Relation between roots and coefficients Questions in English

(A) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^2 - 18x + 9 = 0$.
By the properties of quadratic equations,the product of the roots $\alpha \beta = \frac{c}{a} = \frac{9}{1} = 9$.
The $G.M.$ (Geometric Mean) of two numbers $\alpha$ and $\beta$ is defined as $\sqrt{\alpha \beta}$.
Therefore,$G.M. = \sqrt{9} = 3$.

(D) Let the roots of the quadratic equation $x^2 - 10x + 11 = 0$ be $\alpha$ and $\beta$.
From the properties of quadratic equations,the sum of roots $\alpha + \beta = -(-10)/1 = 10$ and the product of roots $\alpha \beta = 11/1 = 11$.
The Harmonic Mean $(H.M.)$ of two numbers $\alpha$ and $\beta$ is given by the formula $H.M. = \frac{2\alpha \beta}{\alpha + \beta}$.
Substituting the values,we get $H.M. = \frac{2 \times 11}{10} = \frac{22}{10} = \frac{11}{5}$.

(A) Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.
Given that $\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$.
$\alpha + \beta = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{(\alpha \beta)^2}$.
Substituting the values of $\alpha + \beta$ and $\alpha \beta$:
$-\frac{b}{a} = \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} = \frac{b^2 - 2ac}{c^2}$.
$-bc^2 = a(b^2 - 2ac) = ab^2 - 2a^2c$.
$2a^2c = ab^2 + bc^2$.
Dividing both sides by $abc$:
$\frac{2a^2c}{abc} = \frac{ab^2}{abc} + \frac{bc^2}{abc}$.
$\frac{2a}{b} = \frac{b}{c} + \frac{c}{a}$.
This implies that $\frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in $A.P.$ since $2(\text{middle term}) = \text{first term} + \text{third term}$.

(C) For the equation $x^2 - 2ax + b^2 = 0,$ the sum of the roots is $2a.$
Therefore,the $A.M.$ of the roots is $A = \frac{2a}{2} = a.$
For the equation $x^2 - 2bx + a^2 = 0,$ the product of the roots is $a^2.$
Therefore,the $G.M.$ of the roots is $G = \sqrt{a^2} = |a|.$
Assuming $a > 0,$ we get $A = a$ and $G = a.$
Thus,$A = G.$

(C) Given that $A.M. = 8$ and $G.M. = 5$.
Let $\alpha$ and $\beta$ be the roots of the quadratic equation.
The quadratic equation is given by $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.
Since $A.M. = \frac{\alpha + \beta}{2} = 8$,we have $\alpha + \beta = 16$.
Since $G.M. = \sqrt{\alpha\beta} = 5$,we have $\alpha\beta = 25$.
Substituting these values into the equation,we get $x^2 - 16x + 25 = 0$.

(A) For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.
Comparing the given equation $2x^2 + 6x + (\alpha^2 + 1) = 0$ with the standard form,we have $a = 2$ and $c = \alpha^2 + 1$.
The product of the roots is $\frac{\alpha^2 + 1}{2}$.
Given that the product of the roots is $-\alpha$,we have $\frac{\alpha^2 + 1}{2} = -\alpha$.
Multiplying both sides by $2$,we get $\alpha^2 + 1 = -2\alpha$.
Rearranging the terms,we get $\alpha^2 + 2\alpha + 1 = 0$.
This is a perfect square: $(\alpha + 1)^2 = 0$.
Thus,$\alpha = -1$.

(B) Let the roots of the quadratic equation $2x^2 + 3(\lambda - 2)x + \lambda + 4 = 0$ be $\alpha$ and $-\alpha$.
For a quadratic equation $ax^2 + bx + c = 0$,the sum of the roots is given by $-\frac{b}{a}$.
Here,$a = 2$ and $b = 3(\lambda - 2)$.
Sum of the roots: $\alpha + (-\alpha) = -\frac{3(\lambda - 2)}{2}$.
$0 = -\frac{3(\lambda - 2)}{2}$.
This implies $3(\lambda - 2) = 0$.
Therefore,$\lambda - 2 = 0$,which gives $\lambda = 2$.

