Relation between roots and coefficients Questions in English

(B) Let the roots of the quadratic equation $x^2 + px + q = 0$ be $\alpha$ and $\beta$.
Given that the difference between the roots is $|\alpha - \beta| = 1$.
We know that $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.
Substituting the values from the quadratic equation,where $\alpha + \beta = -p$ and $\alpha\beta = q$:
$1^2 = (-p)^2 - 4(q)$
$1 = p^2 - 4q$
$p^2 = 4q + 1$.

(A) Given the quadratic equation $4x^2 + 3x + 7 = 0$.
Comparing this with $ax^2 + bx + c = 0$,we have $a = 4, b = 3, c = 7$.
Sum of roots $\alpha + \beta = -b/a = -3/4$.
Product of roots $\alpha\beta = c/a = 7/4$.
Now,$1/\alpha + 1/\beta = (\alpha + \beta) / (\alpha\beta)$.
Substituting the values,we get $(-3/4) / (7/4) = -3/7$.

(D) Let the roots be $\alpha$ and $2\alpha$.
From the sum of roots: $\alpha + 2\alpha = 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3} \Rightarrow \alpha = -\frac{3a - 1}{3(a^2 - 5a + 3)}$.
From the product of roots: $\alpha \times 2\alpha = 2\alpha^2 = \frac{2}{a^2 - 5a + 3} \Rightarrow \alpha^2 = \frac{1}{a^2 - 5a + 3}$.
Squaring the sum of roots expression: $\alpha^2 = \frac{(3a - 1)^2}{9(a^2 - 5a + 3)^2}$.
Equating the two expressions for $\alpha^2$: $\frac{1}{a^2 - 5a + 3} = \frac{(3a - 1)^2}{9(a^2 - 5a + 3)^2}$.
Since $a^2 - 5a + 3 \neq 0$,we have $9(a^2 - 5a + 3) = (3a - 1)^2$.
$9a^2 - 45a + 27 = 9a^2 - 6a + 1$.
$-39a = -26 \Rightarrow a = \frac{26}{39} = \frac{2}{3}$.

(A) Let the roots of the equation $x^2 + px + q = 0$ be $\alpha$ and $\alpha^2$.
From the relation between roots and coefficients:
Sum of roots: $\alpha + \alpha^2 = -p$
Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = q$
So,$\alpha = q^{1/3}$.
Substituting $\alpha$ into the sum equation:
$q^{1/3} + (q^{1/3})^2 = -p$
$q^{1/3} + q^{2/3} = -p$
Cubing both sides:
$(q^{1/3} + q^{2/3})^3 = (-p)^3$
Using $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$:
$q + q^2 + 3(q^{1/3})(q^{2/3})(q^{1/3} + q^{2/3}) = -p^3$
$q + q^2 + 3q(q^{1/3} + q^{2/3}) = -p^3$
Since $q^{1/3} + q^{2/3} = -p$:
$q + q^2 + 3q(-p) = -p^3$
$p^3 + q^2 - 3pq + q = 0$
$p^3 + q(q - 3p) + q = 0$
$p^3 - q(3p - q) + q = 0$.

(B) For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.
Here,$a = 1$ and $c = 2e^{2\log k} - 1$.
Given that the product of the roots is $7$,we have:
$2e^{2\log k} - 1 = 7$
$2e^{\log k^2} = 8$
Since $e^{\log k^2} = k^2$,we get:
$2k^2 = 8$
$k^2 = 4$
$k = \pm 2$
Since the term $\log k$ is present in the equation,$k$ must be greater than $0$ for the logarithm to be defined.
Therefore,$k = 2$.

(C) Given $\alpha^2 = 5\alpha - 3$ and $\beta^2 = 5\beta - 3$,it implies that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 5x + 3 = 0$.
From the properties of roots,we have:
Sum of roots: $\alpha + \beta = 5$
Product of roots: $\alpha \beta = 3$
We need to find the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.
Sum of new roots: $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} = \frac{5^2 - 2(3)}{3} = \frac{25 - 6}{3} = \frac{19}{3}$.
Product of new roots: $\frac{\alpha}{\beta} \times \frac{\beta}{\alpha} = 1$.
The required quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - \frac{19}{3}x + 1 = 0$.
Multiplying by $3$,we get $3x^2 - 19x + 3 = 0$.

(B) Let the roots of the quadratic equation $2x^2 - 2(p - 2)x - p - 1 = 0$ be $\alpha$ and $\beta$.
From the properties of roots,we have $\alpha + \beta = p - 2$ and $\alpha \beta = \frac{-(p + 1)}{2}$.
Let $S$ be the sum of the squares of the roots: $S = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.
Substituting the values,we get $S = (p - 2)^2 - 2 \left( \frac{-(p + 1)}{2} \right) = (p - 2)^2 + p + 1$.
Expanding this,$S = p^2 - 4p + 4 + p + 1 = p^2 - 3p + 5$.
To minimize $S$,we complete the square: $S = \left( p - \frac{3}{2} \right)^2 + 5 - \frac{9}{4} = \left( p - \frac{3}{2} \right)^2 + \frac{11}{4}$.
The expression $S$ is minimized when $p - \frac{3}{2} = 0$,which gives $p = \frac{3}{2}$.

(D) Given the equation: $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$
Simplify the left side: $\frac{(x + b) - (x + a)}{(x + a)(x + b)} = \frac{1}{x + c}$
$\frac{b - a}{x^2 + (a + b)x + ab} = \frac{1}{x + c}$
Cross-multiply: $x^2 + (a + b)x + ab = (b - a)(x + c)$
$x^2 + (a + b)x + ab = (b - a)x + (b - a)c$
$x^2 + (a + b - b + a)x + ab - (b - a)c = 0$
$x^2 + 2ax + ab - bc + ac = 0$
Since the product of the roots is $0$,the constant term must be $0$:
$ab + ac - bc = 0 \implies a(b + c) = bc \implies a = \frac{bc}{b + c}$
The sum of the roots of a quadratic equation $x^2 + Bx + C = 0$ is $-B$.
Sum of roots $= -2a = -2 \left( \frac{bc}{b + c} \right) = \frac{-2bc}{b + c}$.

