Solution of quadratic equations and Nature of roots Questions in English

(A) Given equation is $x^{10}-3x^8+5x^6-5x^4+3x^2-1=0$.
Let $f(x) = x^{10}-3x^8+5x^6-5x^4+3x^2-1$.
We can factorize the expression by grouping terms:
$f(x) = (x^{10}-1) - 3x^2(x^6-1) + 5x^4(x^2-1)$.
Using the identity $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \dots + b^{n-1})$,we can see that $(x^2-1)$ is a common factor.
$f(x) = (x^2-1)(x^8+x^6+x^4+x^2+1) - 3x^2(x^2-1)(x^4+x^2+1) + 5x^4(x^2-1)$.
$f(x) = (x^2-1)[(x^8+x^6+x^4+x^2+1) - 3x^2(x^4+x^2+1) + 5x^4]$.
$f(x) = (x^2-1)[x^8+x^6+x^4+x^2+1 - 3x^6-3x^4-3x^2 + 5x^4]$.
$f(x) = (x^2-1)(x^8-2x^6+3x^4-2x^2+1)$.
$f(x) = (x^2-1)(x^4-x^2+1)^2$.
Setting $f(x) = 0$,we get $x^2-1=0$ or $(x^4-x^2+1)^2=0$.
$x^2-1=0 \implies x = \pm 1$ ($2$ real roots).
For $x^4-x^2+1=0$,let $t = x^2$. Then $t^2-t+1=0$.
The discriminant $D = (-1)^2 - 4(1)(1) = -3 < 0$.
Thus,$t = \frac{1 \pm i\sqrt{3}}{2}$.
Since $x^2 = t$,each value of $t$ gives $2$ complex roots for $x$.
Since there are $2$ values of $t$,we get $2 \times 2 = 4$ complex roots from the squared term.
However,the term $(x^4-x^2+1)^2$ implies each root is repeated twice,so there are $8$ non-real roots in total.

(A) Given the quadratic equation $x^2+bx+c=0$ with $b=17$.
Since the roots are $-2$ and $-15$,the sum of roots is $-2 + (-15) = -17$.
From the equation $x^2+bx+c=0$,the sum of roots is $-b$.
Thus,$-b = -17$,which matches $b=17$.
The product of roots is $c = (-2) \times (-15) = 30$.
Now,consider the equation with $b=13$ and $c=30$: $x^2+13x+30=0$.
Factoring the quadratic: $x^2+10x+3x+30=0 \implies (x+10)(x+3)=0$.
The roots are $\alpha = -10$ and $\beta = -3$.
Then,$|\alpha-\beta| = |-10 - (-3)| = |-10+3| = |-7| = 7$.
Therefore,the correct option is $A$.

(B) Given the equation $x^2 - 2cx + ab = 0$ has real and unequal roots,its discriminant $D_1 > 0$.
$D_1 = (-2c)^2 - 4(1)(ab) = 4c^2 - 4ab > 0$,which implies $c^2 > ab$.
Now,consider the equation $x^2 - 2(a + b)x + a^2 + b^2 + 2c^2 = 0$.
The discriminant $D_2$ of this equation is:
$D_2 = [-2(a + b)]^2 - 4(1)(a^2 + b^2 + 2c^2)$
$D_2 = 4(a^2 + 2ab + b^2) - 4a^2 - 4b^2 - 8c^2$
$D_2 = 4a^2 + 8ab + 4b^2 - 4a^2 - 4b^2 - 8c^2$
$D_2 = 8ab - 8c^2 = 8(ab - c^2)$.
Since $c^2 > ab$,it follows that $ab - c^2 < 0$.
Therefore,$D_2 < 0$.
Since the discriminant is negative,the roots of the equation are imaginary.

(D) Assertion $(A)$: Let $f(x) = -x^2+3x+1$.
To find the maximum value,we find the critical point by setting $f'(x) = 0$.
$f'(x) = -2x+3 = 0 \Rightarrow x = \frac{3}{2}$.
Since $f''(x) = -2 < 0$,the function has a maximum at $x = \frac{3}{2}$.
The maximum value is $f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) + 1 = -\frac{9}{4} + \frac{9}{2} + 1 = \frac{-9+18+4}{4} = \frac{13}{4}$.
Since the given assertion states the maximum value is $\frac{11}{4}$,the assertion $(A)$ is false.
Reason $(R)$: For $f(x) = ax^2+bx+c$ with $a < 0$,the vertex occurs at $x = -\frac{b}{2a}$.
$f'(x) = 2ax+b = 0 \Rightarrow x = -\frac{b}{2a}$.
Since $f''(x) = 2a < 0$,this point is indeed a maximum.
Thus,the reason $(R)$ is true.

(D) Let $y = \frac{2(x^2+2x-11)}{2x-5}$.
$y(2x-5) = 2x^2+4x-22$.
$2x^2 + x(4-2y) + (5y-22) = 0$.
Since $x$ is real,the discriminant $D \geq 0$.
$D = (4-2y)^2 - 4(2)(5y-22) \geq 0$.
$16 + 4y^2 - 16y - 40y + 176 \geq 0$.
$4y^2 - 56y + 192 \geq 0$.
$y^2 - 14y + 48 \geq 0$.
$(y-6)(y-8) \geq 0$.
Thus,$y \in (-\infty, 6] \cup [8, \infty)$.
Therefore,no value of the expression lies in the interval $(6,8)$.

(B) Given $y = \frac{11 x^2+12 x+6}{x^2+4 x+2}$.
Rearranging the terms,we get:
$y(x^2+4x+2) = 11x^2+12x+6$
$(y-11)x^2 + (4y-12)x + (2y-6) = 0$
Since $x$ is a real number,the discriminant $D$ must be greater than or equal to $0$:
$D = (4y-12)^2 - 4(y-11)(2y-6) \geq 0$
$16(y-3)^2 - 8(y-11)(y-3) \geq 0$
$8(y-3) [2(y-3) - (y-11)] \geq 0$
$8(y-3)(2y-6-y+11) \geq 0$
$8(y-3)(y+5) \geq 0$
This inequality holds when $y \leq -5$ or $y \geq 3$.
Comparing this with the given condition $y < a$ or $y \geq b$,we identify $a = -5$ and $b = 3$.