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

(B) Let the first root be $\alpha$.
Since the second root is the reciprocal of the first,the second root is $\frac{1}{\alpha}$.
For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.
Here,$a = 5$,$b = 13$,and $c = k$.
Therefore,the product of the roots is $\alpha \times \frac{1}{\alpha} = \frac{k}{5}$.
$1 = \frac{k}{5}$.
$k = 5$.

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

(C) Given that $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$.
From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Let the roots of the equation $cx^2 + bx + a = 0$ be $\alpha'$ and $\beta'$.
Then,$\alpha' + \beta' = -\frac{b}{c}$ and $\alpha'\beta' = \frac{a}{c}$.
We observe that $\alpha'\beta' = \frac{a}{c} = \frac{1}{c/a} = \frac{1}{\alpha\beta}$.
Also,$\alpha' + \beta' = -\frac{b}{c} = \frac{-b/a}{c/a} = \frac{\alpha + \beta}{\alpha\beta} = \frac{1}{\beta} + \frac{1}{\alpha}$.
Thus,the roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.

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

(B) Given the quadratic equation $2x^2 + 2(a + b)x + a^2 + b^2 = 0$.
Sum of roots $\alpha + \beta = -\frac{2(a + b)}{2} = -(a + b)$.
Product of roots $\alpha\beta = \frac{a^2 + b^2}{2}$.
Now,$(\alpha + \beta)^2 = (-(a + b))^2 = (a + b)^2 = a^2 + b^2 + 2ab$.
Also,$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (a + b)^2 - 4\left(\frac{a^2 + b^2}{2}\right) = a^2 + b^2 + 2ab - 2a^2 - 2b^2 = 2ab - a^2 - b^2 = -(a - b)^2$.
The required equation is $x^2 - [(\alpha + \beta)^2 + (\alpha - \beta)^2]x + [(\alpha + \beta)^2 \cdot (\alpha - \beta)^2] = 0$.
Sum of new roots: $(a + b)^2 - (a - b)^2 = (a^2 + b^2 + 2ab) - (a^2 + b^2 - 2ab) = 4ab$.
Product of new roots: $(a + b)^2 \cdot (-(a - b)^2) = -((a + b)(a - b))^2 = -(a^2 - b^2)^2$.
Thus,the equation is $x^2 - 4abx - (a^2 - b^2)^2 = 0$.

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

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

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

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

(D) Given the quadratic equation $2x^2 - 2(m^2 + 1)x + m^4 + m^2 + 1 = 0$.
Comparing with $ax^2 + bx + c = 0$,we have $a = 2$,$b = -2(m^2 + 1)$,and $c = m^4 + m^2 + 1$.
Sum of roots: $\alpha + \beta = -\frac{b}{a} = \frac{2(m^2 + 1)}{2} = m^2 + 1$.
Product of roots: $\alpha \beta = \frac{c}{a} = \frac{m^4 + m^2 + 1}{2}$.
We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.
Substituting the values:
$\alpha^2 + \beta^2 = (m^2 + 1)^2 - 2 \left( \frac{m^4 + m^2 + 1}{2} \right)$.
$\alpha^2 + \beta^2 = (m^4 + 2m^2 + 1) - (m^4 + m^2 + 1)$.
$\alpha^2 + \beta^2 = m^4 - m^4 + 2m^2 - m^2 + 1 - 1 = m^2$.

