QUADRATIC EQUATION Questions in English

(D) Given that $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$.
From the properties of quadratic equations,we have $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.
Let the new roots be $\alpha' = 2 + \alpha$ and $\beta' = 2 + \beta$.
The sum of the new roots is $\alpha' + \beta' = (2 + \alpha) + (2 + \beta) = 4 + (\alpha + \beta) = 4 - b/a = (4a - b)/a$.
The product of the new roots is $\alpha'\beta' = (2 + \alpha)(2 + \beta) = 4 + 2(\alpha + \beta) + \alpha\beta = 4 + 2(-b/a) + c/a = (4a - 2b + c)/a$.
The required quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values: $x^2 - ((4a - b)/a)x + (4a - 2b + c)/a = 0$.
Multiplying the entire equation by $a$,we get $ax^2 - (4a - b)x + (4a - 2b + c) = 0$.
This simplifies to $ax^2 + x(b - 4a) + 4a - 2b + c = 0$.

(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,$\alpha + \beta = -b$,$\alpha\beta = c$,$\alpha' + \beta' = -q$,and $\alpha'\beta' = r$.
Given that the ratio of the roots is the same,$\frac{\alpha}{\beta} = \frac{\alpha'}{\beta'}$.
Using the property of componendo and dividendo,$\frac{\alpha + \beta}{\alpha - \beta} = \frac{\alpha' + \beta'}{\alpha' - \beta'}$.
Squaring both sides,$\frac{(\alpha + \beta)^2}{(\alpha - \beta)^2} = \frac{(\alpha' + \beta')^2}{(\alpha' - \beta')^2}$.
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 $b^2(q^2 - 4r) = q^2(b^2 - 4c)$.
$b^2q^2 - 4b^2r = q^2b^2 - 4cq^2$.
$-4b^2r = -4cq^2$,which simplifies to $b^2r = cq^2$.

(C) Let the roots of the quadratic equation $x^2 - x - k = 0$ be $\alpha$ and $\alpha^2$.
From the properties of roots,the sum of roots is $\alpha + \alpha^2 = 1$ and the product of roots is $\alpha \cdot \alpha^2 = \alpha^3 = -k$.
We have $\alpha^2 + \alpha - 1 = 0$. Cubing both sides of $\alpha^2 + \alpha = 1$ is not direct,so we use the identity $(\alpha^2 + \alpha)^3 = 1^3$.
Expanding this,we get $(\alpha^2)^3 + \alpha^3 + 3(\alpha^2)(\alpha)(\alpha^2 + \alpha) = 1$.
Substituting $\alpha^3 = -k$ and $\alpha^2 + \alpha = 1$:
$(-k)^2 + (-k) + 3(-k)(1) = 1$.
$k^2 - k - 3k = 1$.
$k^2 - 4k - 1 = 0$.
Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $k = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$.

(D) Since the coefficients $p$ and $q$ of the quadratic equation ${x^2} + px + q = 0$ are real,complex roots must occur in conjugate pairs.
Given that $3 + 4i$ is a root,its conjugate $3 - 4i$ must also be a root of the equation.
For a quadratic equation ${x^2} + px + q = 0$,the sum of the roots is given by $-p$ and the product of the roots is given by $q$.
Sum of the roots $= (3 + 4i) + (3 - 4i) = 6$.
Therefore,$-p = 6$,which implies $p = -6$.
Product of the roots $= (3 + 4i)(3 - 4i) = {3^2} - {(4i)^2} = 9 - 16({i^2}) = 9 - 16(-1) = 9 + 16 = 25$.
Therefore,$q = 25$.
Thus,the values are $p = -6$ and $q = 25$.

(A) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$.
From the properties of roots,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
According to the problem,the sum of the roots is equal to the sum of the squares of their reciprocals:
$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$
$\alpha + \beta = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$
Substitute $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$:
$-\frac{b}{a} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$
$-\frac{b}{a} = \frac{(-\frac{b}{a})^2 - 2(\frac{c}{a})}{(\frac{c}{a})^2}$
$-\frac{b}{a} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}}$
Multiply both sides by $\frac{c^2}{a^2}$:
$-\frac{b}{a} \cdot \frac{c^2}{a^2} = \frac{b^2}{a^2} - \frac{2c}{a}$
$-\frac{bc^2}{a^3} = \frac{b^2}{a^2} - \frac{2c}{a}$
Divide the entire equation by $\frac{c}{a}$ (assuming $c \neq 0$):
$-\frac{bc}{a^2} = \frac{b^2}{ac} - 2$
Rearranging the terms,we get:
$\frac{b^2}{ac} + \frac{bc}{a^2} = 2$.

(B) Given the quadratic equation $x^2 - 6x + a = 0$.
Since $\alpha$ and $\beta$ are the roots,the sum of roots is $\alpha + \beta = -(-6)/1 = 6$.
We are given the relation $3\alpha + 2\beta = 16$.
We can rewrite this as $2(\alpha + \beta) + \alpha = 16$.
Substituting the value of $\alpha + \beta = 6$ into the equation:
$2(6) + \alpha = 16$
$12 + \alpha = 16$
$\alpha = 4$.
Since $\alpha$ is a root of the equation,it must satisfy $x^2 - 6x + a = 0$:
$(4)^2 - 6(4) + a = 0$
$16 - 24 + a = 0$
$-8 + a = 0$
$a = 8$.

