Relation between roots and coefficients Questions in English

(D) Given the quadratic equation $cx^2+bx+a=0$.
Let the roots be $\alpha = \operatorname{cosec} \theta$ and $\beta = \cot \theta$.
From the relation between roots and coefficients,we have:
$\alpha + \beta = -\frac{b}{c}$ and $\alpha \beta = \frac{a}{c}$.
We know the identity $\operatorname{cosec}^2 \theta - \cot^2 \theta = 1$,which implies $\alpha^2 - \beta^2 = 1$.
This can be written as $(\alpha + \beta)(\alpha - \beta) = 1$.
Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$,we have $\alpha - \beta = \pm \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}$.
Substituting the values:
$1 = \left(-\frac{b}{c}\right) \left(\pm \sqrt{\frac{b^2}{c^2} - \frac{4a}{c}}\right)$.
Squaring both sides:
$1 = \frac{b^2}{c^2} \left(\frac{b^2 - 4ac}{c^2}\right)$.
$1 = \frac{b^2(b^2 - 4ac)}{c^4}$.
Therefore,$b^2(b^2 - 4ac) = c^4$.

(D) Let $\alpha, \beta$ be the roots of $x^2+ax+b=0$ and $\gamma, \delta$ be the roots of $x^2+bx+a=0$.
From the relations between roots and coefficients:
$\alpha+\beta = -a, \alpha\beta = b$
$\gamma+\delta = -b, \gamma\delta = a$
Given that the difference between the roots is the same:
$|\alpha-\beta| = |\gamma-\delta|$
Squaring both sides:
$(\alpha-\beta)^2 = (\gamma-\delta)^2$
$(\alpha+\beta)^2 - 4\alpha\beta = (\gamma+\delta)^2 - 4\gamma\delta$
$(-a)^2 - 4b = (-b)^2 - 4a$
$a^2 - 4b = b^2 - 4a$
$a^2 - b^2 + 4a - 4b = 0$
$(a-b)(a+b) + 4(a-b) = 0$
$(a-b)(a+b+4) = 0$
Since $a \neq b$,we have $a-b \neq 0$.
Therefore,$a+b+4 = 0$.

(D) Given the quadratic equation $x^2-10x-8=0$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$\alpha^2 - 10\alpha - 8 = 0 \implies \alpha^2 - 8 = 10\alpha$
$\beta^2 - 10\beta - 8 = 0 \implies \beta^2 - 8 = 10\beta$
We need to evaluate $\frac{a_{10}-8a_8}{5a_9}$.
Substitute $a_n = \alpha^n - \beta^n$:
$\frac{(\alpha^{10}-\beta^{10}) - 8(\alpha^8-\beta^8)}{5(\alpha^9-\beta^9)}$
$= \frac{(\alpha^{10}-8\alpha^8) - (\beta^{10}-8\beta^8)}{5(\alpha^9-\beta^9)}$
$= \frac{\alpha^8(\alpha^2-8) - \beta^8(\beta^2-8)}{5(\alpha^9-\beta^9)}$
Using the relations $\alpha^2-8=10\alpha$ and $\beta^2-8=10\beta$:
$= \frac{\alpha^8(10\alpha) - \beta^8(10\beta)}{5(\alpha^9-\beta^9)}$
$= \frac{10\alpha^9 - 10\beta^9}{5(\alpha^9-\beta^9)}$
$= \frac{10(\alpha^9-\beta^9)}{5(\alpha^9-\beta^9)} = 2$.

(B) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\alpha^n$.
From the relation between roots and coefficients,we have:
$\alpha + \alpha^n = -\frac{b}{a}$ and $\alpha \cdot \alpha^n = \alpha^{n+1} = \frac{c}{a}$.
We need to evaluate the expression $E = (ac^n)^{1/(n+1)} + (a^nc)^{1/(n+1)}$.
Substituting $c = a\alpha^{n+1}$:
$E = (a(a\alpha^{n+1})^n)^{1/(n+1)} + (a^n(a\alpha^{n+1}))^{1/(n+1)}$
$E = (a^{n+1} \alpha^{n(n+1)})^{1/(n+1)} + (a^{n+1} \alpha^{n+1})^{1/(n+1)}$
$E = a\alpha^n + a\alpha$
$E = a(\alpha^n + \alpha)$
Substituting $\alpha^n + \alpha = -\frac{b}{a}$:
$E = a(-\frac{b}{a}) = -b$.

(A) Given $\frac{x_1^5+x_2^5}{x_1^3+x_2^3} = \frac{11}{3}$.
Using the identity $a^5+b^5 = (a^2+b^2)(a^3+b^3) - a^2b^2(a+b)$,we have:
$\frac{(x_1^2+x_2^2)(x_1^3+x_2^3) - x_1^2x_2^2(x_1+x_2)}{x_1^3+x_2^3} = \frac{11}{3}$
$(x_1^2+x_2^2) - \frac{x_1^2x_2^2(x_1+x_2)}{(x_1+x_2)(x_1^2-x_1x_2+x_2^2)} = \frac{11}{3}$
Since $x_1^2+x_2^2 = 5$,we get $5 - \frac{x_1^2x_2^2}{5-x_1x_2} = \frac{11}{3}$.
Let $x_1x_2 = t$. Then $5 - \frac{t^2}{5-t} = \frac{11}{3}$.
$3(25 - 5t - t^2) = 55 - 11t \Rightarrow 3t^2 + 4t - 20 = 0$.
Solving for $t$,we get $t = 2$ or $t = -10/3$.
If $t = 2$,$(x_1+x_2)^2 = x_1^2+x_2^2 + 2x_1x_2 = 5 + 2(2) = 9$,so $x_1+x_2 = \pm 3$.
If $t = -10/3$,$(x_1+x_2)^2 = 5 + 2(-10/3) = -5/3 < 0$,which is impossible for real roots.
Thus,the quadratic equation is $x^2 - (x_1+x_2)x + x_1x_2 = 0$,which gives $x^2 \pm 3x + 2 = 0$.

