QUADRATIC EQUATION Questions in English

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

(B) Let the roots of the equation $ax^2 + bx + c = 0$ be $p\alpha$ and $q\alpha$.
According to the properties of roots:
Sum of roots: $p\alpha + q\alpha = -\frac{b}{a} \implies \alpha(p + q) = -\frac{b}{a} \implies \alpha = -\frac{b}{a(p + q)}$.
Product of roots: $(p\alpha)(q\alpha) = \frac{c}{a} \implies pq\alpha^2 = \frac{c}{a}$.
Substituting the value of $\alpha$ into the product equation:
$pq \left( -\frac{b}{a(p + q)} \right)^2 = \frac{c}{a}$
$pq \left( \frac{b^2}{a^2(p + q)^2} \right) = \frac{c}{a}$
$pqb^2 = ac(p + q)^2$
$pqb^2 - (p + q)^2ac = 0$.

(D) Given that $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$.
From the properties of roots,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Also,since $\alpha$ and $\beta$ satisfy the equation,$a\alpha^2 + b\alpha + c = 0 \implies a\alpha + b = -\frac{c}{\alpha}$ and $a\beta + b = -\frac{c}{\beta}$.
Substituting these into the expression:
$\frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b} = \frac{\alpha}{-c/\beta} + \frac{\beta}{-c/\alpha} = -\frac{\alpha\beta}{c} - \frac{\beta\alpha}{c} = -\frac{2\alpha\beta}{c}$.
Substituting $\alpha\beta = \frac{c}{a}$:
$-\frac{2(c/a)}{c} = -\frac{2}{a}$.

(C) Let $\alpha$ and $\beta$ be the two roots of the quadratic equation $ax^2 + bx + c = 0$.
According to the properties of roots,the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product of the roots is $\alpha\beta = \frac{c}{a}$.
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 $\alpha^2 + \beta^2 = (-\frac{b}{a})^2 - 2(\frac{c}{a}) = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2 - 2ac}{a^2}$.
Given that the sum of the roots is equal to the sum of their squares,we have $\alpha + \beta = \alpha^2 + \beta^2$.
Substituting the expressions,$-\frac{b}{a} = \frac{b^2 - 2ac}{a^2}$.
Multiplying both sides by $a^2$,we get $-ab = b^2 - 2ac$.
Rearranging the terms,$2ac = b^2 + ab$,which simplifies to $2ac = b(a + b)$.

(A) Given equation: $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$
Multiplying by $(x - \alpha)(x - \beta)$,we get:
$\alpha(x - \beta) + \beta(x - \alpha) = (x - \alpha)(x - \beta)$
$\alpha x - \alpha \beta + \beta x - \alpha \beta = x^2 - \alpha x - \beta x + \alpha \beta$
$(\alpha + \beta)x - 2\alpha \beta = x^2 - (\alpha + \beta)x + \alpha \beta$
Rearranging terms to form a quadratic equation:
$x^2 - 2(\alpha + \beta)x + 3\alpha \beta = 0$
Let the roots be $k$ and $-k$. For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots is $-b/a$.
Sum of roots: $k + (-k) = 0$
From the equation,the sum of roots is $2(\alpha + \beta)$.
Therefore,$2(\alpha + \beta) = 0$,which implies $\alpha + \beta = 0$.

(B) Given the quadratic equation $x^2 - 2x + 3 = 0$.
For this equation,the sum of roots $\alpha + \beta = 2$ and the product of roots $\alpha\beta = 3$.
We need to find the equation whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$.
The sum of the new roots is $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$.
Substituting the values: $\frac{2^2 - 2(3)}{3^2} = \frac{4 - 6}{9} = -\frac{2}{9}$.
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 $(\alpha - \gamma)(\beta - \gamma)(\alpha + \delta)(\beta + \delta)$.
Expanding the first part: $(\alpha - \gamma)(\beta - \gamma) = \alpha \beta - \gamma(\alpha + \beta) + \gamma^2 = 1 + p\gamma + \gamma^2$.
Since $\gamma$ is a root of $x^2 + qx + 1 = 0$,we have $\gamma^2 + q\gamma + 1 = 0$,so $\gamma^2 + 1 = -q\gamma$.
Thus,$1 + p\gamma + \gamma^2 = -q\gamma + p\gamma = \gamma(p - q)$.
Similarly,expanding the second part: $(\alpha + \delta)(\beta + \delta) = \alpha \beta + \delta(\alpha + \beta) + \delta^2 = 1 - p\delta + \delta^2$.
Since $\delta$ is a root of $x^2 + qx + 1 = 0$,we have $\delta^2 + q\delta + 1 = 0$,so $\delta^2 + 1 = -q\delta$.
Thus,$1 - p\delta + \delta^2 = -q\delta - p\delta = -\delta(p + q)$.
Multiplying these results: $\gamma(p - q) \cdot [-\delta(p + q)] = -\gamma \delta (p^2 - q^2) = -1 \cdot (p^2 - q^2) = q^2 - p^2$.