(A) Given the quadratic equation $lx^2 + mx + n = 0$,the sum of roots is $\alpha + \beta = -m/l$ and the product of roots is $\alpha\beta = n/l$.
We need to find the equation with roots $S_1 = \alpha^3\beta$ and $S_2 = \alpha\beta^3$.
The sum of the new roots is $S_1 + S_2 = \alpha\beta(\alpha^2 + \beta^2) = \alpha\beta((\alpha + \beta)^2 - 2\alpha\beta)$.
Substituting the values: $S_1 + S_2 = (n/l)((-m/l)^2 - 2(n/l)) = (n/l)((m^2/l^2) - (2n/l)) = (n/l)((m^2 - 2nl)/l^2) = (n(m^2 - 2nl))/l^3$.
The product of the new roots is $S_1 \cdot S_2 = (\alpha^3\beta)(\alpha\beta^3) = \alpha^4\beta^4 = (\alpha\beta)^4 = (n/l)^4 = n^4/l^4$.
The required quadratic equation is $x^2 - (S_1 + S_2)x + (S_1 \cdot S_2) = 0$.
Substituting the values: $x^2 - [n(m^2 - 2nl)/l^3]x + (n^4/l^4) = 0$.
Multiplying by $l^4$,we get $l^4x^2 - nl(m^2 - 2nl)x + n^4 = 0$.

(B) Let the roots of the equation $x^2 + px + q = 0$ be $\alpha$ and $\beta$.
Given that the difference between the roots is $\alpha - \beta = 1$.
We know that for a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -b/a$ and the product of roots $\alpha\beta = c/a$.
For the given equation,$\alpha + \beta = -p$ and $\alpha\beta = q$.
Using the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$,we substitute the known values:
$(1)^2 = (-p)^2 - 4(q)$.
$1 = p^2 - 4q$.
Therefore,$p^2 = 4q + 1$.

(B) Given the quadratic equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$.
Let the roots be $x_1$ and $x_2$.
From the properties of quadratic equations, the sum of roots is $x_1 + x_2 = \frac{4 + \sqrt{5}}{5 + \sqrt{2}}$.
The product of roots is $x_1 x_2 = \frac{8 + 2\sqrt{5}}{5 + \sqrt{2}} = \frac{2(4 + \sqrt{5})}{5 + \sqrt{2}} = 2(x_1 + x_2)$.
The harmonic mean $(HM)$ of two numbers $x_1$ and $x_2$ is given by $HM = \frac{2x_1 x_2}{x_1 + x_2}$.
Substituting $x_1 x_2 = 2(x_1 + x_2)$ into the formula, we get $HM = \frac{2 \cdot 2(x_1 + x_2)}{x_1 + x_2} = 4$.

(A) Let the roots be $\alpha$ and $\alpha + 1$.
According to the properties of quadratic equations,the sum of the roots is $\alpha + (\alpha + 1) = 2\alpha + 1 = b$.
The product of the roots is $\alpha(\alpha + 1) = c$.
We need to find the value of the discriminant $D = b^2 - 4c$.
Substituting the expressions for $b$ and $c$:
$b^2 - 4c = (2\alpha + 1)^2 - 4(\alpha(\alpha + 1))$
Expanding the terms:
$b^2 - 4c = (4\alpha^2 + 4\alpha + 1) - (4\alpha^2 + 4\alpha)$
Simplifying the expression:
$b^2 - 4c = 4\alpha^2 + 4\alpha + 1 - 4\alpha^2 - 4\alpha = 1$.
Therefore,$b^2 - 4c = 1$.

(A) Given the quadratic equation $Ax^2 + Bx + C = 0$.
From the relationship between roots and coefficients,we have:
Sum of roots: $\alpha + \beta = -\frac{B}{A}$
Product of roots: $\alpha \beta = \frac{C}{A}$
We use the algebraic identity: $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$.
Substituting the values:
$\alpha^3 + \beta^3 = \left(-\frac{B}{A}\right)^3 - 3\left(\frac{C}{A}\right)\left(-\frac{B}{A}\right)$
$= -\frac{B^3}{A^3} + \frac{3BC}{A^2}$
To combine these,find a common denominator $A^3$:
$= \frac{-B^3 + 3ABC}{A^3} = \frac{3ABC - B^3}{A^3}$.

(C) Given the quadratic equation ${x^2} - (1 + {n^2})x + \frac{1}{2}(1 + {n^2} + {n^4}) = 0$.
For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -b/a$ and the product of roots $\alpha \beta = c/a$.
Here,$\alpha + \beta = 1 + {n^2}$ and $\alpha \beta = \frac{1}{2}(1 + {n^2} + {n^4})$.
We know that ${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta$.
Substituting the values:
${\alpha ^2} + {\beta ^2} = {(1 + {n^2})^2} - 2 \times \frac{1}{2}(1 + {n^2} + {n^4})$
${\alpha ^2} + {\beta ^2} = (1 + {n^4} + 2{n^2}) - (1 + {n^2} + {n^4})$
${\alpha ^2} + {\beta ^2} = 1 + {n^4} + 2{n^2} - 1 - {n^2} - {n^4}$
${\alpha ^2} + {\beta ^2} = {n^2}$.