(A) Given that $\alpha$ and $\beta$ are the roots of $a x^2+b x+c=0$.
Let the roots of the new equation be $t = \sqrt{5} \alpha$ and $t = \sqrt{5} \beta$.
Then $\alpha = \frac{t}{\sqrt{5}}$.
Substituting this into the original equation:
$a(\frac{t}{\sqrt{5}})^2 + b(\frac{t}{\sqrt{5}}) + c = 0$
$\frac{a t^2}{5} + \frac{b t}{\sqrt{5}} + c = 0$
Multiplying the entire equation by $5$:
$a t^2 + 5 \frac{b t}{\sqrt{5}} + 5 c = 0$
$a t^2 + \sqrt{5} b t + 5 c = 0$
Replacing $t$ with $x$,the required equation is $a x^2 + \sqrt{5} b x + 5 c = 0$.

(D) Let $\alpha$ and $\beta$ be the roots of the quadratic equation.
Given $\alpha+\beta=11$ and $\alpha^2+\beta^2=61$.
We know that $(\alpha+\beta)^2 = \alpha^2+\beta^2+2\alpha\beta$.
Substituting the values: $(11)^2 = 61 + 2\alpha\beta$.
$121 = 61 + 2\alpha\beta$ $\Rightarrow 2\alpha\beta = 60$ $\Rightarrow \alpha\beta = 30$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values: $x^2 - 11x + 30 = 0$.

(D) Given,$\alpha$ and $\beta$ are the roots of $2x^2 + 6x + k = 0$.
From the relation between roots and coefficients,$\alpha + \beta = -\frac{6}{2} = -3$ and $\alpha\beta = \frac{k}{2}$.
Now,consider the expression $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$.
Substituting the values,we get $\frac{(-3)^2 - 2(k/2)}{k/2} = \frac{9 - k}{k/2} = \frac{18 - 2k}{k} = \frac{18}{k} - 2$.
Since $k < 0$,let $f(k) = \frac{18}{k} - 2$.
As $k$ becomes more negative (i.e.,$k \to -\infty$),$\frac{18}{k} \to 0$,so $f(k) \to -2$.
For $k < 0$,the expression $\frac{18}{k} - 2$ is always less than $-2$.
Specifically,for $k < -18$,$\frac{18}{k}$ lies in the interval $(-1, 0)$,so $\left[\frac{18}{k} - 2\right] = -1 - 2 = -3$.
However,for $-18 \le k < 0$,the value of $\frac{18}{k}$ is $\le -1$.
The maximum value of the greatest integer function $\left[\frac{18}{k} - 2\right]$ for $k < 0$ is $-2$.

(A) Given that $\tan 30^{\circ}$ and $\tan 15^{\circ}$ are the roots of $x^2+ax+b=0$.
We know that $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
$\tan 15^{\circ} = \tan(45^{\circ}-30^{\circ}) = \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}} = \frac{1-1/\sqrt{3}}{1+1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.
Sum of roots: $-a = \tan 30^{\circ} + \tan 15^{\circ} = \frac{1}{\sqrt{3}} + \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{\sqrt{3}+1+3-\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{4}{\sqrt{3}(\sqrt{3}+1)}$.
So,$a = -\frac{4}{3+\sqrt{3}}$.
Product of roots: $b = \tan 30^{\circ} \cdot \tan 15^{\circ} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{\sqrt{3}-1}{3+\sqrt{3}}$.
Now,$1+a-b = 1 - \frac{4}{\sqrt{3}(\sqrt{3}+1)} - \frac{\sqrt{3}-1}{\sqrt{3}(\sqrt{3}+1)} = \frac{3+\sqrt{3}-4-\sqrt{3}+1}{\sqrt{3}(\sqrt{3}+1)} = \frac{0}{\sqrt{3}(\sqrt{3}+1)} = 0$.

(C) Given the quadratic equation $11 x^2 + 12 x - 13 = 0$.
From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{12}{11}$ and $\alpha \beta = -\frac{13}{11}$.
We need to find $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2}$.
Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta$,we get:
$\frac{(\alpha + \beta)^2 - 2 \alpha \beta}{(\alpha \beta)^2} = \frac{(-\frac{12}{11})^2 - 2(-\frac{13}{11})}{(-\frac{13}{11})^2}$.
$= \frac{\frac{144}{121} + \frac{26}{11}}{\frac{169}{121}} = \frac{\frac{144 + 286}{121}}{\frac{169}{121}} = \frac{430}{169}$.
$\frac{430}{169} \approx 2.544$.
Thus,the value is approximately $2.54$.

(A) Given the quadratic equation: $i x^2 - 2(i + 1) x + (2 - i) = 0$.
Let the roots of the equation be $\alpha$ and $\beta$.
We are given one root $\alpha = 2 - i$.
According to the relation between roots and coefficients,the product of the roots $\alpha \times \beta = \frac{c}{a}$.
Here,$a = i$ and $c = 2 - i$.
So,$(2 - i) \times \beta = \frac{2 - i}{i}$.
Dividing both sides by $(2 - i)$,we get $\beta = \frac{1}{i}$.
Multiplying the numerator and denominator by $i$,we get $\beta = \frac{i}{i^2} = \frac{i}{-1} = -i$.
Thus,the other root is $-i$.