(A) Let $r$ be the common ratio of the geometric progression. Then $\beta = \alpha r, \gamma = \alpha r^2, \delta = \alpha r^3$.
From the given equations:
$\alpha + \beta = 1 \Rightarrow \alpha(1 + r) = 1 \dots (i)$
$\alpha \beta = p \Rightarrow \alpha^2 r = p \dots (ii)$
$\gamma + \delta = 4 \Rightarrow \alpha r^2(1 + r) = 4 \dots (iii)$
$\gamma \delta = q \Rightarrow \alpha^2 r^5 = q \dots (iv)$
Dividing $(iii)$ by $(i)$,we get $r^2 = 4$,so $r = \pm 2$.
If $r = 2$,then $\alpha(1 + 2) = 1 \Rightarrow \alpha = 1/3$. Then $p = \alpha^2 r = (1/9)(2) = 2/9$ and $q = \alpha^2 r^5 = (1/9)(32) = 32/9$.
If $r = -2$,then $\alpha(1 - 2) = 1$ $\Rightarrow -\alpha = 1$ $\Rightarrow \alpha = -1$.
Then $p = \alpha^2 r = (-1)^2(-2) = -2$ and $q = \alpha^2 r^5 = (-1)^2(-2)^5 = -32$.
Thus,$(p, q) = (-2, -32)$.

(B) Given the equation $x^2 + px + q = 0$ with roots $\alpha$ and $\beta$.
We know that $\alpha + \beta = -p$ and $\alpha \beta = q$.
Let the new roots be $\alpha' = q/\alpha$ and $\beta' = q/\beta$.
The sum of the new roots is $\alpha' + \beta' = \frac{q}{\alpha} + \frac{q}{\beta} = q \left( \frac{\alpha + \beta}{\alpha \beta} \right) = q \left( \frac{-p}{q} \right) = -p$.
The product of the new roots is $\alpha' \beta' = \left( \frac{q}{\alpha} \right) \left( \frac{q}{\beta} \right) = \frac{q^2}{\alpha \beta} = \frac{q^2}{q} = q$.
The required equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - (-p)x + q = 0$,which simplifies to $x^2 + px + q = 0$.

(B) Given the equation $x^3 + 5x^2 - 7x - 1 = 0$ with roots $\alpha, \beta, \gamma$.
From Vieta's formulas,the product of the roots is $\alpha\beta\gamma = -(-1)/1 = 1$.
We want to find the equation whose roots are $\alpha\beta, \beta\gamma, \gamma\alpha$.
Since $\alpha\beta\gamma = 1$,we can write the roots as:
$\alpha\beta = \frac{\alpha\beta\gamma}{\gamma} = \frac{1}{\gamma}$,
$\beta\gamma = \frac{\alpha\beta\gamma}{\alpha} = \frac{1}{\alpha}$,
$\gamma\alpha = \frac{\alpha\beta\gamma}{\beta} = \frac{1}{\beta}$.
Thus,the roots are $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$.
To find the equation with roots $\frac{1}{x}$,we substitute $x$ with $\frac{1}{x}$ in the original equation:
$(\frac{1}{x})^3 + 5(\frac{1}{x})^2 - 7(\frac{1}{x}) - 1 = 0$.
Multiplying by $x^3$,we get:
$1 + 5x - 7x^2 - x^3 = 0$,
which simplifies to $x^3 + 7x^2 - 5x - 1 = 0$.

(D) Given the quadratic equation $4x^2 - \sqrt{13}x - 7 = 0$.
Comparing this with $ax^2 + bx + c = 0$,we get $a = 4$,$b = -\sqrt{13}$,and $c = -7$.
The sum of roots $\alpha + \beta = -\frac{b}{a} = -\frac{-\sqrt{13}}{4} = \frac{\sqrt{13}}{4}$.
The product of roots $\alpha \beta = \frac{c}{a} = -\frac{7}{4}$.
We know that $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.
Substituting the values: $(\alpha - \beta)^2 = (\frac{\sqrt{13}}{4})^2 - 4(-\frac{7}{4}) = \frac{13}{16} + 7 = \frac{13 + 112}{16} = \frac{125}{16}$.
Therefore,$|\alpha - \beta| = \sqrt{\frac{125}{16}} = \frac{5\sqrt{5}}{4}$.

(C) Let the roots of the equation $2x^2 - 3x + 5 = 0$ be $\alpha$ and $\beta$.
From the properties of quadratic equations,the sum of roots $\alpha + \beta = -(-3)/2 = 3/2$ and the product of roots $\alpha \beta = 5/2$.
The roots of the equation $ax^2 + bx + 2 = 0$ are given as $1/\alpha$ and $1/\beta$.
For the equation $ax^2 + bx + 2 = 0$,the product of roots is $(1/\alpha)(1/\beta) = 1/(\alpha \beta) = 1/(5/2) = 2/5$.
Also,the product of roots for $ax^2 + bx + 2 = 0$ is $c/a = 2/a$.
Equating the two,$2/a = 2/5$,which gives $a = 5$.
The sum of roots for $ax^2 + bx + 2 = 0$ is $(1/\alpha) + (1/\beta) = (\alpha + \beta) / (\alpha \beta) = (3/2) / (5/2) = 3/5$.
Also,the sum of roots is $-b/a = -b/5$.
Equating the two,$-b/5 = 3/5$,which gives $b = -3$.
Thus,$a = 5$ and $b = -3$.