(B) Let the roots of the quadratic equation ${x^2} - 30x + p = 0$ be $\alpha$ and ${\alpha ^2}$.
According to the relationship between roots and coefficients:
Sum of roots: $\alpha + {\alpha ^2} = 30$
Product of roots: $\alpha \cdot {\alpha ^2} = {\alpha ^3} = p$
Solving the sum equation: ${\alpha ^2} + \alpha - 30 = 0$
$(\alpha + 6)(\alpha - 5) = 0$
This gives $\alpha = -6$ or $\alpha = 5$.
If $\alpha = -6$,then $p = {\alpha ^3} = {(-6)^3} = -216$.
If $\alpha = 5$,then $p = {\alpha ^3} = {(5)^3} = 125$.
Thus,the values of $p$ are $125$ and $-216$.

(B) For a quadratic equation of the form $ax^2 + bx + c = 0$,the sum of the roots is given by $\alpha + \beta = -b/a$ and the product of the roots is given by $\alpha \beta = c/a$.
Given the equation $x^2 - 3x + 1 = 0$,we have $a = 1$,$b = -3$,and $c = 1$.
Thus,the sum of the roots $\alpha + \beta = -(-3)/1 = 3$ and the product of the roots $\alpha \beta = 1/1 = 1$.
We need to find the sum of the squares of the roots,which is $\alpha^2 + \beta^2$.
Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$,we substitute the values:
$\alpha^2 + \beta^2 = (3)^2 - 2(1) = 9 - 2 = 7$.
Therefore,the sum of the squares of the roots is $7$.

(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
Given that the sum of the roots is $\alpha + \beta = -1$.
Given that the sum of their reciprocals is $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{6}$.
Simplifying the sum of reciprocals: $\frac{\alpha + \beta}{\alpha \beta} = \frac{1}{6}$.
Substituting $\alpha + \beta = -1$ into the equation: $\frac{-1}{\alpha \beta} = \frac{1}{6}$,which gives $\alpha \beta = -6$.
The standard form of a quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values: $x^2 - (-1)x + (-6) = 0$.
Therefore,the equation is $x^2 + x - 6 = 0$.

(C) Let the roots of the quadratic equation ${x^2} + px + q = 0$ be $\alpha$ and $\beta$.
From the properties of quadratic equations,the sum of the roots is $\alpha + \beta = -p$ and the product of the roots is $\alpha \beta = q$.
According to the problem,the sum of the roots is equal to the sum of their squares:
$\alpha + \beta = {\alpha ^2} + {\beta ^2}$
We know that ${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta$.
Substituting the values of the sum and product of the roots into the equation:
$-p = {(-p)^2} - 2q$
$-p = {p^2} - 2q$
Rearranging the terms,we get:
${p^2} + p = 2q$.

(C) Given that $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 1 = 0$.
From the properties of quadratic equations,the sum of roots $\alpha + \beta = 3$ and the product of roots $\alpha \beta = 1$.
Let the new roots be $x_1 = \frac{1}{\alpha - 2}$ and $x_2 = \frac{1}{\beta - 2}$.
The sum of the new roots $S = x_1 + x_2 = \frac{1}{\alpha - 2} + \frac{1}{\beta - 2} = \frac{\beta - 2 + \alpha - 2}{(\alpha - 2)(\beta - 2)} = \frac{(\alpha + \beta) - 4}{\alpha \beta - 2(\alpha + \beta) + 4}$.
Substituting the values: $S = \frac{3 - 4}{1 - 2(3) + 4} = \frac{-1}{1 - 6 + 4} = \frac{-1}{-1} = 1$.
The product of the new roots $P = x_1 \cdot x_2 = \frac{1}{(\alpha - 2)(\beta - 2)} = \frac{1}{\alpha \beta - 2(\alpha + \beta) + 4}$.
Substituting the values: $P = \frac{1}{1 - 2(3) + 4} = \frac{1}{1 - 6 + 4} = \frac{1}{-1} = -1$.
The required quadratic equation is $x^2 - Sx + P = 0$,which gives $x^2 - (1)x + (-1) = 0$,or $x^2 - x - 1 = 0$.

(B) Let $\alpha$ and $\beta$ be the roots of the equation $ax^2 + bx + c = 0$.
From the properties of roots,we have $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.
The roots of the new equation are $(\alpha - 1)$ and $(\beta - 1)$.
The new equation is $2x^2 + 8x + 2 = 0$,which can be written as $x^2 + 4x + 1 = 0$.
For the new equation,the sum of roots is $(\alpha - 1) + (\beta - 1) = -4/1 = -4$.
Thus,$(\alpha + \beta) - 2 = -4$,which implies $\alpha + \beta = -2$.
Substituting $\alpha + \beta = -b/a$,we get $-b/a = -2$,so $b = 2a$.
The product of the new roots is $(\alpha - 1)(\beta - 1) = 1/1 = 1$.
Expanding this,we get $\alpha\beta - (\alpha + \beta) + 1 = 1$,which simplifies to $\alpha\beta - (\alpha + \beta) = 0$.
Substituting $\alpha\beta = c/a$ and $\alpha + \beta = -2$,we get $c/a - (-2) = 0$,so $c/a + 2 = 0$,which means $c = -2a$.
Since $b = 2a$ and $c = -2a$,we have $b = -c$.