(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.
Multiplying both sides by $(x^2+x+1)$,we get $y(x^2+x+1) = x^2-x+1$.
Rearranging the terms as a quadratic in $x$:
$(y-1)x^2 + (y+1)x + (y-1) = 0$.
Since $x$ is real,the discriminant $D \geq 0$.
$D = (y+1)^2 - 4(y-1)^2 \geq 0$.
$(y+1)^2 - [2(y-1)]^2 \geq 0$.
Using $a^2-b^2 = (a-b)(a+b)$:
$[(y+1) - 2(y-1)][(y+1) + 2(y-1)] \geq 0$.
$(-y+3)(3y-1) \geq 0$.
$(y-3)(3y-1) \leq 0$.
This inequality holds for $\frac{1}{3} \leq y \leq 3$.
Thus,the minimum value of $y$ is $\frac{1}{3}$.

(A) Given roots are $\alpha = \sin^2 18^{\circ}$ and $\beta = \cos^2 36^{\circ}$.
We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.
Sum of roots: $\alpha + \beta = \left(\frac{\sqrt{5}-1}{4}\right)^2 + \left(\frac{\sqrt{5}+1}{4}\right)^2 = \frac{6-2\sqrt{5}}{16} + \frac{6+2\sqrt{5}}{16} = \frac{12}{16} = \frac{3}{4}$.
Product of roots: $\alpha \cdot \beta = \left(\frac{\sqrt{5}-1}{4}\right)^2 \cdot \left(\frac{\sqrt{5}+1}{4}\right)^2 = \left(\frac{(\sqrt{5}-1)(\sqrt{5}+1)}{16}\right)^2 = \left(\frac{5-1}{16}\right)^2 = \left(\frac{4}{16}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - \frac{3}{4}x + \frac{1}{16} = 0$.
Multiplying by $16$,we get $16x^2 - 12x + 1 = 0$.

(D) Let $y = (x-\alpha)(x-\beta)$.
Expanding the expression,we get $y = x^2 - (\alpha+\beta)x + \alpha\beta$.
To find the minimum value,we find the derivative with respect to $x$ and set it to $0$:
$\frac{dy}{dx} = 2x - (\alpha+\beta) = 0$.
This gives $x = \frac{\alpha+\beta}{2}$.
Since the second derivative $\frac{d^2y}{dx^2} = 2 > 0$,the function has a minimum at $x = \frac{\alpha+\beta}{2}$.
Substituting this value into the original expression:
$y_{min} = \left(\frac{\alpha+\beta}{2} - \alpha\right)\left(\frac{\alpha+\beta}{2} - \beta\right)$
$y_{min} = \left(\frac{\beta-\alpha}{2}\right)\left(\frac{\alpha-\beta}{2}\right)$
$y_{min} = -\frac{1}{4}(\alpha-\beta)(\alpha-\beta) = -\frac{1}{4}(\alpha-\beta)^2$.

(B) Given equation: $\frac{x-1}{\sqrt{2x^2-5x+2}} = \frac{41}{60}$.
Squaring both sides:
$\frac{x^2-2x+1}{2x^2-5x+2} = \frac{1681}{3600}$.
Cross-multiplying:
$3600(x^2-2x+1) = 1681(2x^2-5x+2)$.
$3600x^2 - 7200x + 3600 = 3362x^2 - 8405x + 3362$.
$238x^2 + 1205x + 238 = 0$.
Factoring the quadratic equation:
$(34x + 7)(7x + 34) = 0$.
Thus,$x = -\frac{7}{34}$ or $x = -\frac{34}{7}$.
Since $-\frac{1}{2} < \alpha < 0$ and $-\frac{7}{34} \approx -0.205$ while $-\frac{34}{7} \approx -4.857$,the root $\alpha = -\frac{7}{34}$ satisfies the condition.

(A) The given equation is $3x^3 + bx^2 + bx + 3 = 0$.
Factoring the equation: $3(x^3 + 1) + bx(x + 1) = 0$.
$3(x + 1)(x^2 - x + 1) + bx(x + 1) = 0$.
$(x + 1)(3x^2 - 3x + 3 + bx) = 0$.
$(x + 1)(3x^2 + (b - 3)x + 3) = 0$.
One root is $x = -1$. The other two roots are roots of $3x^2 + (b - 3)x + 3 = 0$.
Let $f(x) = 3x^2 + (b - 3)x + 3$.
For $A$: All roots are negative. If $b = 9$,$f(x) = 3x^2 + 6x + 3 = 3(x + 1)^2$. Roots are $-1, -1, -1$. All negative. So $A \rightarrow IV$.
For $B$: Two roots are complex. Discriminant $D < 0 \Rightarrow (b - 3)^2 - 4(3)(3) < 0 \Rightarrow (b - 3)^2 < 36 \Rightarrow -6 < b - 3 < 6 \Rightarrow -3 < b < 9$. So $B \rightarrow II$.
For $C$: Two roots are positive. If $b = -3$,$f(x) = 3x^2 - 6x + 3 = 3(x - 1)^2$. Roots are $-1, 1, 1$. Two positive. So $C \rightarrow V$.
For $D$: All roots are real and distinct. $D > 0$ and $f(-1) \neq 0$. $D = (b - 3)^2 - 36 > 0 \Rightarrow b \in (-\infty, -3) \cup (9, \infty)$. Also $f(-1) = 3 - (b - 3) + 3 = 9 - b \neq 0 \Rightarrow b \neq 9$. So $D \rightarrow III$.
Thus,$A-IV, B-II, C-V, D-III$.