(B) Let the roots of the equation $ax^2 + bx + c = 0$ be $p\alpha$ and $q\alpha$.
From the relationship between roots and coefficients:
$p\alpha + q\alpha = -\frac{b}{a} \implies \alpha(p + q) = -\frac{b}{a} \implies \alpha = -\frac{b}{a(p + q)}$
Also,$p\alpha \cdot q\alpha = \frac{c}{a} \implies pq\alpha^2 = \frac{c}{a}$
Substituting the value of $\alpha$ in the second equation:
$pq \left( -\frac{b}{a(p + q)} \right)^2 = \frac{c}{a}$
$pq \cdot \frac{b^2}{a^2(p + q)^2} = \frac{c}{a}$
Multiplying both sides by $a^2(p + q)^2$:
$pqb^2 = ac(p + q)^2$
$pqb^2 - (p + q)^2ac = 0$

(D) Given that $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
We need to evaluate the expression $E = \frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b}$.
Since $a\alpha^2 + b\alpha + c = 0$,we have $a\alpha + b = -\frac{c}{\alpha}$. Similarly,$a\beta + b = -\frac{c}{\beta}$.
Substituting these into the expression:
$E = \frac{\alpha}{-c/\beta} + \frac{\beta}{-c/\alpha} = -\frac{\alpha\beta}{c} - \frac{\beta\alpha}{c} = -\frac{2\alpha\beta}{c}$.
Substituting $\alpha\beta = \frac{c}{a}$:
$E = -\frac{2(c/a)}{c} = -\frac{2}{a}$.

(C) Let $\alpha$ and $\beta$ be the roots of the equation $ax^2 + bx + c = 0$.
From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
The sum of the squares of the roots is given by $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
Substituting the values,we get $\alpha^2 + \beta^2 = (-\frac{b}{a})^2 - 2(\frac{c}{a}) = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2 - 2ac}{a^2}$.
According to the given condition,the sum of the roots is equal to the sum of their squares:
$\alpha + \beta = \alpha^2 + \beta^2$
$-\frac{b}{a} = \frac{b^2 - 2ac}{a^2}$
Multiplying both sides by $a^2$ (assuming $a \neq 0$):
$-ab = b^2 - 2ac$
$2ac = b^2 + ab$
$2ac = b(a + b)$.

(B) Given $\alpha, \beta$ are roots of $x^2 - 2x + 3 = 0$.
From the relation between roots and coefficients,$\alpha + \beta = 2$ and $\alpha \beta = 3$.
We need to find the equation with roots $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$.
The sum of the new roots is $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{(\alpha \beta)^2} = \frac{2^2 - 2(3)}{3^2} = \frac{4 - 6}{9} = -\frac{2}{9}$.
The product of the new roots is $\frac{1}{\alpha^2} \cdot \frac{1}{\beta^2} = \frac{1}{(\alpha \beta)^2} = \frac{1}{3^2} = \frac{1}{9}$.
The required quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - (-\frac{2}{9})x + \frac{1}{9} = 0$.
Multiplying by $9$,we get $9x^2 + 2x + 1 = 0$.

(A) Given that $\alpha, \beta$ are roots of $x^2 + px + 1 = 0$,we have $\alpha + \beta = -p$ and $\alpha \beta = 1$.
Given that $\gamma, \delta$ are roots of $x^2 + qx + 1 = 0$,we have $\gamma + \delta = -q$ and $\gamma \delta = 1$.
Consider the expression $(\alpha - \gamma)(\beta - \gamma)(\alpha + \delta)(\beta + \delta)$:
$= (\alpha \beta - \gamma(\alpha + \beta) + \gamma^2)(\alpha \beta + \delta(\alpha + \beta) + \delta^2)$
$= (1 + p\gamma + \gamma^2)(1 - p\delta + \delta^2)$.
Since $\gamma$ is a root of $x^2 + qx + 1 = 0$,$\gamma^2 + q\gamma + 1 = 0$,so $1 + \gamma^2 = -q\gamma$.
Similarly,$\delta^2 + 1 = -q\delta$.
Substituting these into the expression:
$= (-q\gamma + p\gamma)(-q\delta - p\delta)$
$= \gamma(p - q) \times \delta(-p - q)$
$= -\gamma \delta (p - q)(p + q)$
$= -1(p^2 - q^2) = q^2 - p^2$.