(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 $\alpha'\beta' = q'$.
We need to evaluate the expression $E = (\alpha - \alpha')^2 + (\beta - \alpha')^2 + (\alpha - \beta')^2 + (\beta - \beta')^2$.
Expanding the terms: $E = (\alpha^2 - 2\alpha\alpha' + \alpha'^2) + (\beta^2 - 2\beta\alpha' + \alpha'^2) + (\alpha^2 - 2\alpha\beta' + \beta'^2) + (\beta^2 - 2\beta\beta' + \beta'^2)$.
Grouping the terms: $E = 2(\alpha^2 + \beta^2) + 2(\alpha'^2 + \beta'^2) - 2\alpha'(\alpha + \beta) - 2\beta'(\alpha + \beta)$.
Using the identity $x^2 + y^2 = (x+y)^2 - 2xy$,we get:
$E = 2[(\alpha + \beta)^2 - 2\alpha\beta] + 2[(\alpha' + \beta')^2 - 2\alpha'\beta'] - 2(\alpha' + \beta')(\alpha + \beta)$.
Substituting the values: $E = 2[p^2 - 2q] + 2[p'^2 - 2q'] - 2(p')(p)$.
$E = 2\{p^2 - 2q + p'^2 - 2q' - pp'\}$.

(A) Let the roots be $\alpha$ and $\alpha^2$.
From the properties of quadratic equations,the sum of roots is $\alpha + \alpha^2 = -\frac{b}{a}$ $(i)$ and the product of roots is $\alpha \cdot \alpha^2 = \alpha^3 = \frac{c}{a}$ $(ii)$.
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 $\alpha + \alpha^2 = -\frac{b}{a}$ into the equation:
$\frac{c}{a} + (\frac{c}{a})^2 + 3(\frac{c}{a})(-\frac{b}{a}) = -\frac{b^3}{a^3}$
$\frac{c}{a} + \frac{c^2}{a^2} - \frac{3bc}{a^2} = -\frac{b^3}{a^3}$
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 Arithmetic Mean $(A.M.)$ of the roots is given by $A = \frac{\alpha + \beta}{2}$,which implies $\alpha + \beta = 2A$.
The Geometric Mean $(G.M.)$ 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 the formula $t^2 - (\text{sum of roots})t + (\text{product of roots}) = 0$.
Substituting the values,we get $t^2 - (2A)t + G^2 = 0$,or $t^2 - 2At + G^2 = 0$.

(C) Given that $\alpha$ and $\beta$ are the roots of the equation $(x - a)(x - b) = c$.
Expanding this,we get $x^2 - (a + b)x + ab = c$,or $x^2 - (a + b)x + (ab - c) = 0$.
From the properties of quadratic equations,the sum of the roots is $\alpha + \beta = a + b$ and the product of the roots is $\alpha \beta = ab - c$.
We need to find the roots of the equation $(x - \alpha)(x - \beta) + c = 0$.
Expanding this,we get $x^2 - (\alpha + \beta)x + \alpha \beta + c = 0$.
Substituting the values $\alpha + \beta = a + b$ and $\alpha \beta = ab - c$ into the equation,we get:
$x^2 - (a + b)x + (ab - c) + c = 0$
$x^2 - (a + b)x + ab = 0$
This is equivalent to $(x - a)(x - b) = 0$.
Therefore,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 quadratic equations,the sum of the roots is $\alpha + \beta = p$ and the product of the roots is $\alpha \beta = 8$.
We are given that the difference of the roots is $|\alpha - \beta| = 2$,which implies $(\alpha - \beta)^2 = 4$.
Using the identity $(\alpha + \beta)^2 - (\alpha - \beta)^2 = 4\alpha \beta$,we substitute the known values:
$p^2 - 2^2 = 4(8)$
$p^2 - 4 = 32$
$p^2 = 36$
$p = \pm 6$.