(D) Let the roots of the quadratic equation $2x^2 - (a + 1)x + (a - 1) = 0$ be $\alpha$ and $\beta$.
From the properties of roots,the sum of roots $\alpha + \beta = \frac{a + 1}{2}$ and the product of roots $\alpha \beta = \frac{a - 1}{2}$.
The difference of the roots is given by $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}$.
Given that the difference of the roots is equal to their product,we have $|\alpha - \beta| = \alpha \beta$.
Squaring both sides,we get $(\alpha - \beta)^2 = (\alpha \beta)^2$.
Substituting the values,$(\alpha + \beta)^2 - 4\alpha \beta = (\alpha \beta)^2$.
$(\frac{a + 1}{2})^2 - 4(\frac{a - 1}{2}) = (\frac{a - 1}{2})^2$.
$\frac{a^2 + 2a + 1}{4} - 2(a - 1) = \frac{a^2 - 2a + 1}{4}$.
Multiplying by $4$,$a^2 + 2a + 1 - 8(a - 1) = a^2 - 2a + 1$.
$a^2 + 2a + 1 - 8a + 8 = a^2 - 2a + 1$.
$-6a + 9 = -2a + 1$.
$8 = 4a$.
$a = 2$.

(B) Since $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $a\alpha^2 + b\alpha + c = 0$ and $a\beta^2 + b\beta + c = 0$.
From $a\alpha^2 + b\alpha + c = 0$,we get $\alpha(a\alpha + b) = -c$,so $(a\alpha + b) = -\frac{c}{\alpha}$.
Similarly,$(a\beta + b) = -\frac{c}{\beta}$.
Substituting these into the expression:
$\frac{1}{(a\alpha + b)^2} + \frac{1}{(a\beta + b)^2} = \frac{1}{(-c/\alpha)^2} + \frac{1}{(-c/\beta)^2} = \frac{\alpha^2}{c^2} + \frac{\beta^2}{c^2} = \frac{\alpha^2 + \beta^2}{c^2}$.
Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$,where $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$:
$\frac{(\alpha + \beta)^2 - 2\alpha\beta}{c^2} = \frac{(-b/a)^2 - 2(c/a)}{c^2} = \frac{b^2/a^2 - 2c/a}{c^2} = \frac{b^2 - 2ac}{a^2c^2}$.

(B) Let the roots of the quadratic equation be $\alpha$ and $\beta$. Given $|\alpha - \beta| = 3$.
For the equation $(k - 2)x^2 - (k - 4)x - 2 = 0$, the sum of roots $\alpha + \beta = \frac{k - 4}{k - 2}$ and the product of roots $\alpha \beta = \frac{-2}{k - 2}$.
We know that $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$.
Substituting the values: $3^2 = \left(\frac{k - 4}{k - 2}\right)^2 - 4\left(\frac{-2}{k - 2}\right)$.
$9 = \frac{(k - 4)^2 + 8(k - 2)}{(k - 2)^2}$.
$9(k - 2)^2 = k^2 - 8k + 16 + 8k - 16$.
$9(k^2 - 4k + 4) = k^2$.
$9k^2 - 36k + 36 = k^2$.
$8k^2 - 36k + 36 = 0$.
Dividing by $4$: $2k^2 - 9k + 9 = 0$.
$2k^2 - 6k - 3k + 9 = 0 \Rightarrow 2k(k - 3) - 3(k - 3) = 0$.
$(2k - 3)(k - 3) = 0$.
Thus, $k = 3$ or $k = 3/2$.

(B) 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 $E = (\alpha - \gamma) (\beta - \gamma) (\alpha + \delta) (\beta + \delta)$.
$E = [\alpha\beta - \gamma(\alpha + \beta) + \gamma^2] [\alpha\beta + \delta(\alpha + \beta) + \delta^2]$.
Substituting the values:
$E = [1 - \gamma(-p) + \gamma^2] [1 + \delta(-p) + \delta^2] = [1 + p\gamma + \gamma^2] [1 - p\delta + \delta^2]$.
Since $\gamma^2 + q\gamma + 1 = 0$,we have $\gamma^2 + 1 = -q\gamma$.
Since $\delta^2 + q\delta + 1 = 0$,we have $\delta^2 + 1 = -q\delta$.
Substituting these:
$E = [-q\gamma + p\gamma] [-q\delta - p\delta] = \gamma(p - q) \cdot \delta(-p - q) = -\gamma\delta(p - q)(p + q) = -1(p^2 - q^2) = q^2 - p^2$.

(A) Let the roots of the equation $x^2 - x - k = 0$ be $\alpha$ and $\alpha^2$.
From the relation between roots and coefficients:
$\alpha + \alpha^2 = 1$
$\alpha \cdot \alpha^2 = -k \implies \alpha^3 = -k$
Since $\alpha^2 = 1 - \alpha$,we substitute this into $\alpha^3 = -k$:
$\alpha \cdot \alpha^2 = -k \implies \alpha(1 - \alpha) = -k$
Alternatively,cube the equation $\alpha^2 + \alpha = 1$:
$(\alpha^2 + \alpha)^3 = 1^3$
$\alpha^6 + \alpha^3 + 3\alpha^2 \cdot \alpha(\alpha^2 + \alpha) = 1$
Since $\alpha^3 = -k$,then $\alpha^6 = k^2$:
$k^2 + (-k) + 3(-k)(1) = 1$
$k^2 - k - 3k = 1$
$k^2 - 4k - 1 = 0$
Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$k = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$

(A) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Consider the equation $cx^2 + bx + a = 0$. Let its roots be $x_1$ and $x_2$.
For this equation,the sum of roots is $x_1 + x_2 = -\frac{b}{c}$ and the product of roots is $x_1x_2 = \frac{a}{c}$.
If we substitute $x = \frac{1}{y}$ into the original equation $ax^2 + bx + c = 0$,we get $a(\frac{1}{y})^2 + b(\frac{1}{y}) + c = 0$.
Multiplying by $y^2$,we get $a + by + cy^2 = 0$,which is $cy^2 + by + a = 0$.
Since $x = \frac{1}{y}$,the roots of the new equation are the reciprocals of the roots of the original equation.
Therefore,the roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.