(B) The length of the intercept on the $X$-axis made by the curve $f(x) = a x^2 + b x + c$ is given by $A = |x_1 - x_2| = \frac{\sqrt{D}}{|a|}$,where $D = b^2 - 4 a c$.
For $f_1(x) = 5 x^2 + 2 x + 1$,$D = 2^2 - 4(5)(1) = 4 - 20 = -16$. Since $D < 0$,the curve does not intersect the $X$-axis,so $A_1$ is not defined (or $0$ in terms of real intercept length).
For $f_2(x) = 5 x^2 + 6 x + 1$,$D = 6^2 - 4(5)(1) = 36 - 20 = 16$. Thus,$A_2 = \frac{\sqrt{16}}{5} = \frac{4}{5} = 0.8$.
For $f_3(x) = x^2 - 7 x + 6$,$D = (-7)^2 - 4(1)(6) = 49 - 24 = 25$. Thus,$A_3 = \frac{\sqrt{25}}{1} = 5$.
For $f_4(x) = 64 x^2 + 48 x + 9$,$D = 48^2 - 4(64)(9) = 2304 - 2304 = 0$. Thus,$A_4 = \frac{\sqrt{0}}{64} = 0$.
Comparing the values: $A_4 = 0$,$A_2 = 0.8$,$A_3 = 5$.
Therefore,$A_4 < A_2 < A_3$ is true.

(D) Given the quadratic equation $(p-3)x^2 + 2(p-3)x + 2p-5 = 0$. For the roots to be real and distinct,the discriminant $D > 0$.
Since the coefficient of $x^2$ is $(p-3)$,we must have $p \neq 3$.
$D = [2(p-3)]^2 - 4(p-3)(2p-5) > 0$
$4(p-3)^2 - 4(p-3)(2p-5) > 0$
Dividing by $4(p-3)$,we get $(p-3) - (2p-5) > 0$ if $p > 3$,or $(p-3) - (2p-5) < 0$ if $p < 3$.
Case $1$: $p > 3$: $-p + 2 > 0 \Rightarrow p < 2$ (Contradiction).
Case $2$: $p < 3$: $-p + 2 < 0 \Rightarrow p > 2$.
Thus,$2 < p < 3$. So,$\alpha = 2$ and $\beta = 3$.
The expression becomes $f(x) = -(2+3)x^2 + (2 \times 3)x + (2-3) = -5x^2 + 6x - 1$.
The extreme value (maximum) of a quadratic $ax^2 + bx + c$ where $a < 0$ is given by $\frac{4ac - b^2}{4a}$.
Here $a = -5, b = 6, c = -1$.
Extreme value $= \frac{4(-5)(-1) - (6)^2}{4(-5)} = \frac{20 - 36}{-20} = \frac{-16}{-20} = \frac{4}{5}$.

(D) Let $f(x) = ax^2 + bx + c$.
Given $f(0) + f(1) = 0$,we have $c + (a + b + c) = 0$,which implies $a + b + 2c = 0$ $(i)$.
Given $f(-2) = 0$,we have $4a - 2b + c = 0$ (ii).
From $(i)$,$b = -a - 2c$. Substituting into (ii):
$4a - 2(-a - 2c) + c = 0$ $\Rightarrow 4a + 2a + 4c + c = 0$ $\Rightarrow 6a + 5c = 0$.
Let $a = 5k$,then $c = -6k$.
Substituting into $(i)$: $5k + b + 2(-6k) = 0$ $\Rightarrow 5k + b - 12k = 0$ $\Rightarrow b = 7k$.
Thus,$f(x) = k(5x^2 + 7x - 6) = k(5x^2 + 10x - 3x - 6) = k(5x(x + 2) - 3(x + 2)) = k(5x - 3)(x + 2)$.
The roots are $x = -2$ and $x = \frac{3}{5}$.
Therefore,$f\left(\frac{3}{5}\right) = 0$.

(A) Let the roots of $x^3-5x^2+4x=0$ be $y_1, y_2, y_3$.
Factoring the equation,we get $x(x^2-5x+4)=0$,so $x(x-1)(x-4)=0$.
The roots are $0, 1, 4$.
Given $(\alpha-2)^2, (\beta-2)^2, (\gamma-2)^2$ are $0, 1, 4$.
Thus,$(\alpha-2)^2=0 \implies \alpha=2$.
$(\beta-2)^2=1 \implies \beta-2 = \pm 1 \implies \beta=3$ or $\beta=1$.
$(\gamma-2)^2=4 \implies \gamma-2 = \pm 2 \implies \gamma=4$ or $\gamma=0$.
By Vieta's formulas for $x^3-Px^2+Qx-R=0$:
$P = \alpha+\beta+\gamma$,$Q = \alpha\beta+\beta\gamma+\gamma\alpha$,$R = \alpha\beta\gamma$.
We want to minimize $P+Q+R$. Note that $P+Q+R = (\alpha+1)(\beta+1)(\gamma+1)-1$.
Testing combinations:
If $\alpha=2, \beta=3, \gamma=4$,then $P=9, Q=26, R=24$,$P+Q+R=59$.
If $\alpha=2, \beta=1, \gamma=0$,then $P=3, Q=2, R=0$,$P+Q+R=5$.
If $\alpha=2, \beta=3, \gamma=0$,then $P=5, Q=6, R=0$,$P+Q+R=11$.
If $\alpha=2, \beta=1, \gamma=4$,then $P=7, Q=14, R=8$,$P+Q+R=29$.
Considering negative roots for $\beta-2$ and $\gamma-2$:
If $\beta=1, \gamma=0$,$P=3, Q=2, R=0$,sum $= 5$.
If $\beta=3, \gamma=0$,$P=5, Q=6, R=0$,sum $= 11$.
If $\beta=1, \gamma=4$,$P=7, Q=14, R=8$,sum $= 29$.
If $\beta=3, \gamma=4$,$P=9, Q=26, R=24$,sum $= 59$.
The least value is $5$.