(C) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$.
Then,$\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.
Given that the sum of the roots is equal to the sum of the squares of their reciprocals:
$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$.
Substituting the values:
$-b/a = \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} = \frac{b^2 - 2ac}{c^2}$.
$-b/a = \frac{b^2 - 2ac}{c^2} \implies -bc^2 = ab^2 - 2a^2c$.
Rearranging gives $2a^2c = ab^2 + bc^2$.
Dividing both sides by $abc$,we get $\frac{2a}{b} = \frac{b}{c} + \frac{c}{a}$.
This implies that $\frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in $A.P.$
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 properties of roots,the sum of roots $\alpha + \beta = -\frac{2b}{a}$ and the product of roots $\alpha \beta = \frac{c}{a}$.
We need to evaluate the expression $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}$.
Taking the common denominator,we get $\frac{\sqrt{\alpha}}{\sqrt{\beta}} + \frac{\sqrt{\beta}}{\sqrt{\alpha}} = \frac{(\sqrt{\alpha})^2 + (\sqrt{\beta})^2}{\sqrt{\alpha \beta}} = \frac{\alpha + \beta}{\sqrt{\alpha \beta}}$.
Substituting the values of $\alpha + \beta$ and $\alpha \beta$:
$= \frac{-2b/a}{\sqrt{c/a}} = \frac{-2b/a}{\sqrt{c}/\sqrt{a}} = -\frac{2b}{a} \times \frac{\sqrt{a}}{\sqrt{c}} = -\frac{2b}{\sqrt{a} \cdot \sqrt{a}} \times \frac{\sqrt{a}}{\sqrt{c}} = -\frac{2b}{\sqrt{ac}}$.

(A) For a quadratic equation with real coefficients,complex roots always occur in conjugate pairs.
Given one root is $\alpha = 7 + 5i$,the other root must be its conjugate $\beta = 7 - 5i$.
The quadratic equation is given by the formula: $x^2 - (\alpha + \beta)x + (\alpha \beta) = 0$.
First,calculate the sum of the roots: $\alpha + \beta = (7 + 5i) + (7 - 5i) = 14$.
Next,calculate the product of the roots: $\alpha \beta = (7 + 5i)(7 - 5i) = 7^2 - (5i)^2 = 49 - 25i^2 = 49 + 25 = 74$.
Substituting these values into the formula,we get: $x^2 - 14x + 74 = 0$.

(B) The given equation is $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.
Multiplying by $r(x+p)(x+q)$,we get $r(x+q) + r(x+p) = (x+p)(x+q)$.
$rx + rq + rx + rp = x^2 + px + qx + pq$.
Rearranging the terms to form a quadratic equation: $x^2 + x(p + q - 2r) + (pq - pr - qr) = 0$.
Let the roots be $\alpha$ and $-\alpha$ as they are equal in magnitude but opposite in sign.
The sum of the roots is $\alpha + (-\alpha) = 0$.
From the quadratic equation,the sum of roots is $-(p + q - 2r) = 0$,which implies $p + q - 2r = 0$,or $r = \frac{p + q}{2}$.
The product of the roots is $\alpha \cdot (-\alpha) = -\alpha^2$.
From the quadratic equation,the product of roots is $pq - r(p + q)$.
Substituting $r = \frac{p + q}{2}$ into the product expression:
Product $= pq - \frac{p + q}{2}(p + q) = pq - \frac{(p + q)^2}{2} = \frac{2pq - (p^2 + 2pq + q^2)}{2} = \frac{2pq - p^2 - 2pq - q^2}{2} = -\frac{p^2 + q^2}{2}$.

(A) Given the quadratic equation $ax^2 + bx + c = 0$.
Let the roots of the equation be $\alpha$ and $\frac{1}{\alpha}$ because they are reciprocal to each other.
According to the properties of quadratic equations,the product of the roots is given by the ratio of the constant term to the coefficient of $x^2$.
Product of roots = $\alpha \cdot \frac{1}{\alpha} = \frac{c}{a}$.
Since $\alpha \cdot \frac{1}{\alpha} = 1$,we have $1 = \frac{c}{a}$.
This implies $a = c$,or $a - c = 0$.

(B) Given that the first root is $\alpha = 2 - \sqrt{3}$.
Since the coefficients of the quadratic equation are assumed to be rational,the irrational roots must occur in conjugate pairs.
Therefore,the second root is $\beta = 2 + \sqrt{3}$.
The sum of the roots is $\alpha + \beta = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$.
The product of the roots is $\alpha \cdot \beta = (2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.
The standard form of a quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - 4x + 1 = 0$.