(C) Given equation is $9x^2 + 6x + 1 = 0$.
For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -b/a$ and the product of roots $\alpha\beta = c/a$.
Here,$\alpha + \beta = -6/9 = -2/3$ and $\alpha\beta = 1/9$.
We need to find the equation whose roots are $1/\alpha$ and $1/\beta$.
The sum of the new roots is $S = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-2/3}{1/9} = -6$.
The product of the new roots is $P = \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{1/9} = 9$.
The required quadratic equation is $x^2 - Sx + P = 0$.
Substituting the values,we get $x^2 - (-6)x + 9 = 0$,which simplifies to $x^2 + 6x + 9 = 0$.

(B) Given that $\alpha$ and $\beta$ are the roots of $6x^2 - 6x + 1 = 0$.
From the properties of roots,$\alpha + \beta = -(-6)/6 = 1$ and $\alpha\beta = 1/6$.
The given expression is $\frac{1}{2}[a + b\alpha + c\alpha^2 + d\alpha^3] + \frac{1}{2}[a + b\beta + c\beta^2 + d\beta^3]$.
This simplifies to $a + \frac{b}{2}(\alpha + \beta) + \frac{c}{2}(\alpha^2 + \beta^2) + \frac{d}{2}(\alpha^3 + \beta^3)$.
Substituting $\alpha + \beta = 1$ and $\alpha\beta = 1/6$:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 1^2 - 2(1/6) = 1 - 1/3 = 2/3$.
$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = 1^3 - 3(1/6)(1) = 1 - 1/2 = 1/2$.
Substituting these values back into the expression:
$= a + \frac{b}{2}(1) + \frac{c}{2}(2/3) + \frac{d}{2}(1/2) = a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4}$.

(A) Given that $\tan \alpha$ and $\tan \beta$ are the roots of $x^2 - px + q = 0$.
From the properties of quadratic equations,we have:
$\tan \alpha + \tan \beta = p$ $(i)$
$\tan \alpha \tan \beta = q$ $(ii)$
Using the formula for the tangent of a sum: $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{p}{1 - q}$.
We know that $\sin^2(\alpha + \beta) = \frac{\tan^2(\alpha + \beta)}{1 + \tan^2(\alpha + \beta)}$.
Substituting the value of $\tan(\alpha + \beta)$:
$\sin^2(\alpha + \beta) = \frac{(\frac{p}{1 - q})^2}{1 + (\frac{p}{1 - q})^2} = \frac{\frac{p^2}{(1 - q)^2}}{\frac{(1 - q)^2 + p^2}{(1 - q)^2}} = \frac{p^2}{p^2 + (1 - q)^2}$.

(A) Given the equation: $\frac{x - m}{mx + 1} = \frac{x + n}{nx + 1}$.
Cross-multiplying,we get: $(x - m)(nx + 1) = (x + n)(mx + 1)$.
Expanding both sides: $nx^2 + x - mnx - m = mx^2 + x + mnx + n$.
Rearranging the terms: $(n - m)x^2 - 2mnx - (m + n) = 0$.
Multiplying by $-1$: $(m - n)x^2 + 2mnx + (m + n) = 0$.
Let the roots be $\alpha$ and $\frac{1}{\alpha}$.
The product of the roots for a quadratic equation $ax^2 + bx + c = 0$ is given by $\frac{c}{a}$.
Therefore,$\alpha \cdot \frac{1}{\alpha} = \frac{m + n}{m - n}$.
$1 = \frac{m + n}{m - n}$.
$m - n = m + n$.
$-n = n$,which implies $2n = 0$,so $n = 0$.

(B) Given the equation $x^2 - 5x + 16 = 0$ with roots $\alpha$ and $\beta$,we have:
$\alpha + \beta = 5$ and $\alpha \beta = 16$.
The roots of the equation $x^2 + px + q = 0$ are $\alpha^2 + \beta^2$ and $\frac{\alpha \beta}{2}$.
First,calculate the sum of the roots:
$(\alpha^2 + \beta^2) + \frac{\alpha \beta}{2} = -p$
Since $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = 5^2 - 2(16) = 25 - 32 = -7$,
$-7 + \frac{16}{2} = -p \Rightarrow -7 + 8 = -p \Rightarrow 1 = -p \Rightarrow p = -1$.
Next,calculate the product of the roots:
$(\alpha^2 + \beta^2) \cdot \frac{\alpha \beta}{2} = q$
$(-7) \cdot \frac{16}{2} = q \Rightarrow -7 \cdot 8 = q \Rightarrow q = -56$.
Thus,$p = -1$ and $q = -56$.

(B) Let $\alpha$ be a root of $x^2 - x + k = 0$. Then,$2\alpha$ is a root of $x^2 - x + 3k = 0$.
Substituting these roots into the respective equations:
$1) \alpha^2 - \alpha + k = 0$
$2) (2\alpha)^2 - (2\alpha) + 3k = 0 \Rightarrow 4\alpha^2 - 2\alpha + 3k = 0$
Multiply equation $(1)$ by $2$: $2\alpha^2 - 2\alpha + 2k = 0$.
Subtract this from equation $(2)$:
$(4\alpha^2 - 2\alpha + 3k) - (2\alpha^2 - 2\alpha + 2k) = 0 - 0$
$2\alpha^2 + k = 0 \Rightarrow \alpha^2 = -k/2$.
Substitute $\alpha^2 = -k/2$ into equation $(1)$:
$-k/2 - \alpha + k = 0 \Rightarrow \alpha = k/2$.
Since $\alpha^2 = (k/2)^2$,we have $-k/2 = k^2/4$.
$k^2/4 + k/2 = 0 \Rightarrow k^2 + 2k = 0$.
$k(k + 2) = 0$,so $k = 0$ or $k = -2$.
Since $k=0$ leads to trivial roots $(0, 0)$,the non-zero value is $k = -2$.