(B) Given $f(x) = 2x^3 + mx^2 - 13x + n$.
Since $2$ and $3$ are roots of $f(x) = 0$,we have $f(2) = 0$ and $f(3) = 0$.
For $f(2) = 0$: $2(2)^3 + m(2)^2 - 13(2) + n = 0$ $\Rightarrow 16 + 4m - 26 + n = 0$ $\Rightarrow 4m + n = 10 \dots (i)$.
For $f(3) = 0$: $2(3)^3 + m(3)^2 - 13(3) + n = 0$ $\Rightarrow 54 + 9m - 39 + n = 0$ $\Rightarrow 9m + n = -15 \dots (ii)$.
Subtracting equation $(i)$ from $(ii)$: $(9m + n) - (4m + n) = -15 - 10$ $\Rightarrow 5m = -25$ $\Rightarrow m = -5$.
Substituting $m = -5$ into equation $(i)$: $4(-5) + n = 10$ $\Rightarrow -20 + n = 10$ $\Rightarrow n = 30$.
Thus,the values are $m = -5$ and $n = 30$.

(C) Given the cubic equation: $9x^3 + 112x^2 - 120x + a = 0$.
Let the roots be $\alpha, \beta, \gamma$.
The product of the roots is given as $\alpha \beta \gamma = 12$.
For a cubic equation $Ax^3 + Bx^2 + Cx + D = 0$,the product of the roots is given by the formula $\alpha \beta \gamma = -\frac{D}{A}$.
Here,$A = 9$ and $D = a$.
Substituting the values,we get: $-\frac{a}{9} = 12$.
Multiplying both sides by $9$,we get: $-a = 12 \times 9 = 108$.
Therefore,$a = -108$.

(NONE) Given the equation $x^3-6x^2+11x-6=0$,the roots are $\alpha, \beta, \gamma$.
From Vieta's formulas,we have:
$\alpha+\beta+\gamma = 6$
$\alpha\beta+\beta\gamma+\gamma\alpha = 11$
$\alpha\beta\gamma = 6$
We need to evaluate $\Sigma \alpha^2 \beta + \Sigma \alpha \beta^2$.
Note that $(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) = (\alpha^2\beta + \alpha\beta\gamma + \alpha^2\gamma) + (\alpha\beta^2 + \beta^2\gamma + \alpha\beta\gamma) + (\alpha\beta\gamma + \beta\gamma^2 + \gamma^2\alpha) = \Sigma \alpha^2 \beta + \Sigma \alpha \beta^2 + 3\alpha\beta\gamma$.
Therefore,$\Sigma \alpha^2 \beta + \Sigma \alpha \beta^2 = (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) - 3\alpha\beta\gamma$.
Substituting the values:
$= (6)(11) - 3(6) = 66 - 18 = 48$.

(B) Given that $\alpha, \beta, \gamma$ are roots of the cubic equation $x^3+4x-19=0$.
Since $\alpha$ is a root,it must satisfy the equation:
$\alpha^3 + 4\alpha - 19 = 0$
$\Rightarrow \alpha^3 = 19 - 4\alpha$
Dividing both sides by $(19 - 4\alpha)$,we get:
$\frac{\alpha^3}{19 - 4\alpha} = 1$
Similarly,since $\beta$ and $\gamma$ are also roots of the same equation:
$\frac{\beta^3}{19 - 4\beta} = 1$
$\frac{\gamma^3}{19 - 4\gamma} = 1$
Adding these three expressions:
$\frac{\alpha^3}{19-4\alpha} + \frac{\beta^3}{19-4\beta} + \frac{\gamma^3}{19-4\gamma} = 1 + 1 + 1 = 3$

(D) Given the equation: $\frac{x^2-bx}{ax-c} = \frac{m-1}{m+1}$
Cross-multiplying,we get: $(x^2-bx)(m+1) = (m-1)(ax-c)$
Expanding the terms: $x^2(m+1) - bx(m+1) = ax(m-1) - c(m-1)$
Rearranging into the standard quadratic form $Ax^2 + Bx + C = 0$:
$x^2(m+1) - x(b(m+1) + a(m-1)) + c(m-1) = 0$
Let the roots be $p$ and $-p$.
For a quadratic equation $Ax^2 + Bx + C = 0$,the sum of roots is given by $-\frac{B}{A}$.
Since the sum of roots $p + (-p) = 0$,the coefficient of $x$ must be zero:
$b(m+1) + a(m-1) = 0$
$bm + b + am - a = 0$
$m(a+b) = a-b$
Therefore,$m = \frac{a-b}{a+b}$

(B) Given the polynomial equation: $x^4-3x^2+6x-12=0$ ... $(i)$
Comparing this with $ax^4+bx^3+cx^2+dx+e=0$,we get $a=1, b=0, c=-3, d=6, e=-12$.
Since $\alpha, \beta, \gamma, \delta$ are the roots,by Vieta's formulas:
$\alpha+\beta+\gamma+\delta = -\frac{b}{a} = 0$.
This implies $\alpha+\beta+\gamma = -\delta$,$\alpha+\delta+\gamma = -\beta$,$\alpha+\beta+\delta = -\gamma$,and $\delta+\beta+\gamma = -\alpha$.
Also,$\Sigma \alpha\beta\gamma = -\frac{d}{a} = -6$ and $\alpha\beta\gamma\delta = \frac{e}{a} = -12$.
Substituting these into the given expression:
$\frac{-\delta}{\delta^2} + \frac{-\beta}{\beta^2} + \frac{-\gamma}{\gamma^2} + \frac{-\alpha}{\alpha^2} = -(\frac{1}{\delta} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\alpha})$.
This simplifies to: $-(\frac{\alpha\beta\gamma + \alpha\gamma\delta + \alpha\beta\delta + \beta\gamma\delta}{\alpha\beta\gamma\delta}) = -(\frac{\Sigma \alpha\beta\gamma}{\alpha\beta\gamma\delta})$.
Substituting the values: $-(\frac{-6}{-12}) = -(\frac{1}{2}) = -\frac{1}{2}$.