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

(A) Given one root $\alpha = \frac{1}{2 + \sqrt{5}}$.
Rationalizing the denominator: $\alpha = \frac{1(2 - \sqrt{5})}{(2 + \sqrt{5})(2 - \sqrt{5})} = \frac{2 - \sqrt{5}}{4 - 5} = \frac{2 - \sqrt{5}}{-1} = \sqrt{5} - 2$.
Since the coefficients of the quadratic equation are typically rational,the other root $\beta$ must be the conjugate of $\alpha$,which is $\beta = -\sqrt{5} - 2$.
Sum of roots $\alpha + \beta = (\sqrt{5} - 2) + (-\sqrt{5} - 2) = -4$.
Product of roots $\alpha \beta = (\sqrt{5} - 2)(-\sqrt{5} - 2) = -(\sqrt{5} - 2)(\sqrt{5} + 2) = -(5 - 4) = -1$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values: $x^2 - (-4)x + (-1) = 0$,which simplifies to $x^2 + 4x - 1 = 0$.

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + x + 1 = 0$.
From the properties of roots,$\alpha + \beta = -1$ and $\alpha \beta = 1$.
Now,$\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ are the roots of $x^2 + px + q = 0$.
The sum of the roots is given by $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -p$.
Simplifying the sum: $\frac{\alpha^2 + \beta^2}{\alpha \beta} = -p$.
Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$,we get:
$\frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} = -p$.
Substituting the values $\alpha + \beta = -1$ and $\alpha \beta = 1$:
$\frac{(-1)^2 - 2(1)}{1} = -p$.
$\frac{1 - 2}{1} = -p$.
$-1 = -p$,which implies $p = 1$.

(C) For the quadratic equation $x^2 + ax + b = 0$,the sum of the roots is $\alpha + \beta = -a$ and the product of the roots is $\alpha\beta = b$.
We use the algebraic identity for the sum of cubes: $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)$.
This can be rewritten in terms of the sum and product of roots as: $\alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta]$.
Substituting the values $\alpha + \beta = -a$ and $\alpha\beta = b$:
$\alpha^3 + \beta^3 = (-a)[(-a)^2 - 3(b)]$
$= (-a)(a^2 - 3b)$
$= -a^3 + 3ab$.

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

(A) For a quadratic equation $ax^2 + bx + c = 0$ to have equal roots,the discriminant $D$ must be equal to $0$.
Here,$a = 1$,$b = 2m$,and $c = m^2 - 2m + 6$.
The discriminant $D = b^2 - 4ac = 0$.
Substituting the values: $(2m)^2 - 4(1)(m^2 - 2m + 6) = 0$.
$4m^2 - 4m^2 + 8m - 24 = 0$.
$8m - 24 = 0$.
$8m = 24$.
$m = 3$.

(B) For a quadratic equation of the form $ax^2 + bx + c = 0$,if the roots are reciprocal to each other,then their product is equal to $1$.
Given the equation $(2k + 1)x^2 - (7k + 3)x + k + 2 = 0$,we have $a = 2k + 1$ and $c = k + 2$.
The product of the roots is given by $\frac{c}{a}$.
Since the roots are reciprocals,$\frac{c}{a} = 1$.
Therefore,$\frac{k + 2}{2k + 1} = 1$.
Multiplying both sides by $(2k + 1)$,we get $k + 2 = 2k + 1$.
Rearranging the terms,$2 - 1 = 2k - k$,which gives $k = 1$.

(B) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $l$ and $2l$.
Using the relationship between roots and coefficients:
Sum of roots: $l + 2l = -\frac{b}{a} \Rightarrow 3l = -\frac{b}{a} \Rightarrow l = -\frac{b}{3a}$.....$(i)$
Product of roots: $l \cdot 2l = \frac{c}{a} \Rightarrow 2l^2 = \frac{c}{a}$.....$(ii)$
Substitute $(i)$ into $(ii)$:
$2\left(-\frac{b}{3a}\right)^2 = \frac{c}{a}$
$2\left(\frac{b^2}{9a^2}\right) = \frac{c}{a}$
$\frac{2b^2}{9a^2} = \frac{c}{a}$
$2b^2 = 9ac$.

(D) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
Given: $\alpha + \beta = 2$ and $\alpha^3 + \beta^3 = 98$.
We know the 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 given values: $98 = 2[2^2 - 3\alpha\beta]$.
$49 = 4 - 3\alpha\beta$.
$3\alpha\beta = 4 - 49 = -45$.
$\alpha\beta = -15$.
The standard form of a quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values: $x^2 - 2x - 15 = 0$.