(A) Let $r$ be the common ratio of the $G.P.$ $\alpha, \beta, \gamma, \delta$. Then $\beta = \alpha r, \gamma = \alpha r^2, \delta = \alpha r^3$.
From the sum and product of roots:
$\alpha + \beta = 1 \Rightarrow \alpha(1 + r) = 1$ $(i)$
$\alpha \beta = p \Rightarrow \alpha^2 r = p$ $(ii)$
$\gamma + \delta = 4 \Rightarrow \alpha r^2(1 + r) = 4$ $(iii)$
$\gamma \delta = q \Rightarrow \alpha^2 r^5 = q$ $(iv)$
Dividing $(iii)$ by $(i)$,we get $r^2 = 4$,so $r = \pm 2$.
If $r = 2$,$\alpha(1+2) = 1 \Rightarrow \alpha = 1/3$ (not an integer).
If $r = -2$,$\alpha(1-2) = 1 \Rightarrow -\alpha = 1 \Rightarrow \alpha = -1$.
Substituting $\alpha = -1$ and $r = -2$ into $(ii)$ and $(iv)$:
$p = (-1)^2(-2) = -2$.
$q = (-1)^2(-2)^5 = -32$.
Thus,$(p, q) = (-2, -32)$.

(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
The $A.M.$ of the roots is given by $\frac{\alpha + \beta}{2} = \frac{8}{5}$,which implies $\alpha + \beta = \frac{16}{5}$ $(i)$.
The $A.M.$ of their reciprocals is given by $\frac{\frac{1}{\alpha} + \frac{1}{\beta}}{2} = \frac{8}{7}$.
This simplifies to $\frac{\alpha + \beta}{2\alpha\beta} = \frac{8}{7}$.
Substituting the value of $\alpha + \beta$ from $(i)$,we get $\frac{16/5}{2\alpha\beta} = \frac{8}{7}$.
$\frac{8}{5\alpha\beta} = \frac{8}{7} \Rightarrow 5\alpha\beta = 7 \Rightarrow \alpha\beta = \frac{7}{5}$ $(ii)$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - (\frac{16}{5})x + \frac{7}{5} = 0$.
Multiplying by $5$,we obtain $5x^2 - 16x + 7 = 0$.

(D) Since the coefficients of the quadratic equation ${x^2} - ax + b = 0$ are real, the complex roots must occur in conjugate pairs.
Given that $1 - i$ is a root, its conjugate $1 + i$ must also be a root of the equation.
For a quadratic equation ${x^2} - (\text{sum of roots})x + (\text{product of roots}) = 0$, we have:
Sum of roots $= (1 - i) + (1 + i) = 2$.
Product of roots $= (1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 2$.
Comparing this with ${x^2} - ax + b = 0$, we get $a = 2$ and $b = 2$.

(C) Given that $3$ is a root of the quadratic equation ${x^2} + kx - 24 = 0.$
Substituting $x = 3$ into the equation:
${3^2} + k(3) - 24 = 0$
$9 + 3k - 24 = 0$
$3k - 15 = 0$
$3k = 15$
$k = 5$
Now,we check which equation is satisfied by $x = 3$ when $k = 5$:
For option $(c)$: ${x^2} - kx + 6 = 0$
Substituting $x = 3$ and $k = 5$:
${3^2} - 5(3) + 6 = 9 - 15 + 6 = 0$
Since the result is $0$,$3$ is a root of the equation ${x^2} - kx + 6 = 0$.

(D) Given that $\alpha^2 - 5\alpha + 3 = 0$ $(i)$ and $\beta^2 - 5\beta + 3 = 0$ $(ii)$.
This implies that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 5x + 3 = 0$.
From the properties of roots,the sum of roots is $\alpha + \beta = 5$ and the product of roots is $\alpha\beta = 3$.
We need to find the equation whose roots are $p = \alpha/\beta$ and $q = \beta/\alpha$.
The sum of the new roots is $p + q = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$.
Since $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 5^2 - 2(3) = 25 - 6 = 19$,we have $p + q = \frac{19}{3}$.
The product of the new roots is $pq = \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1$.
The required quadratic equation is $x^2 - (p + q)x + pq = 0$.
Substituting the values,we get $x^2 - \frac{19}{3}x + 1 = 0$.
Multiplying by $3$,we obtain $3x^2 - 19x + 3 = 0$.

