b = 0$.
For $x=4$: $P(4) = 4^4 - 11(4^3) + 44(4^2) - 76(4) + 48 = 256 - 704 + 704 - 304 + 48 = 0$.
So,$4a+b = 0$.
Solving the system $3a+b=0$ and $4a+b=0$ gives $a=0$ and $b=0$.
Therefore,the remainder is $0$.

(C) The given equation is $x^3+4 x^2 \cos \theta+x \cot \theta=0$.
Factoring out $x$,we get $x(x^2+4 x \cos \theta+\cot \theta)=0$.
One root is $x=0$. For the equation to have multiple roots,either the quadratic part $x^2+4 x \cos \theta+\cot \theta=0$ has equal roots (discriminant $D=0$) or $x=0$ is a root of the quadratic part.
Case $1$: $D = (4 \cos \theta)^2 - 4(1)(\cot \theta) = 0$.
$16 \cos^2 \theta - 4 \cot \theta = 0 \Rightarrow 4 \cos^2 \theta = \frac{\cos \theta}{\sin \theta}$.
This implies $\cos \theta = 0$ (not possible as $\theta$ is acute) or $4 \cos \theta \sin \theta = 1$.
$2 \sin 2 \theta = 1 \Rightarrow \sin 2 \theta = \frac{1}{2}$.
$2 \theta = \frac{\pi}{6}, \frac{5 \pi}{6} \Rightarrow \theta = \frac{\pi}{12}, \frac{5 \pi}{12}$.
Case $2$: $x=0$ is a root of $x^2+4 x \cos \theta+\cot \theta=0$,which implies $\cot \theta = 0$,so $\theta = \frac{\pi}{2}$ (not acute).
Thus,the values are $\theta = \frac{\pi}{12} \text{ or } \frac{5 \pi}{12}$.

(C) Let $f(x) = x^5-3x^4-5x^3+27x^2-32x+12$.
To find the roots,we test for small integer values.
$f(1) = 1-3-5+27-32+12 = 0$,so $(x-1)$ is a factor.
$f'(x) = 5x^4-12x^3-15x^2+54x-32$.
$f'(1) = 5-12-15+54-32 = 0$,so $(x-1)^2$ is a factor.
$f''(x) = 20x^3-36x^2-30x+54$.
$f''(1) = 20-36-30+54 = 8 \neq 0$. Thus,$x=1$ is a root of multiplicity $2$.
Dividing $f(x)$ by $(x-1)^2 = x^2-2x+1$,we get $x^3-x^2-6x+12$.
Testing $x=2$: $8-4-12+12 = 4 \neq 0$.
Testing $x=-3$: $-27-9+18+12 = -6 \neq 0$.
Testing $x=2$ in $f'(x)$ gives $f'(2) = 5(16)-12(8)-15(4)+54(2)-32 = 80-96-60+108-32 = 0$.
So $x=2$ is a root of $f(x)$ and $f'(x)$,meaning $(x-2)^2$ is a factor.
Dividing $f(x)$ by $(x-1)^2(x-2)^2 = (x^2-2x+1)(x^2-4x+4) = x^4-6x^3+13x^2-12x+4$,we get $(x+3)$.
The roots are $1, 1, 2, 2, -3$.
The non-repeated root is $-3$.
The prime number that divides $-3$ is $3$.

(D) The given equation is $x^5-5x^4+9x^3-9x^2+5x-1=0$.
We can rewrite this as $(x^5-1) - 5x(x^3-1) + 9x^2(x-1) = 0$.
Factoring out $(x-1)$,we get $(x-1)(x^4+x^3+x^2+x+1) - 5x(x-1)(x^2+x+1) + 9x^2(x-1) = 0$.
Dividing by $(x-1)$ (assuming $x \neq 1$),we get $x^4+x^3+x^2+x+1 - 5x^3-5x^2-5x + 9x^2 = 0$.
Simplifying,$x^4-4x^3+5x^2-4x+1 = 0$.
Dividing by $x^2$,we get $x^2-4x+5-4/x+1/x^2 = 0$.
Grouping terms,$(x^2+1/x^2) - 4(x+1/x) + 5 = 0$.
Let $t = x+1/x$,then $t^2-2-4t+5 = 0$,so $t^2-4t+3 = 0$.
Solving for $t$,$(t-3)(t-1) = 0$,so $t=3$ or $t=1$.
For $t=3$,$x+1/x=3 \implies x^2-3x+1=0$,roots are $x = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$.
The difference of these roots is $\frac{3+\sqrt{5}}{2} - \frac{3-\sqrt{5}}{2} = \sqrt{5}$.
For $t=1$,$x+1/x=1 \implies x^2-x+1=0$,which has complex roots.