(A) For the quadratic equation $ax^2 + bx + c = 0$,the sum and product of roots are given by:
$\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
We need to find the value of $\alpha\beta^2 + \alpha^2\beta + \alpha\beta$.
Factor out $\alpha\beta$ from the expression:
$\alpha\beta^2 + \alpha^2\beta + \alpha\beta = \alpha\beta(\beta + \alpha + 1)$.
Substitute the values of $(\alpha + \beta)$ and $(\alpha\beta)$:
$= \left(\frac{c}{a}\right) \left(-\frac{b}{a} + 1\right)$
$= \left(\frac{c}{a}\right) \left(\frac{a - b}{a}\right)$
$= \frac{c(a - b)}{a^2}$.

(C) For a quadratic equation of the form $ax^2 + bx + c = 0$,the product of the roots is given by the formula $\frac{c}{a}$.
In the given equation $mx^2 + 6x + (2m - 1) = 0$,we have $a = m$,$b = 6$,and $c = 2m - 1$.
According to the problem,the product of the roots is $-1$.
Therefore,$\frac{2m - 1}{m} = -1$.
Multiplying both sides by $m$,we get $2m - 1 = -m$.
Adding $m$ to both sides,we get $3m - 1 = 0$.
Adding $1$ to both sides,we get $3m = 1$.
Dividing by $3$,we get $m = \frac{1}{3}$.

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

(A) Let the roots be $\alpha = \frac{1}{3 + \sqrt{2}}$ and $\beta = \frac{1}{3 - \sqrt{2}}$.
Rationalizing the roots:
$\alpha = \frac{3 - \sqrt{2}}{(3 + \sqrt{2})(3 - \sqrt{2})} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}$
$\beta = \frac{3 + \sqrt{2}}{(3 - \sqrt{2})(3 + \sqrt{2})} = \frac{3 + \sqrt{2}}{9 - 2} = \frac{3 + \sqrt{2}}{7}$
Sum of roots $(\alpha + \beta) = \frac{3 - \sqrt{2} + 3 + \sqrt{2}}{7} = \frac{6}{7}$
Product of roots $(\alpha \beta) = \frac{(3 - \sqrt{2})(3 + \sqrt{2})}{7 \times 7} = \frac{9 - 2}{49} = \frac{7}{49} = \frac{1}{7}$
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - \frac{6}{7}x + \frac{1}{7} = 0$
Multiplying by $7$,we get $7x^2 - 6x + 1 = 0$.

(B) For the quadratic equation $x^2 - 4x + 1 = 0$,the sum of the roots is $\alpha + \beta = -(-4)/1 = 4$ and the product of the roots is $\alpha \beta = 1/1 = 1$.
Using the algebraic identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$,
Substitute the values: $\alpha^3 + \beta^3 = (4)^3 - 3(1)(4)$.
$\alpha^3 + \beta^3 = 64 - 12 = 52$.

(B) Let the two-digit number be represented as $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
According to the problem,the number is four times the sum of its digits:
$10x + y = 4(x + y)$
$10x + y = 4x + 4y$
$6x = 3y$
$y = 2x$ (Equation $1$)
Also,the number is three times the product of its digits:
$10x + y = 3xy$ (Equation $2$)
Substitute $y = 2x$ into Equation $2$:
$10x + 2x = 3x(2x)$
$12x = 6x^2$
Since $x$ cannot be $0$ for a two-digit number,we divide by $6x$:
$x = 2$
Now,find $y$ using Equation $1$:
$y = 2(2) = 4$
Therefore,the number is $10(2) + 4 = 24$.

(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,it satisfies the equation: $2\alpha^2 - 35\alpha + 2 = 0$,which implies $2\alpha^2 - 35\alpha = -2$.
Dividing by $\alpha$ (assuming $\alpha \neq 0$),we get $2\alpha - 35 = -\frac{2}{\alpha}$.
Similarly,for root $\beta$,we have $2\beta - 35 = -\frac{2}{\beta}$.
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 $\frac{64}{(1)^3} = 64$.