(A) Given that $\alpha, \beta$ are roots of $x^2 - px + q = 0$,we have $\alpha + \beta = p$ and $\alpha\beta = q$.
Given that $\alpha', \beta'$ are roots of $x^2 - p'x + q' = 0$,we have $\alpha' + \beta' = p'$ and $\alpha'\beta' = q'$.
Let $S = (\alpha - \alpha')^2 + (\beta - \alpha')^2 + (\alpha - \beta')^2 + (\beta - \beta')^2$.
Expanding the terms: $S = (\alpha^2 - 2\alpha\alpha' + \alpha'^2) + (\beta^2 - 2\beta\alpha' + \alpha'^2) + (\alpha^2 - 2\alpha\beta' + \beta'^2) + (\beta^2 - 2\beta\beta' + \beta'^2)$.
Grouping the terms: $S = 2(\alpha^2 + \beta^2) + 2(\alpha'^2 + \beta'^2) - 2\alpha'(\alpha + \beta) - 2\beta'(\alpha + \beta)$.
$S = 2(\alpha^2 + \beta^2) + 2(\alpha'^2 + \beta'^2) - 2(\alpha' + \beta')(\alpha + \beta)$.
Using the identities $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = p^2 - 2q$ and $\alpha'^2 + \beta'^2 = (\alpha' + \beta')^2 - 2\alpha'\beta' = p'^2 - 2q'$.
Substituting these values: $S = 2(p^2 - 2q) + 2(p'^2 - 2q') - 2(p')(p)$.
$S = 2\{p^2 - 2q + p'^2 - 2q' - pp'\}$.

(A) Let the roots be $\alpha$ and $\alpha^2$.
From the relation between roots and coefficients:
Sum of roots: $\alpha + \alpha^2 = -\frac{b}{a}$ ..... $(i)$
Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = \frac{c}{a}$ ..... $(ii)$
Cubing equation $(i)$ on both sides:
$(\alpha + \alpha^2)^3 = (-\frac{b}{a})^3$
$\alpha^3 + (\alpha^2)^3 + 3\alpha \cdot \alpha^2(\alpha + \alpha^2) = -\frac{b^3}{a^3}$
Substitute $\alpha^3 = \frac{c}{a}$ and $\alpha + \alpha^2 = -\frac{b}{a}$:
$\frac{c}{a} + (\frac{c}{a})^2 + 3(\frac{c}{a})(-\frac{b}{a}) = -\frac{b^3}{a^3}$
$\frac{c}{a} + \frac{c^2}{a^2} - \frac{3bc}{a^2} = -\frac{b^3}{a^3}$
Multiply the entire equation by $a^3$:
$a^2c + ac^2 - 3abc = -b^3$
Rearranging the terms,we get:
$b^3 + a^2c + ac^2 = 3abc$.

(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
The $A.M.$ (Arithmetic Mean) of the roots is given by $A = \frac{\alpha + \beta}{2}$,which implies $\alpha + \beta = 2A$.
The $G.M.$ (Geometric Mean) of the roots is given by $G = \sqrt{\alpha \beta}$,which implies $\alpha \beta = G^2$.
$A$ quadratic equation with roots $\alpha$ and $\beta$ is given by $t^2 - (\text{sum of roots})t + (\text{product of roots}) = 0$.
Substituting the values,we get $t^2 - (2A)t + G^2 = 0$.

(C) Given that $\alpha$ and $\beta$ are the roots of $(x - a)(x - b) - c = 0$.
Expanding this,we get $x^2 - (a + b)x + ab - c = 0$.
Comparing this with the standard quadratic equation $x^2 - (\alpha + \beta)x + \alpha\beta = 0$,we have:
$\alpha + \beta = a + b$ and $\alpha\beta = ab - c$.
Now,consider the equation $(x - \alpha)(x - \beta) + c = 0$.
Expanding this,we get $x^2 - (\alpha + \beta)x + \alpha\beta + c = 0$.
Substituting the values of $(\alpha + \beta)$ and $\alpha\beta$ into this equation:
$x^2 - (a + b)x + (ab - c) + c = 0$
$x^2 - (a + b)x + ab = 0$
$(x - a)(x - b) = 0$.
Thus,the roots of the equation are $a$ and $b$.