(A) Let $\alpha_1, \beta_1$ be the roots of the equation $x^2 + ax + b = 0$. The difference of the roots is $|\alpha_1 - \beta_1| = \sqrt{a^2 - 4b}$.
Let $\alpha_2, \beta_2$ be the roots of the equation $x^2 + bx + a = 0$. The difference of the roots is $|\alpha_2 - \beta_2| = \sqrt{b^2 - 4a}$.
Given that the differences are equal,we have $\sqrt{a^2 - 4b} = \sqrt{b^2 - 4a}$.
Squaring both sides,we get $a^2 - 4b = b^2 - 4a$.
Rearranging the terms,we get $a^2 - b^2 + 4a - 4b = 0$.
Factoring the expression,we get $(a - b)(a + b) + 4(a - b) = 0$.
Since $a \neq b$,we can divide by $(a - b)$,which gives $(a + b) + 4 = 0$,or $a + b + 4 = 0$.

(C) Given the equation: ${t^2}{x^2} + |x| + 9 = 0$.
For any real number $t$,the term ${t^2}{x^2} \ge 0$ and $|x| \ge 0$.
Therefore,the expression ${t^2}{x^2} + |x| + 9$ must be at least $9$ for any real value of $x$.
Since ${t^2}{x^2} + |x| + 9 \ge 9$,the expression can never equal $0$.
Thus,the equation has no real roots.
Consequently,the product of real roots does not exist.

(A) Let the roots of the quadratic equation $12x^2 - mx + 5 = 0$ be $2k$ and $3k$.
From the relationship between roots and coefficients:
Sum of roots: $2k + 3k = \frac{m}{12} \Rightarrow 5k = \frac{m}{12} \Rightarrow m = 60k$ ... $(i)$
Product of roots: $(2k)(3k) = \frac{5}{12} \Rightarrow 6k^2 = \frac{5}{12} \Rightarrow k^2 = \frac{5}{72} \Rightarrow k = \sqrt{\frac{5}{72}} = \frac{\sqrt{5}}{6\sqrt{2}} = \frac{\sqrt{10}}{12}$.
Substituting the value of $k$ in equation $(i)$:
$m = 60 \times \frac{\sqrt{10}}{12} = 5\sqrt{10}$.
Thus,the correct option is $A$.

(A) Since the coefficients of the quadratic equation ${x^2} + px + q = 0$ are assumed to be rational,if one root is $2 + \sqrt{3}$,the other root must be its conjugate,$2 - \sqrt{3}$.
Sum of the roots $= (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$.
From the equation,the sum of the roots is $-p$.
Therefore,$-p = 4$,which implies $p = -4$.
Product of the roots $= (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.
From the equation,the product of the roots is $q$.
Therefore,$q = 1$.
Thus,the values are $(p, q) = (-4, 1)$.

(C) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $3\alpha$.
From the relationship between roots and coefficients:
Sum of roots: $\alpha + 3\alpha = -\frac{b}{a} \Rightarrow 4\alpha = -\frac{b}{a} \Rightarrow \alpha = -\frac{b}{4a}$.
Product of roots: $\alpha \cdot 3\alpha = \frac{c}{a} \Rightarrow 3\alpha^2 = \frac{c}{a}$.
Substitute the value of $\alpha$ into the product equation:
$3\left(-\frac{b}{4a}\right)^2 = \frac{c}{a}$
$3\left(\frac{b^2}{16a^2}\right) = \frac{c}{a}$
$3b^2 = 16ac$.

(B) Given equation is $3x^2 - 20x + 17 = 0$.
To find the equation whose roots are the reciprocals of the roots of $ax^2 + bx + c = 0$,we replace $x$ with $\frac{1}{x}$.
Substituting $\frac{1}{x}$ in the given equation:
$3(\frac{1}{x})^2 - 20(\frac{1}{x}) + 17 = 0$
$\frac{3}{x^2} - \frac{20}{x} + 17 = 0$
Multiplying the entire equation by $x^2$:
$3 - 20x + 17x^2 = 0$
Rearranging the terms,we get $17x^2 - 20x + 3 = 0$.

(D) Given the quadratic equation $x^2 + 2x + 4 = 0$.
For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -\frac{b}{a}$ and the product of roots $\alpha \beta = \frac{c}{a}$.
Here,$\alpha + \beta = -\frac{2}{1} = -2$ and $\alpha \beta = \frac{4}{1} = 4$.
We need to find the value of $\frac{1}{\alpha^3} + \frac{1}{\beta^3}$.
$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\alpha^3 + \beta^3}{(\alpha \beta)^3}$.
Using the identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta(\alpha + \beta)$:
$\frac{\alpha^3 + \beta^3}{(\alpha \beta)^3} = \frac{(\alpha + \beta)^3 - 3\alpha \beta(\alpha + \beta)}{(\alpha \beta)^3}$.
Substituting the values:
$= \frac{(-2)^3 - 3(4)(-2)}{(4)^3} = \frac{-8 + 24}{64} = \frac{16}{64} = \frac{1}{4}$.

(B) Since the coefficients are real,if $1 + i$ is a root,then its conjugate $1 - i$ must also be a root.
Let the roots be $\alpha = 1 + i$ and $\beta = 1 - i$.
The sum of the roots is $\alpha + \beta = (1 + i) + (1 - i) = 2$.
The product of the roots is $\alpha \beta = (1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 2$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - 2x + 2 = 0$.

(B) Let the two numbers be $x_1$ and $x_2$.
The arithmetic mean is given by $\frac{x_1 + x_2}{2} = 9$,which implies $x_1 + x_2 = 18$.
The geometric mean is given by $\sqrt{x_1 x_2} = 4$,which implies $x_1 x_2 = 4^2 = 16$.
$A$ quadratic equation with roots $x_1$ and $x_2$ is given by the formula: $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - 18x + 16 = 0$.