(C) Given the equation $x^2 + x + 1 = 0$.
Since $\alpha$ and $\alpha^2$ are roots,we know $\alpha + \alpha^2 = -1$ and $\alpha^3 = 1$.
We need to find the equation with roots $\alpha^{31}$ and $\alpha^{62}$.
Sum of roots: $\alpha^{31} + \alpha^{62} = \alpha^{31}(1 + \alpha^{31}) = (\alpha^3)^{10} \cdot \alpha (1 + (\alpha^3)^{10} \cdot \alpha) = \alpha(1 + \alpha) = \alpha + \alpha^2 = -1$.
Product of roots: $\alpha^{31} \cdot \alpha^{62} = \alpha^{93} = (\alpha^3)^{31} = 1^{31} = 1$.
The required quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values: $x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.
Alternatively,since $\alpha$ is a complex cube root of unity $\omega$,then $\alpha^{31} = \omega^{31} = \omega$ and $\alpha^{62} = \omega^{62} = \omega^2$. The equation with roots $\omega$ and $\omega^2$ is $x^2 + x + 1 = 0$.

(B) Given that $p$ and $q$ satisfy the equation $3x^2 = 5x + 2$.
Rearranging the terms,we get the quadratic equation $3x^2 - 5x - 2 = 0$.
Since $p$ and $q$ are the roots of this quadratic equation and $p \neq q$,we can use the relationship between the roots and coefficients of a quadratic equation $ax^2 + bx + c = 0$.
The product of the roots $pq$ is given by the formula $c/a$.
Here,$a = 3$,$b = -5$,and $c = -2$.
Therefore,$pq = -2/3$.

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + bx - c = 0$.
From the properties of roots,we have:
Sum of roots: $\alpha + \beta = -b \implies b = -(\alpha + \beta)$
Product of roots: $\alpha \beta = -c \implies c = -\alpha \beta$
We need to find the equation whose roots are $b$ and $c$.
Sum of new roots: $b + c = -(\alpha + \beta) - \alpha \beta = -[(\alpha + \beta) + \alpha \beta]$
Product of new roots: $bc = [ -(\alpha + \beta) ] \times [ -\alpha \beta ] = (\alpha + \beta)(\alpha \beta)$
The required quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
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 $\alpha + \beta = -b/a$ and $\alpha \beta = c/a$.
Expanding the expression $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$:
$= 1 + (\alpha + \beta) + (\alpha^2 + \beta^2) + \alpha \beta + \alpha \beta (\alpha + \beta) + \alpha^2 \beta^2$
$= 1 + (\alpha + \beta) + ((\alpha + \beta)^2 - 2\alpha \beta) + \alpha \beta + \alpha \beta (\alpha + \beta) + (\alpha \beta)^2$
$= 1 + (\alpha + \beta) + (\alpha + \beta)^2 - \alpha \beta + \alpha \beta (\alpha + \beta) + (\alpha \beta)^2$
Substituting the values:
$= 1 - \frac{b}{a} + \frac{b^2}{a^2} - \frac{c}{a} + \left(\frac{c}{a}\right)\left(-\frac{b}{a}\right) + \frac{c^2}{a^2}$
$= \frac{a^2 - ab + b^2 - ac - bc + c^2}{a^2} = \frac{a^2 + b^2 + c^2 - ab - bc - ca}{a^2}$
$= \frac{1}{2a^2} [(a - b)^2 + (b - c)^2 + (c - a)^2]$
Since $a, b, c$ are distinct,the sum of squares is always positive. Thus,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 sum of roots: $x_1 + x_2 = \frac{\alpha}{\alpha - 1} + \frac{\alpha + 1}{\alpha} = \frac{\alpha^2 + \alpha^2 - 1}{\alpha(\alpha - 1)} = \frac{2\alpha^2 - 1}{\alpha^2 - \alpha} = -\frac{b}{a}$.
From the product of roots: $x_1 x_2 = \frac{\alpha}{\alpha - 1} \cdot \frac{\alpha + 1}{\alpha} = \frac{\alpha + 1}{\alpha - 1} = \frac{c}{a}$.
From $\frac{\alpha + 1}{\alpha - 1} = \frac{c}{a}$,we apply componendo and dividendo: $\frac{(\alpha + 1) + (\alpha - 1)}{(\alpha + 1) - (\alpha - 1)} = \frac{c + a}{c - a} \Rightarrow \frac{2\alpha}{2} = \frac{c + a}{c - a} \Rightarrow \alpha = \frac{c + a}{c - a}$.
Also,$\alpha - 1 = \frac{c + a}{c - a} - 1 = \frac{2a}{c - a}$.
Substituting these into the sum of roots equation: $\frac{2\alpha^2 - 1}{\alpha(\alpha - 1)} = -\frac{b}{a}$.
After algebraic simplification,we find that $(a + b + c)^2 = b^2 - 4ac$.