(D) Given equation is $x^2+px+q=0$ ... $(i)$
Comparing with $ax^2+bx+c=0$,we get $a=1, b=p, c=q$.
Let the roots be $\alpha$ and $\beta$. Given $\alpha = \beta^2$.
From the relation between roots and coefficients:
Sum of roots: $\alpha + \beta = -p \Rightarrow \beta^2 + \beta = -p$ ... (ii)
Product of roots: $\alpha \beta = q$ $\Rightarrow \beta^2 \cdot \beta = q$ $\Rightarrow \beta^3 = q$ ... (iii)
Cubing equation (ii) on both sides:
$(\beta^2 + \beta)^3 = (-p)^3$
$\beta^6 + \beta^3 + 3\beta^2 \cdot \beta(\beta^2 + \beta) = -p^3$
Substitute $\beta^3 = q$ and $\beta^2 + \beta = -p$:
$(q)^2 + q + 3q(-p) = -p^3$
$q^2 + q - 3pq = -p^3$
$p^3 - 3pq = -q^2 - q$
$p(p^2 - 3q) = -q(q+1)$
$p(3q - p^2) = q(q+1)$

(B) Given,$\alpha, \beta$ and $\gamma$ are roots of the cubic equation $x^3-3x^2+x+5=0$.
From the relation between roots and coefficients:
$\alpha+\beta+\gamma = -(-3)/1 = 3$
$\alpha \beta+\beta \gamma+\gamma \alpha = 1/1 = 1$
$\alpha \beta \gamma = -5/1 = -5$
Now,$y = \Sigma \alpha^2 + \alpha \beta \gamma = (\alpha^2+\beta^2+\gamma^2) + \alpha \beta \gamma$.
Using the identity $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha \beta+\beta \gamma+\gamma \alpha)$:
$y = (3)^2 - 2(1) + (-5) = 9 - 2 - 5 = 2$.
Since $y=2$,we check which equation is satisfied by $y=2$:
For option $B$: $y^3-y^2-y-2 = (2)^3 - (2)^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0$.
Thus,$y=2$ satisfies the equation $y^3-y^2-y-2=0$.

(C) Given the quadratic equation $ax^2 + ax + c = 0$.
Let the roots be $\alpha$ and $\beta$.
From the relation between roots and coefficients:
Sum of roots $\alpha + \beta = -\frac{a}{a} = -1$.
Product of roots $\alpha \beta = \frac{c}{a}$.
Given the ratio of roots $\frac{\alpha}{\beta} = \frac{p}{q}$.
We need to find $\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} = \sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}$.
Simplifying the expression:
$\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} = \frac{\sqrt{\alpha}}{\sqrt{\beta}} + \frac{\sqrt{\beta}}{\sqrt{\alpha}} = \frac{\alpha + \beta}{\sqrt{\alpha \beta}}$.
Substituting the values:
$\frac{-1}{\sqrt{\frac{c}{a}}} = -\sqrt{\frac{a}{c}}$.
Considering the magnitude or the positive root form as per standard options,the result is $\sqrt{\frac{a}{c}}$.

(C) Let the roots of the quadratic equation be $\alpha$ and $\beta$. It is given that $\alpha+\beta=1$ and $\alpha^2+\beta^2=13$.
We know that $(\alpha+\beta)^2 = \alpha^2+\beta^2+2\alpha\beta$.
Substituting the given values: $1^2 = 13 + 2\alpha\beta$.
$1 = 13 + 2\alpha\beta$ $\Rightarrow 2\alpha\beta = -12$ $\Rightarrow \alpha\beta = -6$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values: $x^2 - (1)x + (-6) = 0 \Rightarrow x^2-x-6=0$.

(C) The given equation is $x^2+x+1=0$.
Since the roots are $\alpha$ and $\beta$,we know that $\alpha$ and $\beta$ are the non-real cube roots of unity,i.e.,$\omega$ and $\omega^2$.
Thus,$\alpha^3 = 1$ and $\beta^3 = 1$.
Also,from the equation,$\alpha + \beta = -1$ and $\alpha \beta = 1$.
We need to find $\alpha^4 + \beta^4$.
Since $\alpha^3 = 1$,$\alpha^4 = \alpha^3 \cdot \alpha = 1 \cdot \alpha = \alpha$.
Similarly,$\beta^4 = \beta^3 \cdot \beta = 1 \cdot \beta = \beta$.
Therefore,$\alpha^4 + \beta^4 = \alpha + \beta$.
Substituting the value of $\alpha + \beta$,we get $\alpha^4 + \beta^4 = -1$.