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

(C) Given the equation $x^2 - 2x + 3 = 0$,the sum of roots is $\alpha + \beta = 2$ and the product of roots is $\alpha \beta = 3$.
Let $y = \frac{x - 1}{x + 1}$. Then $y(x + 1) = x - 1$,which implies $yx + y = x - 1$.
Rearranging gives $x(y - 1) = -y - 1$,so $x = \frac{-(y + 1)}{y - 1} = \frac{y + 1}{1 - y}$.
Substitute $x = \frac{y + 1}{1 - y}$ into the original equation $x^2 - 2x + 3 = 0$:
$(\frac{y + 1}{1 - y})^2 - 2(\frac{y + 1}{1 - y}) + 3 = 0$.
Multiply by $(1 - y)^2$: $(y + 1)^2 - 2(y + 1)(1 - y) + 3(1 - y)^2 = 0$.
$(y^2 + 2y + 1) - 2(1 - y^2) + 3(1 - 2y + y^2) = 0$.
$y^2 + 2y + 1 - 2 + 2y^2 + 3 - 6y + 3y^2 = 0$.
Combining like terms: $(1 + 2 + 3)y^2 + (2 - 6)y + (1 - 2 + 3) = 0$.
$6y^2 - 4y + 2 = 0$,which simplifies to $3y^2 - 2y + 1 = 0$.
Thus,the required equation is $3x^2 - 2x + 1 = 0$.

(B) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$.
From the properties of quadratic equations,we have:
Sum of roots: $\alpha + \beta = -\frac{b}{a}$
Product of roots: $\alpha \beta = \frac{c}{a}$
According to the problem,the sum of the roots is equal to the sum of the squares of the roots:
$\alpha + \beta = \alpha^2 + \beta^2$
We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.
Substituting the values:
$-\frac{b}{a} = (-\frac{b}{a})^2 - 2(\frac{c}{a})$
$-\frac{b}{a} = \frac{b^2}{a^2} - \frac{2c}{a}$
Multiply the entire equation by $a^2$:
$-ab = b^2 - 2ac$
Rearranging the terms:
$2ac = b^2 + ab$
$2ac = b(a + b)$

(C) Let $\alpha, \beta$ be the roots of $x^2 + bx + c = 0$ and $\alpha', \beta'$ be the roots of $x^2 + qx + r = 0$.
Then $\alpha + \beta = -b, \alpha\beta = c$ and $\alpha' + \beta' = -q, \alpha'\beta' = r$.
Given that the ratio of the roots is equal,we have $\frac{\alpha}{\beta} = \frac{\alpha'}{\beta'}$.
Applying the property of componendo and dividendo,we get $\frac{\alpha + \beta}{\alpha - \beta} = \frac{\alpha' + \beta'}{\alpha' - \beta'}$.
Squaring both sides,we get $\frac{(\alpha + \beta)^2}{(\alpha - \beta)^2} = \frac{(\alpha' + \beta')^2}{(\alpha' - \beta')^2}$.
Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$,we have $\frac{b^2}{b^2 - 4c} = \frac{q^2}{q^2 - 4r}$.
Cross-multiplying gives $b^2(q^2 - 4r) = q^2(b^2 - 4c)$.
$b^2q^2 - 4b^2r = q^2b^2 - 4cq^2$.
$-4b^2r = -4cq^2$.
Therefore,$b^2r = cq^2$.

(C) Given that $\alpha + \beta = -p, \alpha \beta = 1$ and $\gamma + \delta = -q, \gamma \delta = 1$.
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 $x^2 + qx + 1 = 0$,we have $\gamma^2 + q\gamma + 1 = 0$,so $\gamma^2 + 1 = -q\gamma$.
Similarly,$\delta^2 + 1 = -q\delta$.
Substituting these into the expression:
$= (p\gamma - q\gamma)(-p\delta - q\delta) = \gamma(p - q) \cdot \delta(-p - q)$
$= -\gamma \delta (p - q)(p + q) = -(1)(p^2 - q^2) = q^2 - p^2$.

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

(A) Given the equation $x^2 - 5x - 3 = 0$,we have $\alpha + \beta = 5$ and $\alpha\beta = -3$.
Let $y = \frac{1}{2\alpha - 3}$. Then $2\alpha - 3 = \frac{1}{y}$,so $2\alpha = \frac{1}{y} + 3 = \frac{1 + 3y}{y}$,which gives $\alpha = \frac{3y + 1}{2y}$.
Since $\alpha$ is a root of $x^2 - 5x - 3 = 0$,we substitute $\alpha$ into the equation:
$(\frac{3y + 1}{2y})^2 - 5(\frac{3y + 1}{2y}) - 3 = 0$.
Multiply by $4y^2$ to clear the denominator:
$(3y + 1)^2 - 10y(3y + 1) - 12y^2 = 0$.
$9y^2 + 6y + 1 - 30y^2 - 10y - 12y^2 = 0$.
$-33y^2 - 4y + 1 = 0$.
Multiplying by $-1$,we get $33y^2 + 4y - 1 = 0$.
Replacing $y$ with $x$,the required equation is $33x^2 + 4x - 1 = 0$.