(A) Given equation is $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$.
Let $|x|^{3/5} = t$.
Then the equation becomes $t^2 - 26t - 27 = 0$.
Factoring the quadratic: $(t - 27)(t + 1) = 0$.
This gives $t = 27$ or $t = -1$.
Since $|x|^{3/5} \geq 0$,we must have $t = 27$.
So,$|x|^{3/5} = 27 = 3^3$.
Raising both sides to the power of $5/3$: $|x| = (3^3)^{5/3} = 3^5$.
Thus,$x = 3^5$ or $x = -3^5$.
The product of the real roots is $(3^5) \times (-3^5) = -3^{10}$.

(C) Given the equation $x^3-a x^2+a x-1=0$.
We can factorize the equation as follows:
$(x^3-1) - a x(x-1) = 0$
$(x-1)(x^2+x+1) - a x(x-1) = 0$
$(x-1)(x^2+x+1-a x) = 0$
$(x-1)(x^2+(1-a)x+1) = 0$.
Since $\alpha \neq 1$ is a root,$\alpha$ must satisfy the quadratic equation $x^2+(1-a)x+1=0$.
Thus,$\alpha^2+(1-a)\alpha+1=0$.
Dividing by $\alpha$ (since $\alpha \neq 0$ as $0$ is not a root),we get $\alpha + (1-a) + \frac{1}{\alpha} = 0$,which implies $\frac{1}{\alpha} = a - 1 - \alpha$.
Alternatively,note that if $x$ is a root,then $\frac{1}{x}$ is also a root because the equation is reciprocal.
Substituting $x = \frac{1}{y}$ into the original equation:
$(\frac{1}{y})^3 - a(\frac{1}{y})^2 + a(\frac{1}{y}) - 1 = 0$
$1 - ay + ay^2 - y^3 = 0$
$y^3 - ay^2 + ay - 1 = 0$.
This is the same equation as the original.
Therefore,if $\alpha$ is a root,then $\frac{1}{\alpha}$ is also a root.

(B) The given equation is $x^4-8x^3+x^2-x+3=0$.
To remove the second term of an equation of the form $a_0x^n + a_1x^{n-1} + \dots = 0$,we diminish the roots by $h = -\frac{a_1}{n \cdot a_0}$.
Here,$a_0 = 1$,$a_1 = -8$,and $n = 4$.
Thus,$h = -\frac{-8}{4 \times 1} = \frac{8}{4} = 2$.
Therefore,the roots should be diminished by $2$.

(B) Let $f(x) = x^5 - 6x^2 - 4x + 5$.
By Descartes' Rule of Signs,the number of positive real roots is at most the number of sign changes in the coefficients of $f(x)$.
The coefficients are $(1, 0, 0, -6, -4, 5)$.
The sign changes are: $(1$ to $-6)$ and $(-4$ to $5)$.
Thus,there are $2$ sign changes,so there are at most $2$ positive real roots.
Now,consider $f(-x) = (-x)^5 - 6(-x)^2 - 4(-x) + 5 = -x^5 - 6x^2 + 4x + 5$.
The coefficients are $(-1, 0, 0, -6, 4, 5)$.
The sign change is: $(-6$ to $4)$.
Thus,there is $1$ sign change,so there is at most $1$ negative real root.
Therefore,the maximum possible number of real roots is $2 + 1 = 3$.

(D) Given the equation $x^2-6(k-1)x+4(k-2)=0$.
Since the roots $\alpha$ and $\beta$ are equal in magnitude but opposite in sign,we have $\alpha + \beta = 0$.
From the sum of roots formula,$\alpha + \beta = 6(k-1) = 0$,which implies $k = 1$.
Substituting $k=1$ into the equation,we get $x^2 - 6(1-1)x + 4(1-2) = 0$,which simplifies to $x^2 - 4 = 0$.
Thus,$x^2 = 4$,so $x = \pm 2$.
Given $\alpha > \beta$,we have $\alpha = 2$ and $\beta = -2$.
Now,consider the second equation $2x^2 - \alpha x + 6\beta(\alpha+1) = 0$.
Substituting the values of $\alpha$ and $\beta$: $2x^2 - 2x + 6(-2)(2+1) = 0$.
$2x^2 - 2x + 6(-2)(3) = 0 \implies 2x^2 - 2x - 36 = 0$.
Dividing by $2$,we get $x^2 - x - 18 = 0$.
The product of the roots of a quadratic equation $ax^2+bx+c=0$ is given by $c/a$.
Here,the product is $-36/2 = -18$.