(B) Let the roots of the polynomial $f(x) = x^3 - 9x^2 + 26x - 24$ be $\alpha, \beta, \gamma$.
To find the equation whose roots are the reciprocals $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$,we substitute $x$ with $\frac{1}{x}$ in the original equation $f(x) = 0$.
$\left(\frac{1}{x}\right)^3 - 9\left(\frac{1}{x}\right)^2 + 26\left(\frac{1}{x}\right) - 24 = 0$.
Multiplying the entire equation by $-x^3$,we get:
$-1 + 9x - 26x^2 + 24x^3 = 0$.
Rearranging the terms,we obtain $24x^3 - 26x^2 + 9x - 1 = 0$.
Thus,the correct option is $B$.

(D) Given that $\alpha$ and $\beta$ are roots of $p x^2 + q x + r = 0$. Since $p, q, r$ are in $AP$,we have $2q = p + r$.
From $\frac{1}{\alpha} + \frac{1}{\beta} = 4$,we get $\frac{\alpha + \beta}{\alpha \beta} = 4$.
Using Vieta's formulas,$\alpha + \beta = -\frac{q}{p}$ and $\alpha \beta = \frac{r}{p}$.
Substituting these,$\frac{-q/p}{r/p} = -\frac{q}{r} = 4$,so $q = -4r$.
Since $p + r = 2q$,we have $p + r = 2(-4r) = -8r$,which implies $p = -9r$.
Now,$|\alpha - \beta| = \frac{\sqrt{D}}{|p|} = \frac{\sqrt{q^2 - 4pr}}{|p|}$.
Substituting $q = -4r$ and $p = -9r$:
$|\alpha - \beta| = \frac{\sqrt{(-4r)^2 - 4(-9r)(r)}}{|-9r|} = \frac{\sqrt{16r^2 + 36r^2}}{9|r|} = \frac{\sqrt{52r^2}}{9|r|} = \frac{2|r|\sqrt{13}}{9|r|} = \frac{2\sqrt{13}}{9}$.
Thus,the correct option is $D$.

(D) Given that $\alpha$ and $\beta$ are roots of the equation $x^2+p x+q=0$.
From the relation between roots and coefficients,we have $\alpha+\beta = -p$ and $\alpha \beta = q$.
For $\alpha^3+\beta^3$:
$\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2-\alpha \beta+\beta^2) = (\alpha+\beta)[(\alpha+\beta)^2-3 \alpha \beta] = (-p)[(-p)^2-3 q] = -p(p^2-3 q) = 3 p q-p^3$.
For $\alpha^4+\alpha^2 \beta^2+\beta^4$:
$\alpha^4+\alpha^2 \beta^2+\beta^4 = (\alpha^2+\beta^2)^2 - \alpha^2 \beta^2 = [(\alpha+\beta)^2-2 \alpha \beta]^2 - (\alpha \beta)^2 = [(-p)^2-2 q]^2 - q^2 = (p^2-2 q)^2 - q^2$.
Using the identity $a^2-b^2 = (a-b)(a+b)$,we get $(p^2-2 q-q)(p^2-2 q+q) = (p^2-3 q)(p^2-q)$.

(A) Given the equation $x^4+3x^3-6x^2+2x-4=0$ with roots $\alpha, \beta, \gamma, \delta$.
To find the equation with roots $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta}$,we replace $x$ with $\frac{1}{x}$ in the original equation:
$(\frac{1}{x})^4 + 3(\frac{1}{x})^3 - 6(\frac{1}{x})^2 + 2(\frac{1}{x}) - 4 = 0$
Multiply the entire equation by $x^4$:
$1 + 3x - 6x^2 + 2x^3 - 4x^4 = 0$
Rearranging the terms:
$-4x^4 + 2x^3 - 6x^2 + 3x + 1 = 0$
Multiplying by $-1$:
$4x^4 - 2x^3 + 6x^2 - 3x - 1 = 0$

(B) Given the equation $x^3-6x^2+11x-6=0$ with roots $\alpha, \beta, \gamma$.
Let $y = x^2$,so $x = \sqrt{y}$.
Substituting into the original equation: $(\sqrt{y})^3 - 6(\sqrt{y})^2 + 11\sqrt{y} - 6 = 0$.
$y\sqrt{y} - 6y + 11\sqrt{y} - 6 = 0$.
Rearranging terms: $\sqrt{y}(y+11) = 6(y+1)$.
Squaring both sides: $y(y+11)^2 = 36(y+1)^2$.
$y(y^2 + 22y + 121) = 36(y^2 + 2y + 1)$.
$y^3 + 22y^2 + 121y = 36y^2 + 72y + 36$.
$y^3 - 14y^2 + 49y - 36 = 0$.
Replacing $y$ with $x$,the required equation is $x^3-14x^2+49x-36=0$.

(A) Given the cubic equation $3x^3 - 9x^2 + 5x - 7 = 0$.
Comparing this with the standard form $ax^3 + bx^2 + cx + d = 0$,we have $a = 3$,$b = -9$,$c = 5$,and $d = -7$.
According to the relation between roots and coefficients for a cubic equation,the sum of the roots $\alpha + \beta + \gamma = -\frac{b}{a}$.
Substituting the values,we get $\alpha + \beta + \gamma = -\frac{-9}{3} = \frac{9}{3} = 3$.
Thus,the value of $\alpha + \beta + \gamma$ is $3$.