(C) Let $P(x) = x^4-4x^3+3x^2+2x-2$.
By the Rational Root Theorem,possible integer roots are $\pm 1, \pm 2$.
Testing $x=1$: $1-4+3+2-2 = 0$. So,$(x-1)$ is a factor.
Testing $x=-1$: $1+4+3-2-2 = 4 \neq 0$.
Testing $x=2$: $16-32+12+4-2 = -2 \neq 0$.
Testing $x=-2$: $16+32+12-4-2 = 54 \neq 0$.
Since $x=1$ is a root,divide $P(x)$ by $(x-1)$ to get $x^3-3x^2+2$.
Testing $x=1$ again in $x^3-3x^2+2$: $1-3+2 = 0$. So,$(x-1)^2$ is a factor.
Dividing $x^3-3x^2+2$ by $(x-1)$ gives $x^2-2x-2$.
Thus,the roots are $\alpha=1, \beta=1$ and the roots of $x^2-2x-2=0$,which are $\gamma, \delta = 1 \pm \sqrt{3}$.
We need to calculate $\alpha+2\beta+\gamma^2+\delta^2$.
Since $\gamma, \delta$ are roots of $x^2-2x-2=0$,$\gamma+\delta=2$ and $\gamma\delta=-2$.
Then $\gamma^2+\delta^2 = (\gamma+\delta)^2 - 2\gamma\delta = (2)^2 - 2(-2) = 4+4 = 8$.
Substituting the values: $1 + 2(1) + 8 = 1+2+8 = 11$.

(D) Given the reciprocal equation $12x^4-56x^3+89x^2-56x+12=0$.
Dividing by $x^2$,we get $12(x^2+\frac{1}{x^2})-56(x+\frac{1}{x})+89=0$.
Let $u = x+\frac{1}{x}$. Then $x^2+\frac{1}{x^2} = u^2-2$.
Substituting this,we get $12(u^2-2)-56u+89=0$,which simplifies to $12u^2-56u+65=0$.
Solving for $u$ using the quadratic formula: $u = \frac{56 \pm \sqrt{56^2-4(12)(65)}}{2(12)} = \frac{56 \pm \sqrt{3136-3120}}{24} = \frac{56 \pm 4}{24}$.
Thus,$u_1 = \frac{60}{24} = \frac{5}{2}$ and $u_2 = \frac{52}{24} = \frac{13}{6}$.
Since $\alpha\beta=1$,$\alpha+\beta$ is one of the values of $u$. Let $\alpha+\beta = u_1 = \frac{5}{2}$ and $\gamma+\delta = u_2 = \frac{13}{6}$.
Then $\frac{\alpha+\beta}{\gamma+\delta} = \frac{5/2}{13/6} = \frac{5}{2} \times \frac{6}{13} = \frac{15}{13}$.
Since $\frac{15}{13} > 1$,this satisfies the condition.

(D) Given the equation $x^5+x^4-13x^3-13x^2+36x+36=0$.
Factorizing the equation:
$x^4(x+1) - 13x^2(x+1) + 36(x+1) = 0$
$(x+1)(x^4-13x^2+36) = 0$
$(x+1)(x^2-4)(x^2-9) = 0$
$(x+1)(x-2)(x+2)(x-3)(x+3) = 0$
The roots are $-3, -2, -1, 2, 3$.
Given $\alpha < \beta < \gamma < \delta < \varepsilon$,we have:
$\alpha = -3, \beta = -2, \gamma = -1, \delta = 2, \varepsilon = 3$.
Now,calculate $\frac{\varepsilon}{\alpha}+\frac{\delta}{\beta}+\frac{1}{\gamma}$:
$\frac{3}{-3} + \frac{2}{-2} + \frac{1}{-1} = -1 - 1 - 1 = -3$.

(C) Given the cubic equation $x^3 + 3 x^2 - 10 x - 24 = 0$.
By testing integer roots,for $x = 3$: $(3)^3 + 3(3)^2 - 10(3) - 24 = 27 + 27 - 30 - 24 = 0$.
So,$(x - 3)$ is a factor.
Dividing the polynomial by $(x - 3)$,we get $(x - 3)(x^2 + 6 x + 8) = 0$.
Factoring the quadratic part: $(x - 3)(x + 4)(x + 2) = 0$.
The roots are $3, -2, -4$.
Given $\alpha > \beta > \gamma$,we have $\alpha = 3, \beta = -2, \gamma = -4$.
Substitute these into the expression $\alpha^3 + 3 \beta^2 - 10 \gamma - 24$:
$(3)^3 + 3(-2)^2 - 10(-4) - 24 = 27 + 12 + 40 - 24 = 55$.
Given $55 = 11 k$,we find $k = 5$.

(D) Given equation: $12 x^{1/3} - 25 x^{1/6} + 12 = 0$.
Let $t = x^{1/6}$. Then the equation becomes $12 t^2 - 25 t + 12 = 0$.
Factoring the quadratic: $12 t^2 - 16 t - 9 t + 12 = 0 \Rightarrow 4 t(3 t - 4) - 3(3 t - 4) = 0$.
$(4 t - 3)(3 t - 4) = 0$,so $t = \frac{3}{4}$ or $t = \frac{4}{3}$.
Since $\alpha > \beta$ and $x^{1/6} = t$,we have $\alpha^{1/6} = \frac{4}{3}$ and $\beta^{1/6} = \frac{3}{4}$.
We need to find $\sqrt[6]{\frac{\alpha}{\beta}} = \frac{\alpha^{1/6}}{\beta^{1/6}}$.
Substituting the values: $\frac{4/3}{3/4} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$.

(D) Given the equation $f(x) = 16x^4 + 16x^3 - 4x - 1 = 0$. Since it has a multiple root,let $\alpha$ be the multiple root. Then $f(\alpha) = 0$ and $f'(\alpha) = 0$.
$f'(x) = 64x^3 + 48x^2 - 4$. Setting $f'(\alpha) = 0$ gives $16\alpha^3 + 12\alpha^2 - 1 = 0$.
From $f(\alpha) = 16\alpha^4 + 16\alpha^3 - 4\alpha - 1 = 0$,we substitute $16\alpha^3 = 1 - 12\alpha^2$ to get $16\alpha^4 + (1 - 12\alpha^2) - 4\alpha - 1 = 0$,which simplifies to $16\alpha^4 - 12\alpha^2 - 4\alpha = 0$.
Factoring gives $4\alpha(4\alpha^3 - 3\alpha - 1) = 0$. Since $\alpha \neq 0$,we solve $4\alpha^3 - 3\alpha - 1 = 0$,which factors as $(\alpha - 1)(2\alpha + 1)^2 = 0$.
Testing $\alpha = 1$ in $f(x)$ gives $16+16-4-1 \neq 0$. Thus,$\alpha = -\frac{1}{2}$ is the multiple root.
Dividing $f(x)$ by $(x + \frac{1}{2})^2 = (x^2 + x + \frac{1}{4})$,we get $16x^2 - 4 = 0$,so $x = \pm \frac{1}{2}$.
The roots are $\alpha = -\frac{1}{2}, \beta = -\frac{1}{2}, \gamma = -\frac{1}{2}, \delta = \frac{1}{2}$.
Then $\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} = (-2)^4 + (-2)^4 + (-2)^4 + (2)^4 = 16 + 16 + 16 + 16 = 64$.