(C) Let the roots of the quadratic equation $x^2 - px + 8 = 0$ be $\alpha$ and $\beta$.
From the properties of roots,we have $\alpha + \beta = p$ and $\alpha \beta = 8$.
Given that the difference of the roots is $|\alpha - \beta| = 2$,so $(\alpha - \beta)^2 = 4$.
We know the identity $(\alpha + \beta)^2 - (\alpha - \beta)^2 = 4\alpha \beta$.
Substituting the known values: $p^2 - 2^2 = 4(8)$.
$p^2 - 4 = 32$.
$p^2 = 36$.
Therefore,$p = \pm 6$.

(C) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$.
Then,$\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.
According to the problem,$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$.
This implies $\alpha + \beta = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$.
Substituting the values,we get $-b/a = \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} = \frac{b^2 - 2ac}{c^2}$.
Thus,$-bc^2 = a(b^2 - 2ac) = ab^2 - 2a^2c$.
Rearranging,$2a^2c = ab^2 + bc^2$.
Dividing both sides by $abc$,we get $\frac{2a}{b} = \frac{b}{c} + \frac{c}{a}$.
This shows that $\frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in $A.P.$
Therefore,their reciprocals $\frac{a}{c}, \frac{b}{a}, \frac{c}{b}$ are in $H.P.$

(C) Given the quadratic equation $ax^2 + 2bx + c = 0$.
From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{2b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Now,consider the expression $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}$.
Taking the common denominator,we get $\frac{\sqrt{\alpha} \cdot \sqrt{\alpha} + \sqrt{\beta} \cdot \sqrt{\beta}}{\sqrt{\alpha\beta}} = \frac{\alpha + \beta}{\sqrt{\alpha\beta}}$.
Substituting the values of $\alpha + \beta$ and $\alpha\beta$,we get $\frac{-\frac{2b}{a}}{\sqrt{\frac{c}{a}}} = \frac{-\frac{2b}{a}}{\frac{\sqrt{c}}{\sqrt{a}}} = -\frac{2b}{a} \cdot \frac{\sqrt{a}}{\sqrt{c}} = -\frac{2b}{\sqrt{a} \cdot \sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{c}} = -\frac{2b}{\sqrt{ac}}$.

(A) Given the quadratic equation $ax^2 + bx + c = 0$.
Let the roots of the equation be $\alpha$ and $\frac{1}{\alpha}$ as they are reciprocal to each other.
The product of the roots for a quadratic equation $ax^2 + bx + c = 0$ is given by the formula $\frac{c}{a}$.
Therefore,$\alpha \times \frac{1}{\alpha} = \frac{c}{a}$.
$1 = \frac{c}{a}$.
This implies $a = c$,or $a - c = 0$.

(B) Given that $\alpha, \beta$ are the roots of $Ax^2 + Bx + C = 0$.
From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{B}{A}$ and $\alpha\beta = \frac{C}{A}$.
Given that $\alpha^2, \beta^2$ are the roots of $x^2 + px + q = 0$.
From the relation between roots and coefficients,we have $\alpha^2 + \beta^2 = -p$ and $\alpha^2\beta^2 = q$.
We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
Substituting the values,we get $-p = (-\frac{B}{A})^2 - 2(\frac{C}{A})$.
$-p = \frac{B^2}{A^2} - \frac{2C}{A} = \frac{B^2 - 2AC}{A^2}$.
Therefore,$p = -\frac{B^2 - 2AC}{A^2} = \frac{2AC - B^2}{A^2}$.

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + x + 1 = 0$,we have $\alpha + \beta = -1$ and $\alpha \beta = 1$.
The roots of the equation $x^2 + px + q = 0$ are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.
By the relation between roots and coefficients,the sum of the roots is $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -p$.
Thus,$-p = \frac{\alpha^2 + \beta^2}{\alpha \beta}$.
Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$,we get:
$-p = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} = \frac{(-1)^2 - 2(1)}{1} = \frac{1 - 2}{1} = -1$.
Therefore,$-p = -1$,which implies $p = 1$.