(B) Let the roots of the equation $ax^2 + 2bx + c = 0$ be $\alpha$ and $\beta$ such that $\frac{\alpha}{\beta} = \frac{m}{n}$.
From the properties of quadratic equations,the sum of roots is $\alpha + \beta = -\frac{2b}{a}$ and the product of roots is $\alpha\beta = \frac{c}{a}$.
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 $\frac{(-2b/a)^2}{c/a} = \frac{4b^2/a^2}{c/a} = \frac{4b^2}{ac} = \frac{(m+n)^2}{mn}$.
Thus,$\frac{b^2}{ac} = \frac{(m+n)^2}{4mn}$.
Similarly,for the equation $px^2 + 2qx + r = 0$,if the ratio of roots is $\frac{m}{n}$,then $\frac{q^2}{pr} = \frac{(m+n)^2}{4mn}$.
Equating the two expressions,we get $\frac{b^2}{ac} = \frac{q^2}{pr}$.

(C) Given the quadratic equation $x^2 + bx - c = 0$ with $b, c > 0$.
Let the roots of the equation be $\alpha$ and $\beta$.
According to the properties of quadratic equations,the sum of the roots is $\alpha + \beta = -b$ and the product of the roots is $\alpha \beta = -c$.
Since $b > 0$,it follows that $\alpha + \beta = -b < 0$.
Since $c > 0$,it follows that $\alpha \beta = -c < 0$.
If the product of the roots $(\alpha \beta)$ is negative,it implies that one root must be positive and the other must be negative.
Therefore,the roots are of opposite sign.

(B) The given quadratic equation is ${x^2} - (p + 1)x + pq = 0$.
Since $p$ and $q$ are the roots of this equation,we can use the relationship between roots and coefficients.
According to Vieta's formulas,the sum of the roots is equal to the negative of the coefficient of $x$ divided by the coefficient of ${x^2}$.
Therefore,$p + q = -(-(p + 1)) / 1$.
This simplifies to $p + q = p + 1$.
Subtracting $p$ from both sides,we get $q = 1$.

(C) For the equation $ax^2 + bx + c = 0$,the discriminant is $D_1 = b^2 - 4ac$. The difference of the roots is given by $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \left( -\frac{b}{a} \right)^2 - 4\left( \frac{c}{a} \right) = \frac{b^2 - 4ac}{a^2}$.
For the equation $Ax^2 + Bx + C = 0$,the roots are $\alpha - k$ and $\beta - k$. The difference of these roots is $(\alpha - k) - (\beta - k) = \alpha - \beta$. Thus,the square of the difference of the roots is $(\alpha - \beta)^2$.
Using the coefficients of the second equation,the square of the difference of the roots is $\frac{B^2 - 4AC}{A^2}$.
Equating the two expressions for $(\alpha - \beta)^2$,we get $\frac{b^2 - 4ac}{a^2} = \frac{B^2 - 4AC}{A^2}$.
Rearranging the terms,we find $\frac{B^2 - 4AC}{b^2 - 4ac} = \frac{A^2}{a^2} = \left( \frac{A}{a} \right)^2$.

(A) For the quadratic equation $x^2 + px + q = 0$,the sum of the roots is given by $p + q = -p$ and the product of the roots is given by $pq = q$.
From the product of roots equation: $pq - q = 0 \Rightarrow q(p - 1) = 0$.
This implies either $q = 0$ or $p = 1$.
Case $1$: If $q = 0$,then the sum of roots equation $p + q = -p$ becomes $p + 0 = -p$,which implies $2p = 0$,so $p = 0$. Thus,$(p, q) = (0, 0)$.
Case $2$: If $p = 1$,then the sum of roots equation $p + q = -p$ becomes $1 + q = -1$,which implies $q = -2$. Thus,$(p, q) = (1, -2)$.
Comparing with the given options,the correct pair is $p = 1, q = -2$.

(A) Given the quadratic equation $ix^2 - 2(i + 1)x + (2 - i) = 0$.
Let the roots be $\alpha$ and $\beta$. We are given $\beta = 2 - i$.
The product of the roots for a quadratic equation $ax^2 + bx + c = 0$ is given by $\alpha \cdot \beta = \frac{c}{a}$.
Here,$a = i$ and $c = 2 - i$.
Therefore,$\alpha \cdot (2 - i) = \frac{2 - i}{i}$.
Dividing both sides by $(2 - i)$,we get $\alpha = \frac{1}{i}$.
Since $\frac{1}{i} = -i$,the other root is $\alpha = -i$.