(D) Given $x^3+3x^2-9x+\lambda = (x-\alpha)^2(x-\beta) = (x^2-2\alpha x+\alpha^2)(x-\beta) = x^3 - (2\alpha+\beta)x^2 + (2\alpha\beta+\alpha^2)x - \alpha^2\beta$.
Comparing coefficients:
$2\alpha+\beta = -3$ $(i)$
$2\alpha\beta+\alpha^2 = -9$ (ii)
$-\alpha^2\beta = \lambda$ (iii)
From $(i)$,$\beta = -3-2\alpha$.
Substitute into (ii): $2\alpha(-3-2\alpha) + \alpha^2 = -9$ $\Rightarrow -6\alpha - 4\alpha^2 + \alpha^2 = -9$ $\Rightarrow 3\alpha^2 + 6\alpha - 9 = 0$ $\Rightarrow \alpha^2 + 2\alpha - 3 = 0$.
Solving for $\alpha$: $(\alpha+3)(\alpha-1) = 0$,so $\alpha = 1$ or $\alpha = -3$.
If $\alpha = 1$,then $\beta = -3-2(1) = -5$. From (iii),$\lambda = -\alpha^2\beta = -(1)^2(-5) = 5$.
If $\alpha = -3$,then $\beta = -3-2(-3) = 3$. From (iii),$\lambda = -\alpha^2\beta = -(-3)^2(3) = -27$.
Thus,the values of $\lambda$ are $-27$ and $5$.

(D) Given the equation $x^4-4x^3-7x^2+22x+24=0$.
By testing roots,we find the factors:
$(x+1)(x+2)(x-3)(x-4)=0$.
The roots are $x_1 = -1, x_2 = -2, x_3 = 3, x_4 = 4$.
We check the condition: the sum of two roots equals the sum of the other two.
$3 + (-1) = 2$ and $4 + (-2) = 2$.
The condition is satisfied.
The sum of the cubes of the roots is:
$(-1)^3 + (-2)^3 + (3)^3 + (4)^3 = -1 - 8 + 27 + 64 = 82$.

(C) Given the cubic equation: $x^3+4x^2-9x-36=0$
Factorizing the equation:
$x^2(x+4)-9(x+4)=0$
$(x^2-9)(x+4)=0$
$(x-3)(x+3)(x+4)=0$
The roots are $x = -4, -3, 3$.
Given the condition $\alpha < \beta < \gamma$,we have $\alpha = -4$,$\beta = -3$,and $\gamma = 3$.
Now,calculate $\alpha+2\beta+3\gamma$:
$\alpha+2\beta+3\gamma = -4 + 2(-3) + 3(3)$
$= -4 - 6 + 9$
$= -10 + 9 = -1$.

(D) Let $f(x) = x^5-6x^4+11x^3-2x^2-12x+8$.
By testing small integer roots,we find that $x=2$ is a root.
Using synthetic division or polynomial division,we factor the expression:
$f(x) = (x-2)^3(x^2-1) = (x-2)^3(x-1)(x+1)$.
The roots are $x=2$ (with multiplicity $3$),$x=1$,and $x=-1$.
Since $\alpha$ is a multiple root,$\alpha=2$.
Substituting $\alpha=2$ into the expression $3\alpha^2-2\alpha+1$:
$3(2)^2-2(2)+1 = 3(4)-4+1 = 12-4+1 = 9$.

(C) Given the cubic equation $x^3+2x^2-x-2=0$.
Factorizing the equation:
$x^2(x+2)-1(x+2)=0$
$(x^2-1)(x+2)=0$
$(x-1)(x+1)(x+2)=0$
Thus,the roots are $\alpha=1, \beta=-1, \gamma=-2$.
We need to calculate $\alpha^6+\beta^6+\gamma^6$:
$\alpha^6+\beta^6+\gamma^6 = (1)^6 + (-1)^6 + (-2)^6$
$= 1 + 1 + 64$
$= 66$

(A) The given equation is $x^4+x^3-4x^2+x+1=0$.
Dividing by $x^2$,we get $(x^2 + \frac{1}{x^2}) + (x + \frac{1}{x}) - 4 = 0$.
Let $y = x + \frac{1}{x}$,then $y^2 - 2 + y - 4 = 0 \Rightarrow y^2 + y - 6 = 0$.
$(y+3)(y-2) = 0$,so $y = 2$ or $y = -3$.
If $x + \frac{1}{x} = 2$,then $x=1, 1$.
If $x + \frac{1}{x} = -3$,then $x^2 + 3x + 1 = 0$,so $x = \frac{-3 \pm \sqrt{5}}{2}$.
Let the roots be $r_1, r_2, r_3, r_4$. The sum of roots taken two at a time is $\sum r_i r_j = -4$.
When roots are diminished by $h$,the new roots are $r_i - h$.
The new coefficient of $x^2$ is $\sum (r_i - h)(r_j - h) = 0$.
This expands to $\sum r_i r_j - 3h \sum r_i + 6h^2 = 0$.
From the original equation,$\sum r_i = -1$ and $\sum r_i r_j = -4$.
Substituting these,$-4 - 3h(-1) + 6h^2 = 0 \Rightarrow 6h^2 + 3h - 4 = 0$.
The roots of this quadratic in $h$ are $\alpha$ and $\beta$.
Thus,$\alpha + \beta = -\frac{3}{6} = -\frac{1}{2}$ and $\alpha \beta = -\frac{4}{6} = -\frac{2}{3}$.
Then $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta = (-\frac{1}{2})^2 - 4(-\frac{2}{3}) = \frac{1}{4} + \frac{8}{3} = \frac{3+32}{12} = \frac{35}{12}$.
Therefore,$12(\alpha - \beta)^2 = 12 \times \frac{35}{12} = 35$.