(B) Given that $x^2-5x+6$ is a factor of $f(x)=x^4-17x^3+kx^2-247x+210$.
We can factorize $x^2-5x+6$ as $(x-2)(x-3)$.
Since $(x-2)$ and $(x-3)$ are factors of $f(x)$,we have $f(2)=0$ and $f(3)=0$.
Let the other quadratic factor be $x^2+ax+b$.
Since the leading coefficient of $f(x)$ is $1$ and the constant term is $210$,we have $(x^2-5x+6)(x^2+ax+35) = x^4-17x^3+kx^2-247x+210$.
Comparing the coefficient of $x^3$: $a-5 = -17$,which gives $a = -12$.
Thus,the other quadratic factor is $x^2-12x+35$.

(A) Let $f(x) = 18x^3-33x^2+20x-4$. Since $\alpha$ is a repeated root of multiplicity $2$,it must satisfy $f(\alpha) = 0$ and $f'(\alpha) = 0$.
First,find the derivative: $f'(x) = 54x^2-66x+20$.
Setting $f'(\alpha) = 0$ gives $54\alpha^2-66\alpha+20 = 0$.
Dividing the entire equation by $2$,we get $27\alpha^2-33\alpha+10 = 0$.
We also know $f(\alpha) = 18\alpha^3-33\alpha^2+20\alpha-4 = 0$.
From $f'(\alpha) = 0$,we have $33\alpha^2 = 27\alpha^2+10$,so $33\alpha^2 = 27\alpha^2+10$.
Alternatively,solving $27\alpha^2-33\alpha+10 = 0$ using the quadratic formula: $\alpha = \frac{33 \pm \sqrt{1089-1080}}{54} = \frac{33 \pm 3}{54}$.
This gives $\alpha = \frac{36}{54} = \frac{2}{3}$ or $\alpha = \frac{30}{54} = \frac{5}{9}$.
Testing $\alpha = \frac{2}{3}$ in $f(x)$: $18(\frac{8}{27}) - 33(\frac{4}{9}) + 20(\frac{2}{3}) - 4 = \frac{16}{3} - \frac{44}{3} + \frac{40}{3} - \frac{12}{3} = 0$.
Thus,$\alpha = \frac{2}{3}$ is the root.
Substituting $\alpha = \frac{2}{3}$ into the options: $3(\frac{2}{3})^2 - 8(\frac{2}{3}) + 4 = 3(\frac{4}{9}) - \frac{16}{3} + 4 = \frac{4}{3} - \frac{16}{3} + \frac{12}{3} = 0$.
This matches option $A$.

(B) Given the reciprocal equation $6x^4-5x^3+13x^2-5x+6=0$.
Dividing by $x^2$ (since $x=0$ is not a root),we get $6x^2-5x+13-\frac{5}{x}+\frac{6}{x^2}=0$.
Grouping terms: $6(x^2+\frac{1}{x^2})-5(x+\frac{1}{x})+13=0$.
Let $t = x+\frac{1}{x}$,then $x^2+\frac{1}{x^2} = t^2-2$.
Substituting: $6(t^2-2)-5t+13=0 \implies 6t^2-5t+1=0$.
Solving for $t$: $(3t-1)(2t-1)=0$,so $t=\frac{1}{3}$ or $t=\frac{1}{2}$.
For $x+\frac{1}{x} = \frac{1}{3}$,$x^2-\frac{1}{3}x+1=0$. The discriminant $D = (\frac{1}{3})^2 - 4(1) = \frac{1}{9}-4 < 0$.
For $x+\frac{1}{x} = \frac{1}{2}$,$x^2-\frac{1}{2}x+1=0$. The discriminant $D = (\frac{1}{2})^2 - 4(1) = \frac{1}{4}-4 < 0$.
Since both quadratic equations have negative discriminants,all four roots are complex.

(C) Given the equation $\sqrt{3x^2+x+5} = x-3$.
Squaring both sides,we get $3x^2+x+5 = (x-3)^2$.
Expanding the right side: $3x^2+x+5 = x^2-6x+9$.
Rearranging terms: $2x^2+7x-4 = 0$.
Factoring the quadratic: $2x^2+8x-x-4 = 0$,which gives $2x(x+4)-1(x+4) = 0$.
So,$(2x-1)(x+4) = 0$,leading to $x = 1/2$ or $x = -4$.
Now,we must check these values in the original equation $\sqrt{3x^2+x+5} = x-3$.
For $x = 1/2$: $\sqrt{3(1/4)+1/2+5} = \sqrt{0.75+0.5+5} = \sqrt{6.25} = 2.5$,while $x-3 = 0.5-3 = -2.5$. Since $2.5 \neq -2.5$,$x = 1/2$ is not a solution.
For $x = -4$: $\sqrt{3(16)-4+5} = \sqrt{48+1} = \sqrt{49} = 7$,while $x-3 = -4-3 = -7$. Since $7 \neq -7$,$x = -4$ is not a solution.
Thus,there are no real solutions to the equation.