(A) Given $\alpha + \beta = -\frac{b}{a}$,$\alpha \beta = \frac{c}{a}$ and $\gamma + \delta = -\frac{q}{p}$,$\gamma \delta = \frac{r}{p}$.
Since $\alpha, \beta, \gamma, \delta$ are in arithmetic progression,the common difference is constant.
Thus,$\beta - \alpha = \delta - \gamma$.
Squaring both sides,we get $(\beta - \alpha)^2 = (\delta - \gamma)^2$.
Using the identity $(x - y)^2 = (x + y)^2 - 4xy$,we have $(\alpha + \beta)^2 - 4\alpha \beta = (\gamma + \delta)^2 - 4\gamma \delta$.
Substituting the values,we get $\frac{b^2}{a^2} - \frac{4c}{a} = \frac{q^2}{p^2} - \frac{4r}{p}$.
This simplifies to $\frac{b^2 - 4ac}{a^2} = \frac{q^2 - 4pr}{p^2}$.
Since $D_1 = b^2 - 4ac$ and $D_2 = q^2 - 4pr$,we have $\frac{D_1}{a^2} = \frac{D_2}{p^2}$.
Therefore,$\frac{D_1}{D_2} = \frac{a^2}{p^2}$.

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 - 6x - 2 = 0$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$\alpha^2 - 6\alpha - 2 = 0 \implies \alpha^2 - 2 = 6\alpha$
$\beta^2 - 6\beta - 2 = 0 \implies \beta^2 - 2 = 6\beta$
We need to evaluate $\frac{a_{10} - 2a_8}{2a_9}$.
Substituting $a_n = \alpha^n - \beta^n$:
$\frac{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)}{2(\alpha^9 - \beta^9)}$
$= \frac{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)}{2(\alpha^9 - \beta^9)}$
Using the relations $\alpha^2 - 2 = 6\alpha$ and $\beta^2 - 2 = 6\beta$:
$= \frac{\alpha^8(6\alpha) - \beta^8(6\beta)}{2(\alpha^9 - \beta^9)}$
$= \frac{6\alpha^9 - 6\beta^9}{2(\alpha^9 - \beta^9)}$
$= \frac{6(\alpha^9 - \beta^9)}{2(\alpha^9 - \beta^9)} = \frac{6}{2} = 3$.

(B) Given the roots of $x^2 - 5x + 16 = 0$ are $\alpha$ and $\beta$.
From the properties of quadratic equations,$\alpha + \beta = 5$ and $\alpha\beta = 16$.
For the equation $x^2 + px + q = 0$,the sum of roots is $-p = (\alpha^2 + \beta^2) + \frac{\alpha\beta}{2}$.
Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$,we get:
$-p = (5^2 - 2(16)) + \frac{16}{2} = (25 - 32) + 8 = -7 + 8 = 1$.
Thus,$p = -1$.
The product of roots is $q = (\alpha^2 + \beta^2) \times \frac{\alpha\beta}{2}$.
$q = (25 - 32) \times \frac{16}{2} = (-7) \times 8 = -56$.
Therefore,$p = -1$ and $q = -56$.

(B) Given $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.
Consider the expression $E = (1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$.
Expanding the product:
$E = 1 + (\alpha + \beta) + (\alpha^2 + \beta^2) + \alpha \beta + \alpha \beta(\alpha + \beta) + (\alpha \beta)^2$
Using $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$:
$E = 1 + (\alpha + \beta) + ((\alpha + \beta)^2 - 2\alpha \beta) + \alpha \beta + \alpha \beta(\alpha + \beta) + (\alpha \beta)^2$
Substituting the values of $\alpha + \beta$ and $\alpha \beta$:
$E = 1 - \frac{b}{a} + (\frac{b^2}{a^2} - 2\frac{c}{a}) + \frac{c}{a} + \frac{c}{a}(-\frac{b}{a}) + \frac{c^2}{a^2}$
$E = \frac{a^2 - ab + b^2 - ac - bc + c^2}{a^2} = \frac{a^2 + b^2 + c^2 - ab - bc - ca}{a^2}$
$E = \frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2a^2}$
Since the sum of squares is non-negative and $a \neq 0$,the expression is positive.

(A) Let the roots be $3\alpha, 2\alpha, \beta$. (Since the ratio of two roots is $3 : 2$.)
From the sum of roots: $3\alpha + 2\alpha + \beta = -(-9)/1 = 9$
$\Rightarrow 5\alpha + \beta = 9 \dots (i)$
From the sum of product of roots taken two at a time: $(3\alpha)(2\alpha) + (2\alpha)(\beta) + (3\alpha)(\beta) = 14$
$\Rightarrow 6\alpha^2 + 5\alpha\beta = 14 \dots (ii)$
From the product of roots: $(3\alpha)(2\alpha)(\beta) = -24$
$\Rightarrow 6\alpha^2\beta = -24$
$\Rightarrow \alpha^2\beta = -4 \dots (iii)$
From $(i)$,$\beta = 9 - 5\alpha$. Substituting this into $(ii)$:
$6\alpha^2 + 5\alpha(9 - 5\alpha) = 14$
$6\alpha^2 + 45\alpha - 25\alpha^2 = 14$
$-19\alpha^2 + 45\alpha - 14 = 0$
$19\alpha^2 - 45\alpha + 14 = 0$
$(19\alpha - 7)(\alpha - 2) = 0$
So,$\alpha = 2$ or $\alpha = 7/19$.
If $\alpha = 2$,then $\beta = 9 - 5(2) = -1$.
Check with $(iii)$: $(2)^2(-1) = 4(-1) = -4$. This satisfies the equation.
Thus,the roots are $3(2), 2(2), -1$,which are $6, 4, -1$.