(B) Given that $1+\sqrt{2}$ and $2-i$ are roots of the polynomial equation with rational coefficients,the conjugate roots $1-\sqrt{2}$ and $2+i$ must also be roots.
Let the roots be $\alpha_1 = 1+\sqrt{2}, \alpha_2 = 1-\sqrt{2}, \alpha_3 = 2-i, \alpha_4 = 2+i$.
Using Vieta's formulas:
$-b = \sum \alpha_i = (1+\sqrt{2}) + (1-\sqrt{2}) + (2-i) + (2+i) = 2 + 4 = 6 \Rightarrow b = -6$.
$c = \sum \alpha_i \alpha_j = (1+\sqrt{2})(1-\sqrt{2}) + (1+\sqrt{2})(2-i) + (1+\sqrt{2})(2+i) + (1-\sqrt{2})(2-i) + (1-\sqrt{2})(2+i) + (2-i)(2+i) = -1 + 4 + 4 + 4 + 4 + 5 = 20$.
Wait,recalculating $c$: $\alpha_1 \alpha_2 = -1$,$\alpha_3 \alpha_4 = 5$,$(\alpha_1+\alpha_2)(\alpha_3+\alpha_4) = 2 \times 4 = 8$. So $c = -1 + 5 + 8 = 12$.
$-d = \sum \alpha_i \alpha_j \alpha_k = \alpha_1 \alpha_2 (\alpha_3+\alpha_4) + \alpha_3 \alpha_4 (\alpha_1+\alpha_2) = -1(4) + 5(2) = -4 + 10 = 6$ $\Rightarrow d = -6$.
The equation $bx^2+cx+d=0$ becomes $-6x^2+12x-6=0$,which simplifies to $x^2-2x+1=0$.
This is $(x-1)^2=0$,so the roots are $1, 1$,which are real and equal.

(B) Given equation is $x^5-3x^4+5x^3-5x^2+3x-1=0$.
We can rewrite the equation by grouping terms: $(x^5-1) - 3x(x^3-1) + 5x^2(x-1) = 0$.
Factoring out $(x-1)$: $(x-1)(x^4+x^3+x^2+x+1) - 3x(x-1)(x^2+x+1) + 5x^2(x-1) = 0$.
$(x-1)[(x^4+x^3+x^2+x+1) - 3x(x^2+x+1) + 5x^2] = 0$.
$(x-1)[x^4+x^3+x^2+x+1 - 3x^3-3x^2-3x + 5x^2] = 0$.
$(x-1)(x^4-2x^3+3x^2-2x+1) = 0$.
For the quartic part,divide by $x^2$ (since $x=0$ is not a root): $x^2-2x+3-\frac{2}{x}+\frac{1}{x^2} = 0$.
$(x^2+\frac{1}{x^2}) - 2(x+\frac{1}{x}) + 3 = 0$.
Let $y = x+\frac{1}{x}$,then $y^2-2 - 2y + 3 = 0$,so $y^2-2y+1 = 0$.
$(y-1)^2 = 0$,which gives $y=1$.
$x+\frac{1}{x} = 1 \implies x^2-x+1 = 0$.
The roots of $x^2-x+1=0$ are complex.
The only real root is $x=1$.
The distinct roots are $1, \frac{1+i\sqrt{3}}{2}, \frac{1-i\sqrt{3}}{2}$.
The sum of these distinct roots is $1 + (\frac{1+i\sqrt{3}}{2} + \frac{1-i\sqrt{3}}{2}) = 1 + 1 = 2$.

(A) Let the roots of the given equation $x^4-2 x^3+6 x-21=0$ be $\alpha, \beta, \gamma, \delta$. We want to find the equation whose roots are $\alpha^2, \beta^2, \gamma^2, \delta^2$.
Let $y = x^2$,then $x = \sqrt{y}$.
The given equation can be written as $x^4-2 x^3+6 x-21=0$.
Rearranging terms: $x^4-21 = 2 x^3-6 x = 2 x(x^2-3)$.
Substituting $x^2 = y$: $y^2-21 = 2 x(y-3)$.
Squaring both sides: $(y^2-21)^2 = 4 x^2(y-3)^2$.
Since $x^2 = y$,we have $(y^2-21)^2 = 4 y(y-3)^2$.
Expanding both sides: $y^4-42 y^2+441 = 4 y(y^2-6 y+9)$.
$y^4-42 y^2+441 = 4 y^3-24 y^2+36 y$.
Rearranging to standard form: $y^4-4 y^3-18 y^2-36 y+441=0$.
Replacing $y$ with $x$,the required equation is $x^4-4 x^3-18 x^2-36 x+441=0$.

(C) Given equation: $3x^3+4x^2-5x-2=0$ $(i)$.
Let the roots of $(i)$ be $\alpha, \beta, \gamma$.
The roots are diminished by $h$,so the new roots are $t = x - h$,which implies $x = t + h$.
Substituting $x = t + h$ into $(i)$:
$3(t+h)^3 + 4(t+h)^2 - 5(t+h) - 2 = 0$.
Expanding the terms:
$3(t^3 + 3t^2h + 3th^2 + h^3) + 4(t^2 + 2th + h^2) - 5(t+h) - 2 = 0$.
$3t^3 + (9h + 4)t^2 + (9h^2 + 8h - 5)t + (3h^3 + 4h^2 - 5h - 2) = 0$.
Since the $x^2$ (or $t^2$) term is absent,its coefficient must be zero:
$9h + 4 = 0 \Rightarrow h = -\frac{4}{9}$.
Now,find the roots of the original equation $3x^3+4x^2-5x-2=0$.
By inspection,$x=1$ is a root $(3+4-5-2=0)$.
Dividing by $(x-1)$,we get $(x-1)(3x^2+7x+2) = 0$.
$(x-1)(3x+1)(x+2) = 0$.
The roots are $x_1 = 1, x_2 = -\frac{1}{3}, x_3 = -2$.
The new roots are $x_i - h = x_i - (-\frac{4}{9}) = x_i + \frac{4}{9}$.
New roots: $1 + \frac{4}{9} = \frac{13}{9}$,$-\frac{1}{3} + \frac{4}{9} = \frac{1}{9}$,and $-2 + \frac{4}{9} = -\frac{14}{9}$.