(A) Given the quadratic equation $6x^2 - 5x + 1 = 0$.
For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -b/a$ and the product of roots $\alpha \beta = c/a$.
Here,$\alpha + \beta = -(-5)/6 = 5/6$ and $\alpha \beta = 1/6$.
We use the identity $\tan^{-1} \alpha + \tan^{-1} \beta = \tan^{-1} \left( \frac{\alpha + \beta}{1 - \alpha \beta} \right)$.
Substituting the values: $\tan^{-1} \left( \frac{5/6}{1 - 1/6} \right) = \tan^{-1} \left( \frac{5/6}{5/6} \right) = \tan^{-1}(1)$.
Since $\tan^{-1}(1) = \pi / 4$,the final value is $\pi / 4$.

(D) Given the quadratic equation $x^2 - px + q = 0$.
Let $a$ and $b$ be the roots of the equation.
According to the relationship between roots and coefficients:
Sum of roots: $a + b = -(\text{coefficient of } x) / (\text{coefficient of } x^2) = -(-p) / 1 = p$.
Product of roots: $ab = (\text{constant term}) / (\text{coefficient of } x^2) = q / 1 = q$.
Now,we need to find the value of $\frac{1}{a} + \frac{1}{b}$.
$\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$.
Substituting the values of $(a + b)$ and $(ab)$,we get $\frac{p}{q}$.

(D) Let the roots of the given equation $x^2 + px + q = 0$ be $\alpha$ and $\alpha^2$.
From the relationship between roots and coefficients:
Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = q$
Sum of roots: $\alpha + \alpha^2 = -p$
Now,cube both sides of the sum equation: $(\alpha + \alpha^2)^3 = (-p)^3$
Using the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$:
$\alpha^3 + (\alpha^2)^3 + 3\alpha \cdot \alpha^2(\alpha + \alpha^2) = -p^3$
Substitute $\alpha^3 = q$ and $\alpha + \alpha^2 = -p$ into the equation:
$q + q^2 + 3q(-p) = -p^3$
Rearranging the terms,we get:
$p^3 + q^2 + q - 3pq = 0$
$p^3 + q^2 + q(1 - 3p) = 0$

(A) Given,$3$ is a root of the equation ${x^2} + ax + 3 = 0$.
Substituting $x = 3$ into the equation: ${3^2} + a(3) + 3 = 0$.
$9 + 3a + 3 = 0$ implies $3a = -12$,so $a = -4$.
Now,consider the equation ${x^2} + ax + b = 0$. Substituting $a = -4$,we get ${x^2} - 4x + b = 0$.
Let the roots of this equation be $\alpha$ and $3\alpha$ (since one root is three times the other).
From the sum of roots formula,$\alpha + 3\alpha = -(-4)/1 = 4$.
$4\alpha = 4$,which gives $\alpha = 1$.
The roots are $1$ and $3(1) = 3$.
From the product of roots formula,$\alpha \cdot 3\alpha = b/1$.
$1 \cdot 3 = b$,so $b = 3$.

(D) Given that $\alpha + \beta$,$\alpha^2 + \beta^2$,and $\alpha^3 + \beta^3$ are in $G.P.$
Therefore,$(\alpha^2 + \beta^2)^2 = (\alpha + \beta)(\alpha^3 + \beta^3)$.
Using the relations $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$,we have:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = b^2/a^2 - 2c/a = (b^2 - 2ac)/a^2$.
$\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) = (-b/a)(b^2/a^2 - 3c/a) = (-b^3 + 3abc)/a^3$.
Substituting these into the $G.P.$ condition:
$((b^2 - 2ac)/a^2)^2 = (-b/a)((-b^3 + 3abc)/a^3)$.
$(b^2 - 2ac)^2 / a^4 = (b^4 - 3abc^2)/a^4$ (Correction: $(b^4 - 3ab^2c)/a^4$ is incorrect,let's simplify: $(b^2 - 2ac)^2 = b^4 - 3ab^2c$ is wrong. Correct expansion: $b^4 + 4a^2c^2 - 4ab^2c = b^4 - 3ab^2c$).
$4a^2c^2 - ab^2c = 0$.
$ac(4ac - b^2) = 0$.
Since $a \neq 0$,we have $c(4ac - b^2) = 0$,which implies $c\Delta = 0$.

(B) Given the quadratic equation $2x^2 - (p + 1)x + (p - 1) = 0$.
Let $\alpha$ and $\beta$ be the roots of the equation.
From the relationship between roots and coefficients:
Sum of roots: $\alpha + \beta = \frac{p + 1}{2}$
Product of roots: $\alpha \beta = \frac{p - 1}{2}$
Given the condition: $\alpha - \beta = \alpha \beta$.
Squaring both sides: $(\alpha - \beta)^2 = (\alpha \beta)^2$.
We know that $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$.
Substituting the values: $(\frac{p + 1}{2})^2 - 4(\frac{p - 1}{2}) = (\frac{p - 1}{2})^2$.
$\frac{(p + 1)^2}{4} - 2(p - 1) = \frac{(p - 1)^2}{4}$.
Multiply by $4$: $(p + 1)^2 - 8(p - 1) = (p - 1)^2$.
$p^2 + 2p + 1 - 8p + 8 = p^2 - 2p + 1$.
$-6p + 9 = -2p + 1$.
$8 = 4p$.
$p = 2$.