(D) For the equation $x^2 - 5x + 16 = 0$,the sum of roots $\alpha + \beta = 5$ and the product of roots $\alpha\beta = 16$.
We need to find the roots of the second equation $x^2 + px + q = 0$.
The first root is $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (5)^2 - 2(16) = 25 - 32 = -7$.
The second root is $\frac{\alpha\beta}{2} = \frac{16}{2} = 8$.
The roots of $x^2 + px + q = 0$ are $-7$ and $8$.
Sum of roots: $-7 + 8 = 1 = -p \implies p = -1$.
Product of roots: $(-7)(8) = -56 = q \implies q = -56$.
Thus,$p = -1$ and $q = -56$.

(B) For the quadratic equations,we have:
$\alpha_1 + \alpha_2 = -\frac{b}{a}, \alpha_1\alpha_2 = \frac{c}{a}$
$\beta_1 + \beta_2 = -\frac{q}{p}, \beta_1\beta_2 = \frac{r}{p}$
The system of linear equations $\alpha_1y + \alpha_2z = 0$ and $\beta_1y + \beta_2z = 0$ has a non-zero solution if the determinant of the coefficient matrix is zero:
$\begin{vmatrix} \alpha_1 & \alpha_2 \\ \beta_1 & \beta_2 \end{vmatrix} = 0$
$\Rightarrow \alpha_1\beta_2 - \alpha_2\beta_1 = 0 \Rightarrow \frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} = k$ (say)
Using the property of ratios,$\frac{\alpha_1 + \alpha_2}{\beta_1 + \beta_2} = \frac{\alpha_1 - \alpha_2}{\beta_1 - \beta_2} = k$
Also,$\frac{\alpha_1\alpha_2}{\beta_1\beta_2} = k^2$
Therefore,$\left( \frac{\alpha_1 + \alpha_2}{\beta_1 + \beta_2} \right)^2 = \frac{\alpha_1\alpha_2}{\beta_1\beta_2}$
Substituting the values:
$\left( \frac{-b/a}{-q/p} \right)^2 = \frac{c/a}{r/p}$
$\frac{b^2 p^2}{a^2 q^2} = \frac{cp}{ar}$
$b^2 p^2 ar = a^2 q^2 cp$
$b^2 pr = q^2 ac$

(A) Since $\alpha, \beta, \gamma, \delta$ are in geometric progression,let the common ratio be $k$. Then $\beta = \alpha k$,$\gamma = \alpha k^2$,and $\delta = \alpha k^3$.
For the equation $ax^2 + 2bx + c = 0$,the sum of roots $\alpha + \beta = -\frac{2b}{a}$ and product $\alpha \beta = \frac{c}{a}$.
For the equation $px^2 + 2qx + r = 0$,the sum of roots $\gamma + \delta = -\frac{2q}{p}$ and product $\gamma \delta = \frac{r}{p}$.
Since $\alpha, \beta, \gamma, \delta$ are in $GP$,$\frac{\beta}{\alpha} = \frac{\delta}{\gamma} = k$.
Also,$\frac{\alpha + \beta}{\alpha - \beta} = \frac{1+k}{1-k}$ and $\frac{\gamma + \delta}{\gamma - \delta} = \frac{\alpha k^2 + \alpha k^3}{\alpha k^2 - \alpha k^3} = \frac{1+k}{1-k}$.
Thus,$\frac{(\alpha + \beta)^2}{\alpha \beta} = \frac{(\alpha + \alpha k)^2}{\alpha^2 k} = \frac{\alpha^2(1+k)^2}{\alpha^2 k} = \frac{(1+k)^2}{k}$.
Similarly,$\frac{(\gamma + \delta)^2}{\gamma \delta} = \frac{(\alpha k^2 + \alpha k^3)^2}{\alpha^2 k^5} = \frac{\alpha^2 k^4(1+k)^2}{\alpha^2 k^5} = \frac{(1+k)^2}{k}$.
Therefore,$\frac{(\alpha + \beta)^2}{\alpha \beta} = \frac{(\gamma + \delta)^2}{\gamma \delta}$.
Substituting the coefficients: $\frac{(-2b/a)^2}{c/a} = \frac{(-2q/p)^2}{r/p}$ $\Rightarrow \frac{4b^2/a^2}{c/a} = \frac{4q^2/p^2}{r/p}$ $\Rightarrow \frac{4b^2}{ac} = \frac{4q^2}{pr}$.
This simplifies to $\frac{b^2}{ac} = \frac{q^2}{pr}$,which is equivalent to $\frac{ac}{b^2} = \frac{pr}{q^2}$.

(A) Given that $\tan \alpha$ and $\tan \beta$ are roots of $x^2 - px + q = 0$.
From the properties of roots,$\tan \alpha + \tan \beta = p$ and $\tan \alpha \tan \beta = q$.
Using the formula for $\tan(\alpha + \beta)$:
$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{p}{1 - q}$.
We know that $\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta}$.
Substituting $\theta = \alpha + \beta$:
$\sin^2(\alpha + \beta) = \frac{\tan^2(\alpha + \beta)}{1 + \tan^2(\alpha + \beta)} = \frac{(\frac{p}{1 - q})^2}{1 + (\frac{p}{1 - q})^2}$.
Simplifying the expression:
$= \frac{p^2}{(1 - q)^2 + p^2} = \frac{p^2}{p^2 + (1 - q)^2}$.