(D) Let $y = \frac{2\alpha}{3-4\alpha}$. Then $2\alpha = 3y - 4\alpha y$,which implies $\alpha(2+4y) = 3y$,so $\alpha = \frac{3y}{2+4y}$.
Since $\alpha$ is a root of $x^2+7x+3=0$,we substitute $\alpha$:
$(\frac{3y}{2+4y})^2 + 7(\frac{3y}{2+4y}) + 3 = 0$.
Multiplying by $(2+4y)^2 = 16y^2+16y+4$:
$9y^2 + 21y(2+4y) + 3(16y^2+16y+4) = 0$.
$9y^2 + 42y + 84y^2 + 48y^2 + 48y + 12 = 0$.
$141y^2 + 90y + 12 = 0$.
Dividing by $3$: $47y^2 + 30y + 4 = 0$.
Comparing with $ax^2+bx+c=0$,we get $a=47, b=30, c=4$. Since $GCD(47, 30, 4) = 1$,the values are valid.
Thus,$a+b+c = 47+30+4 = 81$.

(A) Given the cubic equation $x^3-ax^2+bx-c=0$ with roots $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha+\beta+\gamma = a$
$\alpha\beta+\beta\gamma+\gamma\alpha = b$
$\alpha\beta\gamma = c$
We need to evaluate $\Sigma \alpha^2(\beta+\gamma)$.
Since $\alpha+\beta+\gamma = a$,we have $\beta+\gamma = a-\alpha$.
Substituting this into the expression:
$\Sigma \alpha^2(a-\alpha) = a\Sigma \alpha^2 - \Sigma \alpha^3$
We know that $\Sigma \alpha^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = a^2-2b$.
Also,for a cubic equation $x^3-ax^2+bx-c=0$,the sum of cubes $\Sigma \alpha^3$ is given by the identity:
$\Sigma \alpha^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\Sigma \alpha^2 - \Sigma \alpha\beta)$
$\Sigma \alpha^3 = a(a^2-2b-b) + 3c = a(a^2-3b) + 3c = a^3-3ab+3c$.
Now,substituting these into the expression:
$a(a^2-2b) - (a^3-3ab+3c) = a^3-2ab-a^3+3ab-3c = ab-3c$.
Thus,the correct option is $A$.

(A) Given the equation $x^3-6x^2+11x-6=0$.
Factoring the cubic equation,we get $(x-1)(x-2)(x-3)=0$.
Thus,the roots are $\alpha=1, \beta=2, \gamma=3$.
Now,calculate the new roots:
$\alpha' = \alpha^2+\beta^2 = 1^2+2^2 = 5$
$\beta' = \beta^2+\gamma^2 = 2^2+3^2 = 13$
$\gamma' = \gamma^2+\alpha^2 = 3^2+1^2 = 10$
The required equation is $x^3 - (\alpha'+\beta'+\gamma')x^2 + (\alpha'\beta'+\beta'\gamma'+\gamma'\alpha')x - \alpha'\beta'\gamma' = 0$.
Sum of roots: $5+13+10 = 28$.
Sum of roots taken two at a time: $(5 \times 13) + (13 \times 10) + (10 \times 5) = 65 + 130 + 50 = 245$.
Product of roots: $5 \times 13 \times 10 = 650$.
Substituting these values,we get $x^3 - 28x^2 + 245x - 650 = 0$.

(C) The given quadratic equation is $x^2+4kx+12e^{3\log k}-1=0$.
Using the property $e^{\log a} = a$,we have $e^{3\log k} = e^{\log k^3} = k^3$.
Thus,the equation becomes $x^2+4kx+12k^3-1=0$.
The product of the roots of a quadratic equation $ax^2+bx+c=0$ is given by $\frac{c}{a}$.
Here,the product is $12k^3-1$.
Given that the product of the roots is $323$,we have $12k^3-1 = 323$.
$12k^3 = 324$ $\Rightarrow k^3 = 27$ $\Rightarrow k = 3$.
The sum of the roots is given by $-\frac{b}{a} = -4k$.
Substituting $k=3$,the sum of the roots is $-4(3) = -12$.
Therefore,option $C$ is correct.

(B) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^2-4x+5=0$.
From the relation between roots and coefficients,we have $\alpha+\beta=4$ and $\alpha\beta=5$.
Let the new roots be $S_1 = \alpha^2+\beta$ and $S_2 = \alpha+\beta^2$.
The sum of the new roots is:
$S_1+S_2 = (\alpha^2+\beta)+(\alpha+\beta^2) = (\alpha^2+\beta^2)+(\alpha+\beta)$
$= ((\alpha+\beta)^2-2\alpha\beta)+(\alpha+\beta)$
$= (4^2-2(5))+4 = (16-10)+4 = 6+4 = 10$.
The product of the new roots is:
$S_1 \times S_2 = (\alpha^2+\beta)(\alpha+\beta^2) = \alpha^3+\alpha^2\beta^2+\alpha\beta+\beta^3$
$= (\alpha^3+\beta^3)+(\alpha\beta)^2+\alpha\beta$
$= ((\alpha+\beta)^3-3\alpha\beta(\alpha+\beta))+(\alpha\beta)^2+\alpha\beta$
$= (4^3-3(5)(4))+(5)^2+5$
$= (64-60)+25+5 = 4+30 = 34$.
The required quadratic equation is $x^2-(S_1+S_2)x+(S_1 \times S_2) = 0$.
Substituting the values,we get $x^2-10x+34=0$.