(B) Given,one root is $-1+i$.
Since the coefficients of the polynomial are real,the complex conjugate $-1-i$ must also be a root.
Thus,$(x - (-1+i))(x - (-1-i)) = (x+1-i)(x+1+i) = (x+1)^2 - i^2 = x^2+2x+2$ is a factor of the given equation.
Dividing $x^4+4x^3+5x^2+2x-2$ by $x^2+2x+2$,we get the quotient $x^2+2x-1$.
Setting $x^2+2x-1 = 0$,we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$x = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$.
Therefore,the real roots are $-1 \pm \sqrt{2}$.

(D) Given the equation $\sqrt{\frac{5x}{x-2}} + \sqrt{\frac{x-2}{5x}} = \frac{29}{10}$.
Let $y = \sqrt{\frac{5x}{x-2}}$. Then the equation becomes $y + \frac{1}{y} = \frac{29}{10}$.
Multiplying by $10y$,we get $10y^2 - 29y + 10 = 0$.
Factoring the quadratic: $10y^2 - 25y - 4y + 10 = 0 \Rightarrow 5y(2y - 5) - 2(2y - 5) = 0$.
So,$(5y - 2)(2y - 5) = 0$,which gives $y = \frac{2}{5}$ or $y = \frac{5}{2}$.
Case $1$: $\sqrt{\frac{5x}{x-2}} = \frac{2}{5}$ $\Rightarrow \frac{5x}{x-2} = \frac{4}{25}$ $\Rightarrow 125x = 4x - 8$ $\Rightarrow 121x = -8$ $\Rightarrow x = -\frac{8}{121}$.
Case $2$: $\sqrt{\frac{5x}{x-2}} = \frac{5}{2}$ $\Rightarrow \frac{5x}{x-2} = \frac{25}{4}$ $\Rightarrow \frac{x}{x-2} = \frac{5}{4}$ $\Rightarrow 4x = 5x - 10$ $\Rightarrow x = 10$.
Since $\alpha > \beta$,we have $\alpha = 10$ and $\beta = -\frac{8}{121}$.
Now,calculate $\sqrt{\alpha^2 - 11^4 \beta^2} = \sqrt{10^2 - 11^4 \left(-\frac{8}{121}\right)^2} = \sqrt{100 - 11^4 \cdot \frac{64}{11^4}} = \sqrt{100 - 64} = \sqrt{36} = 6$.

(A) Let $f(x) = x^5-8x^4+25x^3-38x^2+28x-8$.
Since $\alpha$ is a root of multiplicity $3$,we have $f(\alpha) = 0$,$f'(\alpha) = 0$,and $f''(\alpha) = 0$.
Calculating the derivatives:
$f'(x) = 5x^4-32x^3+75x^2-76x+28$
$f''(x) = 20x^3-96x^2+150x-76$
Testing $x = 2$:
$f(2) = 32-128+200-152+56-8 = 0$
$f'(2) = 80-256+300-152+28 = 0$
$f''(2) = 160-384+300-76 = 0$
$f'''(2) = 60(4)-192(2)+150 = 240-384+150 = 6 \neq 0$.
Thus,$\alpha = 2$ is a root of multiplicity $3$.
Therefore,$\alpha^2-5\alpha+6 = (2)^2-5(2)+6 = 4-10+6 = 0$.

(D) Given equation: $6x^6-25x^5+31x^4-31x^2+25x-6=0$
Rearranging terms: $6(x^6-1) - 25x(x^4-1) + 31x^2(x^2-1) = 0$
Factoring $(x^2-1)$: $(x^2-1)[6(x^4+x^2+1) - 25x(x^2+1) + 31x^2] = 0$
$(x^2-1)[6x^4 - 25x^3 + 37x^2 - 25x + 6] = 0$
Dividing the second bracket by $x^2$: $x^2(x^2-1)[6(x^2+\frac{1}{x^2}) - 25(x+\frac{1}{x}) + 37] = 0$
Let $y = x+\frac{1}{x}$,then $x^2+\frac{1}{x^2} = y^2-2$.
$6(y^2-2) - 25y + 37 = 0 \Rightarrow 6y^2 - 25y + 25 = 0$
$(2y-5)(3y-5) = 0 \Rightarrow y = \frac{5}{2}, \frac{5}{3}$
For $x+\frac{1}{x} = \frac{5}{2}$,$x=2, \frac{1}{2}$ (real roots).
For $x+\frac{1}{x} = \frac{5}{3}$,$3x^2-5x+3=0$. The roots are complex.
Sum of these complex roots $\alpha+\beta = -\frac{b}{a} = -(\frac{-5}{3}) = \frac{5}{3}$.

(C) Given equation: $x^3-6x^2+6x-2=0$.
Checking option $(a)$: Putting $x=-1$,we get $(-1)^3-6(-1)^2+6(-1)-2 = -1-6-6-2 = -15 \neq 0$.
Checking option $(b)$: Putting $x=2$,we get $(2)^3-6(2)^2+6(2)-2 = 8-24+12-2 = -6 \neq 0$.
Checking option $(c)$: Let $x = \frac{2^{1/3}}{2^{1/3}-1}$.
Applying componendo and dividendo: $\frac{x+1}{x-1} = \frac{2^{1/3}+2^{1/3}-1}{2^{1/3}-2^{1/3}+1} = 2 \cdot 2^{1/3}-1$.
$\frac{x+1}{x-1} + 1 = 2 \cdot 2^{1/3}$ $\Rightarrow \frac{2x}{x-1} = 2 \cdot 2^{1/3}$ $\Rightarrow \frac{x}{x-1} = 2^{1/3}$.
Cubing both sides: $\frac{x^3}{(x-1)^3} = 2
$ $\Rightarrow x^3 = 2(x^3-3x^2+3x-1)
$ $\Rightarrow x^3 = 2x^3-6x^2+6x-2
$ $\Rightarrow x^3-6x^2+6x-2 = 0$.
Thus,option $(c)$ is the correct root.