(C) The discriminant of the first equation is $D_1 = b^2 - 4ac$. The difference of roots is $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2}$.
For the second equation,the roots are $\alpha - k$ and $\beta - k$. The difference of roots is $(\alpha - k - (\beta - k))^2 = (\alpha - \beta)^2 = \frac{b^2 - 4ac}{a^2}$.
Also,for the second equation,the difference of roots squared is $\frac{B^2 - 4AC}{A^2}$.
Equating the two,we get $\frac{b^2 - 4ac}{a^2} = \frac{B^2 - 4AC}{A^2}$.
Therefore,$\frac{B^2 - 4AC}{b^2 - 4ac} = \frac{A^2}{a^2} = \left( \frac{A}{a} \right)^2$.

(B) Let the quadratic equation be $x^2 + bx + c = 0$.
Sachin made a mistake in the constant term $c$,so his roots $(4, 3)$ are correct for the coefficient $b$. The sum of roots is $-(b) = 4 + 3 = 7$,so $b = -7$.
Rahul made a mistake in the coefficient $b$,so his roots $(3, 2)$ are correct for the constant term $c$. The product of roots is $c = 3 \times 2 = 6$.
The correct quadratic equation is $x^2 - 7x + 6 = 0$.
Factoring the equation: $x^2 - 6x - x + 6 = 0 \implies x(x - 6) - 1(x - 6) = 0 \implies (x - 6)(x - 1) = 0$.
The correct roots are $x = 6$ and $x = 1$.

(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 $S_2 = \beta + \frac{1}{\alpha}$.
Sum of roots $= (\alpha + \beta) + (\frac{1}{\beta} + \frac{1}{\alpha}) = (\alpha + \beta) + \frac{\alpha + \beta}{\alpha\beta} = -\frac{b}{a} + \frac{-b/a}{c/a} = -\frac{b}{a} - \frac{b}{c} = -\frac{b(a+c)}{ac}$.
Product of roots $= (\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}) = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = 2 + \frac{c}{a} + \frac{a}{c} = \frac{2ac + c^2 + a^2}{ac} = \frac{(a+c)^2}{ac}$.
The required 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 $acx^2 + (a+c)bx + (a+c)^2 = 0$.

(D) For the first equation $x^2 - 2ax + b^2 = 0$,let the roots be $\alpha_1$ and $\beta_1$. The sum of the roots is $\alpha_1 + \beta_1 = 2a$. The arithmetic mean $(AM)$ of the roots is $\frac{\alpha_1 + \beta_1}{2} = \frac{2a}{2} = a$.
For the second equation $x^2 - 2bx + a^2 = 0$,let the roots be $\alpha_2$ and $\beta_2$. The sum of the roots is $\alpha_2 + \beta_2 = 2b$. The arithmetic mean $(AM)$ of the roots is $\frac{\alpha_2 + \beta_2}{2} = \frac{2b}{2} = b$.
The product of the roots of the second equation is $\alpha_2 \beta_2 = a^2$. The geometric mean $(GM)$ of the roots is $\sqrt{\alpha_2 \beta_2} = \sqrt{a^2} = |a|$.
Comparing the results,the $AM$ of the roots of the first equation is $a$,which is not necessarily equal to the $AM$ of the roots of the second equation $(b)$ or the $GM$ of the roots of the second equation $(|a|)$. However,if we look at the options,the arithmetic mean of the first equation is $a$. The arithmetic mean of the second equation is $b$. Thus,the correct choice is $D$.

(D) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$a\alpha^2 + b\alpha + c = 0 \implies a\alpha^2 + b\alpha = -c \implies \alpha(a\alpha + b) = -c \implies a\alpha + b = -c/\alpha$.
Similarly,$a\beta^2 + b\beta + c = 0 \implies a\beta^2 + b\beta = -c \implies \beta(a\beta + b) = -c \implies a\beta + b = -c/\beta$.
Now,substitute these into the expression:
$\frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b} = \frac{\alpha}{-c/\beta} + \frac{\beta}{-c/\alpha}$.
$= -\frac{\alpha\beta}{c} - \frac{\alpha\beta}{c} = -\frac{2\alpha\beta}{c}$.
Since the product of roots $\alpha\beta = c/a$,we substitute this value:
$= -\frac{2}{c} \times \frac{c}{a} = -\frac{2}{a}$.

(D) Given the quadratic equation $x^2 + px + q = 0$ with roots $\alpha = \tan 30^\circ$ and $\beta = \tan 15^\circ$.
From the properties of roots,the sum of roots is $\alpha + \beta = -p$ and the product of roots is $\alpha \beta = q$.
We know that $\tan(30^\circ + 15^\circ) = \tan 45^\circ = 1$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$1 = \frac{-p}{1 - q}$.
This implies $1 - q = -p$,or $p - q = -1$,which can be rewritten as $p + q = 1$ is incorrect; rather $1 - q = -p$ $\Rightarrow p - q = -1$ $\Rightarrow q - p = 1$.
We need to find the value of $2 + q - p$.
Substituting $q - p = 1$,we get $2 + (1) = 3$.

(A) Let the quadratic equation be $ax^{2} + bx + c = 0$.
Sachin made a mistake in the constant term,so the sum of the roots is correct.
Sum of roots $= 4 + 3 = 7$.
Rahul made a mistake in the coefficient of $x$,so the product of the roots is correct.
Product of roots $= 3 \times 2 = 6$.
The correct quadratic equation is $x^{2} - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^{2} - 7x + 6 = 0$.
Factoring the equation: $x^{2} - 6x - x + 6 = 0 \implies x(x - 6) - 1(x - 6) = 0 \implies (x - 6)(x - 1) = 0$.
Thus,the correct roots are $6$ and $1$.