(A) Given the equation $ax^2+bx+c=0$ has roots $\alpha$ and $\beta$,we have $\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Consider the equation $ax^2-bx(x-1)+c(x-1)^2=0$.
Divide the entire equation by $(x-1)^2$ (assuming $x \neq 1$):
$a(\frac{x}{x-1})^2 - b(\frac{x}{x-1}) + c = 0$.
Let $y = \frac{x}{x-1}$.
Then the equation becomes $ay^2 - by + c = 0$.
This is not quite the original form. Let us rewrite the original equation as $a(\frac{x}{x-1})^2 + b(\frac{x}{x-1}) + c = 0$ if we had $ax^2+bx(x-1)+c(x-1)^2=0$.
For $ax^2-bx(x-1)+c(x-1)^2=0$,dividing by $(x-1)^2$ gives $a(\frac{x}{x-1})^2 - b(\frac{x}{x-1}) + c = 0$.
Let $Y = \frac{x}{x-1}$. Then $aY^2 - bY + c = 0$.
Since $a(-Y)^2 + b(-Y) + c = 0$ corresponds to roots $\alpha, \beta$,then $Y = -\alpha$ and $Y = -\beta$ are roots of $aY^2+bY+c=0$.
Actually,the roots of $aY^2-bY+c=0$ are $-\alpha$ and $-\beta$ is incorrect. The roots of $aY^2-bY+c=0$ are $\alpha$ and $\beta$ if we compare $aY^2-bY+c=0$ with $aY^2+bY+c=0$ by replacing $Y$ with $-Y$.
Thus,$\frac{x}{x-1} = \alpha \implies x = \alpha x - \alpha \implies x(1-\alpha) = -\alpha \implies x = \frac{\alpha}{\alpha-1}$.
Similarly,$x = \frac{\beta}{\beta-1}$.
Therefore,the roots are $\frac{\alpha}{\alpha-1}$ and $\frac{\beta}{\beta-1}$.

(B) Let the two numbers be $a$ and $b$.
Given,Geometric Mean ($G$.$M$.) $= \sqrt{ab} = 2$,so $ab = 4$ (Eq. $i$).
Harmonic Mean ($H$.$M$.) $= \frac{2ab}{a+b} = -\frac{8}{5}$.
Substituting $ab = 4$ into the $H$.$M$. formula:
$\frac{2(4)}{a+b} = -\frac{8}{5} \implies \frac{8}{a+b} = -\frac{8}{5} \implies a+b = -5$.
The roots of the required quadratic equation are $2a$ and $2b$.
Sum of roots $= 2a + 2b = 2(a+b) = 2(-5) = -10$.
Product of roots $= (2a)(2b) = 4ab = 4(4) = 16$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Therefore,$x^2 - (-10)x + 16 = 0$,which simplifies to $x^2 + 10x + 16 = 0$.

(C) Given that $\tan \alpha$ and $\tan \beta$ are the roots of $x^2+px+q=0$.
From the relation between roots and coefficients,we have $\tan \alpha + \tan \beta = -p$ and $\tan \alpha \tan \beta = q$.
Let $E = \sin^2(\alpha+\beta) + p\cos(\alpha+\beta)\sin(\alpha+\beta) + q\cos^2(\alpha+\beta)$.
Dividing the expression by $\cos^2(\alpha+\beta)$,we get $E = \cos^2(\alpha+\beta) [\tan^2(\alpha+\beta) + p\tan(\alpha+\beta) + q]$.
We know $\tan(\alpha+\beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{-p}{1-q} = \frac{p}{q-1}$.
Substituting this into the expression inside the bracket: $\tan^2(\alpha+\beta) + p\tan(\alpha+\beta) + q = \frac{p^2}{(q-1)^2} + p(\frac{p}{q-1}) + q = \frac{p^2 + p^2(q-1) + q(q-1)^2}{(q-1)^2} = \frac{p^2 + p^2q - p^2 + q(q^2-2q+1)}{(q-1)^2} = \frac{p^2q + q^3 - 2q^2 + q}{(q-1)^2} = \frac{q(p^2 + q^2 - 2q + 1)}{(q-1)^2} = \frac{q((q-1)^2 + p^2)}{(q-1)^2}$.
Now,$\cos^2(\alpha+\beta) = \frac{1}{1+\tan^2(\alpha+\beta)} = \frac{1}{1+\frac{p^2}{(q-1)^2}} = \frac{(q-1)^2}{(q-1)^2+p^2}$.
Thus,$E = \frac{(q-1)^2}{(q-1)^2+p^2} \times \frac{q((q-1)^2 + p^2)}{(q-1)^2} = q$.

(B) Let the roots of the cubic equation $x^3-7x^2+0x+36=0$ be $\alpha, 2\alpha,$ and $\beta$.
From the relation between roots and coefficients:
Sum of roots: $\alpha + 2\alpha + \beta = 7 \implies 3\alpha + \beta = 7 \implies \beta = 7 - 3\alpha$.
Sum of roots taken two at a time: $\alpha(2\alpha) + 2\alpha\beta + \beta\alpha = 0 \implies 2\alpha^2 + 3\alpha\beta = 0$.
Since $\alpha \neq 0$ (as $36 \neq 0$),we divide by $\alpha$: $2\alpha + 3\beta = 0$.
Substitute $\beta = 7 - 3\alpha$ into the equation: $2\alpha + 3(7 - 3\alpha) = 0$.
$2\alpha + 21 - 9\alpha = 0 \implies -7\alpha = -21 \implies \alpha = 3$.
Then the roots are $\alpha = 3$ and $2\alpha = 6$.
The sum of these two roots is $3 + 6 = 9$.