(C) Given the quadratic equation $x^2 + ax + b = 0$.
By the relation between roots and coefficients,the sum of roots $\alpha + \beta = -a$ and the product of roots $\alpha \beta = b$.
We know the algebraic identity $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)$.
This can be rewritten as $\alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha \beta]$.
Substituting the values,we get $\alpha^3 + \beta^3 = (-a)[(-a)^2 - 3(b)]$.
$\alpha^3 + \beta^3 = (-a)(a^2 - 3b) = -a^3 + 3ab$.

(C) Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + px + q = 0$.
From the properties of quadratic equations,we have $\alpha + \beta = -p$ and $\alpha \beta = q$.
Given that the sum of the roots is three times their difference: $\alpha + \beta = 3(\alpha - \beta)$.
Substituting $\alpha + \beta = -p$,we get $-p = 3(\alpha - \beta)$,which implies $\alpha - \beta = -\frac{p}{3}$.
We know the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$.
Substituting the values,we get $(-\frac{p}{3})^2 = (-p)^2 - 4q$.
$\frac{p^2}{9} = p^2 - 4q$.
$4q = p^2 - \frac{p^2}{9} = \frac{8p^2}{9}$.
$36q = 8p^2$,which simplifies to $2p^2 = 9q$.

(B) Let the roots of the quadratic equation be $\alpha$ and $\frac{1}{\alpha}$.
For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.
Here,$a = 2k + 1$,$b = -(7k + 3)$,and $c = k + 2$.
Since the roots are reciprocal,their product is $\alpha \times \frac{1}{\alpha} = 1$.
Therefore,$\frac{c}{a} = 1 \Rightarrow \frac{k + 2}{2k + 1} = 1$.
Solving for $k$: $k + 2 = 2k + 1$.
$2 - 1 = 2k - k$.
$k = 1$.

(B) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $l$ and $2l$.
Sum of roots: $l + 2l = -\frac{b}{a}$ $\Rightarrow 3l = -\frac{b}{a}$ $\Rightarrow l = -\frac{b}{3a}$ $(i)$
Product of roots: $l \times 2l = \frac{c}{a} \Rightarrow 2l^2 = \frac{c}{a}$ $(ii)$
Substituting $(i)$ into $(ii)$:
$2\left(-\frac{b}{3a}\right)^2 = \frac{c}{a}$
$2\left(\frac{b^2}{9a^2}\right) = \frac{c}{a}$
$\frac{2b^2}{9a^2} = \frac{c}{a}$
$2b^2 = 9ac$
Alternatively,for a quadratic equation $ax^2 + bx + c = 0$ with roots in ratio $m:n$,the condition is $mnb^2 = (m+n)^2ac$. Here $m=1, n=2$,so $1 \times 2 \times b^2 = (1+2)^2ac \Rightarrow 2b^2 = 9ac$.

(D) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
Given $\alpha + \beta = 2$ and $\alpha^3 + \beta^3 = 98$.
We know that $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)$.
This can be written as $\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta)$.
Substituting the given values: $98 = 2((2)^2 - 3\alpha\beta)$.
$49 = 4 - 3\alpha\beta$.
$3\alpha\beta = 4 - 49 = -45$.
$\alpha\beta = -15$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Therefore,$x^2 - 2x - 15 = 0$.

(A) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,we have:
$\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
We need to find the value of $\alpha\beta^2 + \alpha^2\beta + \alpha\beta$.
Factoring out $\alpha\beta$,we get:
$\alpha\beta(\beta + \alpha) + \alpha\beta = \alpha\beta(\alpha + \beta + 1)$.
Substituting the values of $\alpha + \beta$ and $\alpha\beta$:
$= \frac{c}{a} \left( -\frac{b}{a} + 1 \right)$
$= \frac{c}{a} \left( \frac{a - b}{a} \right)$
$= \frac{c(a - b)}{a^2}$.