(C) Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$,we have $\alpha + \beta = 6$ and $\alpha\beta = -2$.
Also,$\alpha^2 = 6\alpha + 2$ and $\beta^2 = 6\beta + 2$.
Given $a_n = \alpha^n - \beta^n$,we evaluate the expression:
$\frac{a_{10} - 2a_8}{2a_9} = \frac{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)}{2(\alpha^9 - \beta^9)}$
$= \frac{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)}{2(\alpha^9 - \beta^9)}$
Since $\alpha^2 - 2 = 6\alpha$ and $\beta^2 - 2 = 6\beta$,we substitute these values:
$= \frac{\alpha^8(6\alpha) - \beta^8(6\beta)}{2(\alpha^9 - \beta^9)}$
$= \frac{6(\alpha^9 - \beta^9)}{2(\alpha^9 - \beta^9)}$
$= \frac{6}{2} = 3$.

(A) Given the equation $ax^2 + bx + c = 0$. Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
The sum of the reciprocals of their squares is $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$.
Substituting the values: $\frac{(-\frac{b}{a})^2 - 2(\frac{c}{a})}{(\frac{c}{a})^2} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{b^2 - 2ac}{c^2}$.
According to the given condition,$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$,so $-\frac{b}{a} = \frac{b^2 - 2ac}{c^2}$.
Cross-multiplying gives $-bc^2 = a(b^2 - 2ac) = ab^2 - 2a^2c$.
Rearranging the terms,we get $2a^2c = ab^2 + bc^2$.
Dividing by $a^2c^2$ is not necessary here; we observe that $2(ca^2) = bc^2 + ab^2$. This is the condition for $bc^2, ca^2, ab^2$ to be in $A.P.$

(A) Let the roots of $lx^2 + nx + n = 0$ be $\alpha$ and $\beta$.
Given that $\frac{\alpha}{\beta} = \frac{p}{q}$.
From the properties of roots,$\alpha + \beta = -\frac{n}{l}$ and $\alpha\beta = \frac{n}{l}$.
We need to evaluate $\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}}$.
Substituting $\frac{p}{q} = \frac{\alpha}{\beta}$,we get $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta} = \frac{\alpha + \beta}{\sqrt{\alpha\beta}} + \sqrt{\alpha\beta}$.
Substituting the values of $\alpha + \beta$ and $\alpha\beta$:
$= \frac{-n/l}{\sqrt{n/l}} + \sqrt{n/l} = -\sqrt{\frac{n}{l}} + \sqrt{\frac{n}{l}} = 0$.

(D) Let the correct equation be $x^2 + px + q = 0$ $(i)$.
The first student uses a wrong value of $p$ but the correct value of $q$. The roots found are $2$ and $6$.
The product of roots is $2 \times 6 = 12$. Since the constant term $q$ is correct,$q = 12$.
The second student uses a wrong value of $q$ but the correct value of $p$. The roots found are $2$ and $-9$.
The sum of roots is $2 + (-9) = -7$. Since the coefficient of $x$ $(p)$ is correct,$-p = -7$,which means $p = 7$.
Substituting $p = 7$ and $q = 12$ into the original equation $(i)$,we get $x^2 + 7x + 12 = 0$.
Factoring the quadratic equation: $x^2 + 4x + 3x + 12 = 0$ $\Rightarrow x(x + 4) + 3(x + 4) = 0$ $\Rightarrow (x + 3)(x + 4) = 0$.
Thus,the roots are $x = -3$ and $x = -4$.

(B) Given that $\alpha_1, \alpha_2$ are the roots of $ax^2 + bx + c = 0$.
So,$\alpha_1 + \alpha_2 = -\frac{b}{a}$ and $\alpha_1 \alpha_2 = \frac{c}{a}$.
Now,$\beta_1, \beta_2$ are the roots of $px^2 + qx + r = 0$.
So,$\beta_1 + \beta_2 = -\frac{q}{p}$ and $\beta_1 \beta_2 = \frac{r}{p}$.
The system $\alpha_1 y + \alpha_2 z = 0$ and $\beta_1 y + \beta_2 z = 0$ has a non-zero solution if the determinant of the coefficient matrix is zero:
$\alpha_1 \beta_2 - \alpha_2 \beta_1 = 0 \implies \frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} = k$ (say).
Then $\alpha_1 = k \beta_1$ and $\alpha_2 = k \beta_2$.
From the sum and product of roots:
$\frac{\alpha_1 + \alpha_2}{\beta_1 + \beta_2} = \frac{k(\beta_1 + \beta_2)}{\beta_1 + \beta_2} = k = \frac{-b/a}{-q/p} = \frac{bp}{aq}$.
Also,$\frac{\alpha_1 \alpha_2}{\beta_1 \beta_2} = \frac{k^2 \beta_1 \beta_2}{\beta_1 \beta_2} = k^2 = \frac{c/a}{r/p} = \frac{cp}{ar}$.
Equating $k^2$:
$\left(\frac{bp}{aq}\right)^2 = \frac{cp}{ar} \implies \frac{b^2 p^2}{a^2 q^2} = \frac{cp}{ar}$.
Simplifying:
$b^2 p^2 ar = cp a^2 q^2 \implies b^2 pr = q^2 ac$.

(D) Let the correct quadratic equation be $x^2 + bx + c = 0$ (since the coefficient of $x^2$ is $1$).
For the first student,the constant term was copied incorrectly,but the coefficient of $x$ $(b)$ was copied correctly. The roots obtained were $3$ and $2$. Thus,the sum of roots is $-b = 3 + 2 = 5$,which implies $b = -5$.
For the second student,the constant term $(c)$ and the coefficient of $x^2$ $(a=1)$ were copied correctly. Given $c = -6$ and $a = 1$,the product of the roots is $c/a = -6/1 = -6$.
The correct quadratic equation is $x^2 - 5x - 6 = 0$.
Factoring the equation: $x^2 - 6x + x - 6 = 0 \implies x(x - 6) + 1(x - 6) = 0 \implies (x - 6)(x + 1) = 0$.
Therefore,the correct roots are $6$ and $-1$.