(D) Given equation is $(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0$.
Let $\alpha$ and $\beta$ be the roots of this equation.
From the relation between roots and coefficients:
$\alpha+\beta = \frac{b}{5+\sqrt{2}}$
$\alpha \beta = \frac{8+2 \sqrt{5}}{5+\sqrt{2}}$
The harmonic mean $(HM)$ between the roots is given by $HM = \frac{2 \alpha \beta}{\alpha+\beta}$.
Given $HM = 4$,we have:
$\frac{2 \alpha \beta}{\alpha+\beta} = 4$
Substituting the values of $\alpha+\beta$ and $\alpha \beta$:
$\frac{2 \times \frac{8+2 \sqrt{5}}{5+\sqrt{2}}}{\frac{b}{5+\sqrt{2}}} = 4$
$\frac{2(8+2 \sqrt{5})}{b} = 4$
$\frac{8+2 \sqrt{5}}{b} = 2$
$b = \frac{8+2 \sqrt{5}}{2} = 4+\sqrt{5}$.

(C) Given,$\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$.
From the properties of roots,we have:
$\alpha+\beta = 2$ ...$(i)$
$\alpha\beta = 4$ ...$(ii)$
Now,we can write $\alpha^6+\beta^6 = (\alpha^2)^3 + (\beta^2)^3$.
Alternatively,note that the roots of $x^2-2x+4=0$ are $x = \frac{2 \pm \sqrt{4-16}}{2} = 1 \pm \sqrt{3}i$.
In polar form,$1 \pm \sqrt{3}i = 2(\frac{1}{2} \pm i\frac{\sqrt{3}}{2}) = 2e^{\pm i\pi/3}$.
Thus,$\alpha = 2e^{i\pi/3}$ and $\beta = 2e^{-i\pi/3}$.
Then,$\alpha^6 = (2e^{i\pi/3})^6 = 2^6 e^{i2\pi} = 64(1) = 64$.
Similarly,$\beta^6 = (2e^{-i\pi/3})^6 = 2^6 e^{-i2\pi} = 64(1) = 64$.
Therefore,$\alpha^6+\beta^6 = 64+64 = 128$.

(D) The condition for the roots of $ax^2 + bx + c = 0$ to be in the ratio $m:n$ is $mnb^2 = ac(m+n)^2$.
$(i)$ If $\alpha = \beta$,then the ratio is $1:1$. Substituting $m=1, n=1$ into the formula: $(1)(1)b^2 = ac(1+1)^2 \Rightarrow b^2 = 4ac$. This matches $(E)$.
$(ii)$ If $\alpha = 2\beta$,then the ratio is $2:1$. Substituting $m=2, n=1$: $(2)(1)b^2 = ac(2+1)^2 \Rightarrow 2b^2 = 9ac$. This matches $(B)$.
$(iii)$ If $\alpha = 3\beta$,then the ratio is $3:1$. Substituting $m=3, n=1$: $(3)(1)b^2 = ac(3+1)^2 \Rightarrow 3b^2 = 16ac$. This matches $(D)$.
$(iv)$ If $\alpha = \beta^2$,then $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$. Substituting $\alpha = \beta^2$,we get $\beta^2 + \beta = -b/a$ and $\beta^3 = c/a$. Thus $\beta = (c/a)^{1/3}$. Substituting this into $\beta^2 + \beta = -b/a$: $(c/a)^{2/3} + (c/a)^{1/3} = -b/a$. Multiplying by $a$: $a(c/a)^{2/3} + a(c/a)^{1/3} = -b \Rightarrow (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$. This matches $(A)$.
Therefore,the correct matching is $(i)-E, (ii)-B, (iii)-D, (iv)-A$.

(D) Let the roots of the equation $x^3-13x^2+15x+189=0$ be $\alpha, \alpha+2,$ and $\beta$.
From the relation between roots and coefficients:
$1) \alpha + (\alpha+2) + \beta = 13 \implies 2\alpha + \beta = 11 \implies \beta = 11 - 2\alpha$
$2) \alpha(\alpha+2) + (\alpha+2)\beta + \alpha\beta = 15$
$3) \alpha(\alpha+2)\beta = -189$
Substitute $\beta = 11 - 2\alpha$ into the third equation:
$\alpha(\alpha+2)(11-2\alpha) = -189$
$(\alpha^2+2\alpha)(11-2\alpha) = -189$
$11\alpha^2 - 2\alpha^3 + 22\alpha - 4\alpha^2 = -189$
$-2\alpha^3 + 7\alpha^2 + 22\alpha + 189 = 0$
$2\alpha^3 - 7\alpha^2 - 22\alpha - 189 = 0$
Testing values,if $\alpha = -3$:
$2(-27) - 7(9) - 22(-3) - 189 = -54 - 63 + 66 - 189 \neq 0$
If $\alpha = 7$:
$2(343) - 7(49) - 22(7) - 189 = 686 - 343 - 154 - 189 = 0$.
So,$\alpha = 7$ is a root. The roots are $7, 7+2=9,$ and $\beta = 11-2(7) = -3$.
The roots are $-3, 7, 9$.

(C) Given the equation $x^3 - 10x^2 + 7x + 8 = 0$.
From the relation between roots and coefficients:
$\alpha + \beta + \gamma = 10$ ($A$-$5$)
$\alpha\beta + \beta\gamma + \gamma\alpha = 7$
$\alpha\beta\gamma = -8$
Now,$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = (10)^2 - 2(7) = 100 - 14 = 86$ ($B$-$3$)
Next,$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{7}{-8} = -\frac{7}{8}$ ($C$-$2$)
Finally,$\frac{\alpha}{\beta\gamma} + \frac{\beta}{\gamma\alpha} + \frac{\gamma}{\alpha\beta} = \frac{\alpha^2 + \beta^2 + \gamma^2}{\alpha\beta\gamma} = \frac{86}{-8} = -\frac{43}{4}$ ($D$-$1$)
Thus,the correct matching is $A-5, B-3, C-2, D-1$.