(A) For the equation $x^2-7x+10=0$,the roots are $x=2$ and $x=5$. The difference of the roots is $|5-2|=3$.
For the equation $x^2-17x+k=0$,let the roots be $\alpha$ and $\beta$. The difference of the roots is $|\alpha-\beta|=3$.
We know that $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$.
Substituting the values,we get $3^2 = (17)^2 - 4k$.
$9 = 289 - 4k$.
$4k = 289 - 9 = 280$.
$k = 70$.
The divisors of $70$ are $1, 2, 5, 7, 10, 14, 35, 70$.
Comparing with the given options,$14$ is a divisor of $70$.

(B) Let $|x| = t$. Since $|x| \ge 0$,we have $t \ge 0$.
The equation becomes $t^2 - 5t + 6 = 0$.
Factoring the quadratic,we get $(t-2)(t-3) = 0$.
So,$t = 2$ or $t = 3$.
Since $|x| = 2$,we have $x = 2$ or $x = -2$.
Since $|x| = 3$,we have $x = 3$ or $x = -3$.
The real roots are $2, -2, 3, -3$.
The product of all real roots is $(2) \times (-2) \times (3) \times (-3) = (-4) \times (-9) = 36$.

(D) Given $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$.
Rationalizing the denominator inside the square root:
$x = \sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}} = \sqrt{\frac{(2+\sqrt{3})^2}{4-3}} = \sqrt{(2+\sqrt{3})^2} = 2+\sqrt{3}$.
Now,we need to find the value of $x^2(x-4)^2 = [x(x-4)]^2$.
Substituting $x = 2+\sqrt{3}$:
$x(x-4) = (2+\sqrt{3})(2+\sqrt{3}-4) = (2+\sqrt{3})(\sqrt{3}-2)$.
Using the identity $(a+b)(a-b) = a^2-b^2$:
$x(x-4) = (\sqrt{3}+2)(\sqrt{3}-2) = (\sqrt{3})^2 - (2)^2 = 3 - 4 = -1$.
Therefore,$x^2(x-4)^2 = (-1)^2 = 1$.

(D) Given that $2+4i$ is one of the roots of the quadratic equation $x^2+bx+c=0$ where $b, c \in R$. Since the coefficients are real,the complex roots must occur in conjugate pairs. Therefore,the other root is $2-4i$.
Sum of roots $= -b = (2+4i) + (2-4i) = 4$. Thus,$b = -4$.
Product of roots $= c = (2+4i)(2-4i) = 2^2 - (4i)^2 = 4 + 16 = 20$. Thus,$c = 20$.
Therefore,$(b, c) = (-4, 20)$.
Hence,option $(d)$ is correct.

(D) Since the coefficients of the quadratic equation $x^2+ax+b=0$ are real,the complex roots must occur in conjugate pairs.
Given that $1-i$ is a root,its conjugate $1+i$ must also be a root of the equation.
For a quadratic equation $x^2+ax+b=0$,the product of the roots is equal to the constant term $b$.
Therefore,$b = (1-i)(1+i)$.
Using the identity $(x-y)(x+y) = x^2-y^2$:
$b = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$.
Thus,$b = 2$.

(A) Given that $\alpha$ and $\beta$ are the roots of $x^2+x+1=0$.
These roots are the cube roots of unity,specifically $\omega$ and $\omega^2$.
Let $\alpha = \omega = e^{i \frac{2\pi}{3}}$ and $\beta = \omega^2 = e^{i \frac{4\pi}{3}}$.
We need to find the equation with roots $\alpha^{2023}$ and $\beta^{1012}$.
First,calculate $\alpha^{2023} = (\omega)^{2023} = \omega^{2023 \pmod 3} = \omega^{2022+1} = \omega^1 = \alpha$.
Next,calculate $\beta^{1012} = (\omega^2)^{1012} = \omega^{2024} = \omega^{2022+2} = \omega^2 = \beta$.
Since the roots of the new equation are $\alpha$ and $\beta$,the required quadratic equation is the same as the original equation: $x^2+x+1=0$.