(A) For a quadratic equation of the form $ax^2 + bx + c = 0$,if the roots are reciprocals of each other,then their product is equal to $1$.
Let the roots be $\alpha$ and $1/\alpha$.
The product of the roots is given by the formula $\frac{c}{a}$.
Here,$a = 5$,$b = -7$,and $c = k$.
Therefore,$\alpha \cdot \frac{1}{\alpha} = \frac{k}{5}$.
$1 = \frac{k}{5}$.
Multiplying both sides by $5$,we get $k = 5$.

(B) For a quadratic equation of the form $ax^2 + bx + c = 0$,the sum of the roots is $\alpha + \beta = -b/a$ and the product of the roots is $\alpha\beta = c/a$.
Given the equation $x^2 - 7x + 6 = 0$,we have $a = 1$,$b = -7$,and $c = 6$.
Therefore,the sum of the roots is $\alpha + \beta = -(-7)/1 = 7$.
The product of the roots is $\alpha\beta = 6/1 = 6$.
We need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
By taking the common denominator,we get $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$.
Substituting the values,we get $\frac{7}{6}$.

(B) Given the quadratic equation $x^2 - 2x + 4 = 0$.
For this equation,the sum of roots $\alpha + \beta = 2$ and the product of roots $\alpha \beta = 4$.
We know that $\alpha$ and $\beta$ satisfy the equation,so $\alpha^2 = 2\alpha - 4$ and $\beta^2 = 2\beta - 4$.
Using the recurrence relation $S_n = \alpha^n + \beta^n$,we have $S_n - 2S_{n-1} + 4S_{n-2} = 0$.
For $n=1$,$S_1 = \alpha + \beta = 2$.
For $n=2$,$S_2 = (\alpha + \beta)^2 - 2\alpha \beta = (2)^2 - 2(4) = 4 - 8 = -4$.
For $n=3$,$S_3 = 2S_2 - 4S_1 = 2(-4) - 4(2) = -8 - 8 = -16$.
For $n=4$,$S_4 = 2S_3 - 4S_2 = 2(-16) - 4(-4) = -32 + 16 = -16$.
For $n=5$,$S_5 = 2S_4 - 4S_3 = 2(-16) - 4(-16) = -32 + 64 = 32$.
Thus,$\alpha^5 + \beta^5 = 32$.

(A) Given equations are $a(p + q)^2 + 2bpq + c = 0$ and $a(p + r)^2 + 2bpr + c = 0$.
These equations imply that $q$ and $r$ are the roots of the quadratic equation in $x$:
$a(p + x)^2 + 2bpx + c = 0$.
Expanding the equation: $a(p^2 + 2px + x^2) + 2bpx + c = 0$.
$ax^2 + 2apx + ap^2 + 2bpx + c = 0$.
Rearranging in terms of $x$: $ax^2 + 2x(ap + bp) + (ap^2 + c) = 0$.
$ax^2 + 2px(a + b) + (ap^2 + c) = 0$.
For a quadratic equation $Ax^2 + Bx + C = 0$,the product of roots is given by $\frac{C}{A}$.
Here,$A = a$,$B = 2p(a + b)$,and $C = ap^2 + c$.
Therefore,the product of roots $qr = \frac{ap^2 + c}{a} = p^2 + \frac{c}{a}$.

(D) Given the quadratic equation $(a + b - 2c)x^2 - (2a - b - c)x + (a - 2b + c) = 0$.
Let the coefficients be $A = (a + b - 2c)$,$B = -(2a - b - c)$,and $C = (a - 2b + c)$.
Calculate the sum of the coefficients: $A + B + C = (a + b - 2c) - (2a - b - c) + (a - 2b + c) = a + b - 2c - 2a + b + c + a - 2b + c = (a - 2a + a) + (b + b - 2b) + (-2c + c + c) = 0 + 0 + 0 = 0$.
Since the sum of the coefficients is $0$,one root of the equation must be $x = 1$.
Let the other root be $\alpha$. The product of the roots is given by $\frac{C}{A} = \frac{a - 2b + c}{a + b - 2c}$.
Since $1 \cdot \alpha = \frac{a - 2b + c}{a + b - 2c}$,the roots are $1$ and $\frac{a - 2b + c}{a + b - 2c}$.
Comparing this with the given options,none of the options $A$,$B$,or $C$ match the result. Therefore,the correct answer is $D$.