(C) Given equation: $(x+1)^4+(x+3)^4=8$.
Let $x+2=y$,then $x+1=y-1$ and $x+3=y+1$.
The equation becomes $(y-1)^4+(y+1)^4=8$.
Expanding using binomial theorem: $(y^4-4y^3+6y^2-4y+1)+(y^4+4y^3+6y^2+4y+1)=8$.
$2y^4+12y^2+2=8$ $\Rightarrow 2y^4+12y^2-6=0$ $\Rightarrow y^4+6y^2-3=0$.
Let $t=y^2$,then $t^2+6t-3=0$.
$t = \frac{-6 \pm \sqrt{36-4(1)(-3)}}{2} = \frac{-6 \pm \sqrt{48}}{2} = -3 \pm 2\sqrt{3}$.
Since $t=y^2$ must be non-negative for real $y$,we take $y^2 = 2\sqrt{3}-3$.
Substituting back $y=x+2$: $(x+2)^2 = 2\sqrt{3}-3$.
$x^2+4x+4 = 2\sqrt{3}-3 \Rightarrow x^2+4x+(7-2\sqrt{3}) = 0$.
This is a quadratic equation in $x$ of the form $ax^2+bx+c=0$.
The product of the roots is $\frac{c}{a} = \frac{7-2\sqrt{3}}{1} = 7-2\sqrt{3}$.

(A) Given equation: $x^4-10x^3+37x^2-60x+36=0$.
Since the roots are $\alpha, \alpha, \beta, \beta$,by Vieta's formulas:
Sum of roots: $2\alpha + 2\beta = 10 \Rightarrow \alpha + \beta = 5$ $(i)$.
Product of roots: $\alpha^2\beta^2 = 36 \Rightarrow \alpha\beta = 6$ $(ii)$.
From $(i)$,$\beta = 5 - \alpha$. Substituting into $(ii)$: $\alpha(5 - \alpha) = 6 \Rightarrow \alpha^2 - 5\alpha + 6 = 0$.
Solving for $\alpha$: $(\alpha - 2)(\alpha - 3) = 0$,so $\alpha = 2$ or $\alpha = 3$.
Since $\alpha < \beta$,we have $\alpha = 2$ and $\beta = 3$.
Now,calculate $2\alpha + 3\beta - 2\alpha\beta$:
$2(2) + 3(3) - 2(2)(3) = 4 + 9 - 12 = 1$.

(B) Given equation: $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$
Rearranging the terms: $(x^2+\frac{1}{x^2}) - 8(x+\frac{1}{x}) + 9 = 0$
Let $t = x+\frac{1}{x}$. Then $t^2 = x^2+2+\frac{1}{x^2}$,so $x^2+\frac{1}{x^2} = t^2-2$.
Substituting into the equation: $(t^2-2) - 8t + 9 = 0$
$t^2 - 8t + 7 = 0$
$(t-7)(t-1) = 0$
Case $1$: $t = 7$ $\Rightarrow x+\frac{1}{x} = 7$ $\Rightarrow x^2-7x+1 = 0$. The discriminant $D = (-7)^2 - 4(1)(1) = 45 > 0$,so there are two real roots. The product of these roots is $c/a = 1$.
Case $2$: $t = 1$ $\Rightarrow x+\frac{1}{x} = 1$ $\Rightarrow x^2-x+1 = 0$. The discriminant $D = (-1)^2 - 4(1)(1) = -3 < 0$,so there are no real roots.
Thus,the product of all real roots is $1$.

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

(C) Given,$f(x)=2 x^4-13 x^2+a x+b$ is divisible by $x^2-3 x+2 = (x-2)(x-1)$.
Since $f(x)$ is divisible by $(x-2)$ and $(x-1)$,we must have $f(2)=0$ and $f(1)=0$.
For $f(2)=0$:
$2(2)^4-13(2)^2+a(2)+b=0$
$2(16)-13(4)+2a+b=0$
$32-52+2a+b=0$
$2a+b=20$ ... $(i)$
For $f(1)=0$:
$2(1)^4-13(1)^2+a(1)+b=0$
$2-13+a+b=0$
$a+b=11$ ... $(ii)$
Subtracting Eq. $(ii)$ from Eq. $(i)$:
$(2a+b)-(a+b)=20-11$
$a=9$
Substituting $a=9$ in Eq. $(ii)$:
$9+b=11$
$b=2$
Thus,$(a, b) = (9, 2)$.

(A) Given equation is $x^3-3x-2=0$.
Testing $x=-1$:
$(-1)^3-3(-1)-2 = -1+3-2 = 0$.
Since $x=-1$ is a root,$(x+1)$ is a factor.
Dividing $x^3-3x-2$ by $(x+1)$ using synthetic division or polynomial division:
$x^3-3x-2 = (x+1)(x^2-x-2)$.
Factoring the quadratic part:
$x^2-x-2 = (x+1)(x-2)$.
Thus,the equation becomes $(x+1)(x+1)(x-2)=0$.
The roots are $x = -1, -1, 2$.

(B) Given the equation $x^4-2x^3+2x-1=0$.
Since $1$ is a root of order $3$,$(x-1)^3$ is a factor of the polynomial.
We can write the polynomial as $(x-1)^3(x-k) = 0$,where $k$ is the fourth root.
Expanding $(x-1)^3 = x^3-3x^2+3x-1$.
Multiplying by $(x-k)$: $(x^3-3x^2+3x-1)(x-k) = x^4 - kx^3 - 3x^3 + 3kx^2 + 3x^2 - 3kx - x + k = 0$.
$x^4 - (k+3)x^3 + (3k+3)x^2 - (3k+1)x + k = 0$.
Comparing this with the original equation $x^4-2x^3+0x^2+2x-1=0$:
From the constant term,$k = -1$.
Therefore,the other root is $-1$.