(A) Given that $p$ and $q$ are roots of the quadratic equation $3x^2 - 5x - 2 = 0$.
From the properties of roots,the sum of roots is $p + q = -(-5)/3 = 5/3$ and the product of roots is $pq = -2/3$.
We need to find the equation whose roots are $\alpha = 3p - 2q$ and $\beta = 3q - 2p$.
Sum of roots $(\alpha + \beta) = (3p - 2q) + (3q - 2p) = p + q = 5/3$.
Product of roots $(\alpha \beta) = (3p - 2q)(3q - 2p) = 9pq - 6p^2 - 6q^2 + 4pq = 13pq - 6(p^2 + q^2)$.
Since $p^2 + q^2 = (p + q)^2 - 2pq = (5/3)^2 - 2(-2/3) = 25/9 + 4/3 = 37/9$.
Product of roots $= 13(-2/3) - 6(37/9) = -26/3 - 74/3 = -100/3$.
The required quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - (5/3)x - 100/3 = 0$,which simplifies to $3x^2 - 5x - 100 = 0$.

(A) Given the expression $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots \infty}}}$.
Since the expression is infinite,we can substitute the inner part with $x$,so $x = \sqrt{1 + x}$.
Squaring both sides,we get $x^2 = 1 + x$.
Rearranging the terms,we obtain the quadratic equation $x^2 - x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 1, b = -1, c = -1$,we get:
$x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$.
Since $x$ represents a square root of a positive sum,$x$ must be greater than $0$.
Therefore,we discard the negative root $\frac{1 - \sqrt{5}}{2}$ because it is less than $0$.
Thus,$x = \frac{1 + \sqrt{5}}{2}$.

(C) Given equation: $|x^2| + |x| - 6 = 0$.
Since $|x^2| = |x|^2$,the equation can be written as $|x|^2 + |x| - 6 = 0$.
Let $|x| = t$,where $t \ge 0$.
The equation becomes $t^2 + t - 6 = 0$.
Factoring the quadratic: $(t + 3)(t - 2) = 0$.
This gives $t = -3$ or $t = 2$.
Since $t = |x| \ge 0$,we reject $t = -3$.
Thus,$|x| = 2$,which implies $x = 2$ or $x = -2$.
The roots are $2$ and $-2$.
The sum of the roots is $2 + (-2) = 0$.

(C) The standard quadratic formula for the equation $ax^2 + bx + c = 0$ is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Let us check option $(c)$: $\frac{2c}{-b \mp \sqrt{b^2 - 4ac}}$.
Rationalizing the denominator by multiplying the numerator and denominator by $(-b \pm \sqrt{b^2 - 4ac})$:
$\frac{2c(-b \pm \sqrt{b^2 - 4ac})}{(-b \mp \sqrt{b^2 - 4ac})(-b \pm \sqrt{b^2 - 4ac})} = \frac{2c(-b \pm \sqrt{b^2 - 4ac})}{b^2 - (b^2 - 4ac)} = \frac{2c(-b \pm \sqrt{b^2 - 4ac})}{4ac} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Thus,option $(c)$ is equivalent to the standard quadratic formula.

(C) Let $\alpha$ be the common root of the equations $2x^2 + 3x + 5\lambda = 0$ and $x^2 + 2x + 3\lambda = 0$.
Then,$2\alpha^2 + 3\alpha + 5\lambda = 0$ --- $(1)$
And $\alpha^2 + 2\alpha + 3\lambda = 0$ --- $(2)$
Multiply equation $(2)$ by $2$: $2\alpha^2 + 4\alpha + 6\lambda = 0$ --- $(3)$
Subtract equation $(1)$ from equation $(3)$:
$(2\alpha^2 + 4\alpha + 6\lambda) - (2\alpha^2 + 3\alpha + 5\lambda) = 0$
$\alpha + \lambda = 0 \implies \alpha = -\lambda$
Substitute $\alpha = -\lambda$ into equation $(2)$:
$(-\lambda)^2 + 2(-\lambda) + 3\lambda = 0$
$\lambda^2 - 2\lambda + 3\lambda = 0$
$\lambda^2 + \lambda = 0$
$\lambda(\lambda + 1) = 0$
Therefore,$\lambda = 0$ or $\lambda = -1$.

(A) For the equation $x^2 + \lambda x + \mu = 0$ to have equal roots,the discriminant must be zero,i.e.,$D = \lambda^2 - 4\mu = 0$,which implies $\lambda^2 = 4\mu$.
Given that $x = 2$ is a root of the equation $x^2 + \lambda x - 12 = 0$,we substitute $x = 2$ into the equation:
$2^2 + \lambda(2) - 12 = 0$
$4 + 2\lambda - 12 = 0$
$2\lambda - 8 = 0$
$2\lambda = 8$
$\lambda = 4$
Now,substitute $\lambda = 4$ into the relation $\lambda^2 = 4\mu$:
$4^2 = 4\mu$
$16 = 4\mu$
$\mu = 4$
Therefore,$(\lambda, \mu) = (4, 4)$.