(C) For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.
Given the equation $mx^2 + 6x + (2m - 1) = 0$,we have $a = m$ and $c = 2m - 1$.
The product of the roots is given as $-1$.
Therefore,$\frac{2m - 1}{m} = -1$.
Multiplying both sides by $m$,we get $2m - 1 = -m$.
Adding $m$ to both sides,we get $3m - 1 = 0$.
Thus,$3m = 1$,which gives $m = \frac{1}{3}$.

(A) Given the equation $x^2 + ax + b = 0$ with roots $p$ and $q$.
From the relation between roots and coefficients,we have:
$p + q = -a$ and $pq = b$.
We need to find the equation whose roots are $p^2q$ and $pq^2$.
Let the new roots be $\alpha = p^2q$ and $\beta = pq^2$.
Sum of the new roots: $\alpha + \beta = p^2q + pq^2 = pq(p + q) = (b)(-a) = -ab$.
Product of the new roots: $\alpha \beta = (p^2q)(pq^2) = p^3q^3 = (pq)^3 = b^3$.
The required quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values: $x^2 - (-ab)x + b^3 = 0$,which simplifies to $x^2 + abx + b^3 = 0$.

(B) Given the quadratic equation $x^2 - 4x + 1 = 0$.
Comparing with $ax^2 + bx + c = 0$,we have $a=1, b=-4, c=1$.
Sum of roots: $\alpha + \beta = -b/a = -(-4)/1 = 4$.
Product of roots: $\alpha \beta = c/a = 1/1 = 1$.
We know the identity: $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$.
Substituting the values: $\alpha^3 + \beta^3 = (4)^3 - 3(1)(4)$.
$\alpha^3 + \beta^3 = 64 - 12 = 52$.

(B) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $2x^2 - 35x + 2 = 0$.
From the properties of roots,the product of roots $\alpha \beta = \frac{c}{a} = \frac{2}{2} = 1$.
Since $\alpha$ is a root,$2\alpha^2 - 35\alpha + 2 = 0$,which implies $2\alpha^2 - 35\alpha = -2$.
Dividing by $\alpha$,we get $2\alpha - 35 = -\frac{2}{\alpha}$.
Similarly,for root $\beta$,$2\beta - 35 = -\frac{2}{\beta}$.
Now,substitute these into the expression $(2\alpha - 35)^3 \cdot (2\beta - 35)^3$:
$(2\alpha - 35)^3 \cdot (2\beta - 35)^3 = \left(-\frac{2}{\alpha}\right)^3 \cdot \left(-\frac{2}{\beta}\right)^3$
$= \left(-\frac{8}{\alpha^3}\right) \cdot \left(-\frac{8}{\beta^3}\right)$
$= \frac{64}{(\alpha \beta)^3}$
Since $\alpha \beta = 1$,the expression becomes $\frac{64}{1^3} = 64$.

(B) Given that $3p^2 = 5p + 2$ and $3q^2 = 5q + 2$ with $p \ne q$.
This implies that $p$ and $q$ are the distinct roots of the quadratic equation $3x^2 - 5x - 2 = 0$.
For a quadratic equation of the form $ax^2 + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.
Here,$a = 3$,$b = -5$,and $c = -2$.
Therefore,the product of the roots $pq = \frac{-2}{3} = -\frac{2}{3}$.

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + bx - c = 0$.
By the relation between roots and coefficients:
$\alpha + \beta = -b$ and $\alpha \beta = -c$.
This implies $b = -(\alpha + \beta)$ and $c = -\alpha \beta$.
We need to find the equation whose roots are $b$ and $c$.
Sum of the roots = $b + c = -(\alpha + \beta) - \alpha \beta = -[(\alpha + \beta) + \alpha \beta]$.
Product of the roots = $bc = [ -(\alpha + \beta) ] \times [ -\alpha \beta ] = \alpha \beta(\alpha + \beta)$.
The required quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values:
$x^2 - (-[(\alpha + \beta) + \alpha \beta])x + \alpha \beta(\alpha + \beta) = 0$.
$x^2 + [(\alpha + \beta) + \alpha \beta]x + \alpha \beta(\alpha + \beta) = 0$.

(B) Given $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Expanding the expression $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$:
$= 1 + (\alpha + \beta) + (\alpha^2 + \beta^2) + \alpha\beta + \alpha\beta(\alpha + \beta) + (\alpha\beta)^2$
$= 1 + (\alpha + \beta) + ((\alpha + \beta)^2 - 2\alpha\beta) + \alpha\beta + \alpha\beta(\alpha + \beta) + (\alpha\beta)^2$
$= 1 + (\alpha + \beta) + (\alpha + \beta)^2 - \alpha\beta + \alpha\beta(\alpha + \beta) + (\alpha\beta)^2$
Substituting the values:
$= 1 - \frac{b}{a} + \frac{b^2}{a^2} - \frac{c}{a} - \frac{bc}{a^2} + \frac{c^2}{a^2}$
$= \frac{a^2 - ab + b^2 - ac - bc + c^2}{a^2}$
$= \frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2a^2}$
Since $a, b, c$ are distinct,the numerator is always positive and $2a^2 > 0$,so the expression is always positive.

(A) Let the roots be $x_1 = \frac{\alpha}{\alpha - 1}$ and $x_2 = \frac{\alpha + 1}{\alpha}$.
From the relation between roots and coefficients,the sum of roots is $x_1 + x_2 = -\frac{b}{a}$ and the product of roots is $x_1 x_2 = \frac{c}{a}$.
Product of roots: $\frac{\alpha}{\alpha - 1} \times \frac{\alpha + 1}{\alpha} = \frac{\alpha + 1}{\alpha - 1} = \frac{c}{a}$.
By componendo and dividendo,$\frac{(\alpha + 1) + (\alpha - 1)}{(\alpha + 1) - (\alpha - 1)} = \frac{c + a}{c - a} \implies \frac{2\alpha}{2} = \frac{c + a}{c - a} \implies \alpha = \frac{c + a}{c - a}$.
Sum of roots: $\frac{\alpha^2 + \alpha^2 - 1}{\alpha(\alpha - 1)} = \frac{2\alpha^2 - 1}{\alpha^2 - \alpha} = -\frac{b}{a}$.
Substituting $\alpha$ and simplifying,we find that $(a + b + c)^2 = b^2 - 4ac$.

(B) Let the roots of $ax^2 + 2bx + c = 0$ be $\alpha$ and $\beta$ such that $\frac{\alpha}{\beta} = \frac{m}{n}$.
Then $\alpha + \beta = -\frac{2b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Using the relation $\frac{(\alpha + \beta)^2}{\alpha\beta} = \frac{(\frac{m}{n} + 1)^2}{\frac{m}{n}} = \frac{(m+n)^2}{mn}$,we get $\frac{4b^2/a^2}{c/a} = \frac{(m+n)^2}{mn}$,which simplifies to $\frac{4b^2}{ac} = \frac{(m+n)^2}{mn}$.
Similarly,for $px^2 + 2qx + r = 0$,we have $\frac{4q^2}{pr} = \frac{(m+n)^2}{mn}$.
Equating the two,we get $\frac{4b^2}{ac} = \frac{4q^2}{pr}$,which implies $\frac{b^2}{ac} = \frac{q^2}{pr}$.

(C) Given the quadratic equation $x^2 + bx - c = 0$ with $b, c > 0$.
Let the roots be $\alpha$ and $\beta$.
According to the relation between roots and coefficients:
Sum of roots $\alpha + \beta = -b$.
Since $b > 0$,then $-b < 0$,so $\alpha + \beta < 0$.
Product of roots $\alpha \beta = -c$.
Since $c > 0$,then $-c < 0$,so $\alpha \beta < 0$.
Since the product of the roots is negative,the roots must have opposite signs.

(B) The given equation is $x^2 - (p + 1)x + pq = 0$.
Since $p$ and $q$ are the roots of the equation,the sum of the roots is given by $p + q = -(\text{coefficient of } x) / (\text{coefficient of } x^2)$.
Therefore,$p + q = p + 1$.
Subtracting $p$ from both sides,we get $q = 1$.