Solution of quadratic equations and Nature of roots Questions in English

(C) To find the root,we substitute the given options into the equation $f(x) = 2x^5 - 14x^4 + 31x^3 - 64x^2 + 19x + 130 = 0$.
For $x = 5$:
$f(5) = 2(5)^5 - 14(5)^4 + 31(5)^3 - 64(5)^2 + 19(5) + 130$
$f(5) = 2(3125) - 14(625) + 31(125) - 64(25) + 95 + 130$
$f(5) = 6250 - 8750 + 3875 - 1600 + 95 + 130$
$f(5) = (6250 + 3875 + 95 + 130) - (8750 + 1600)$
$f(5) = 10350 - 10350 = 0$
Since $f(5) = 0$,$x = 5$ is a root of the equation.

(B) Given equation is $x^3 - 3x + 2 = 0$.
By testing $x = 1$,we get $1^3 - 3(1) + 2 = 1 - 3 + 2 = 0$.
So,$(x - 1)$ is a factor.
Dividing $x^3 - 3x + 2$ by $(x - 1)$,we get $x^2(x - 1) + x(x - 1) - 2(x - 1) = 0$.
$(x - 1)(x^2 + x - 2) = 0$.
Factoring the quadratic part: $(x - 1)(x + 2)(x - 1) = 0$.
Thus,$(x - 1)^2(x + 2) = 0$.
The roots are $1, 1, -2$.

(A) Given equation is $x^4 - 2x^3 + x - 380 = 0$.
By testing integer roots,we find that $x = 5$ and $x = -4$ are roots.
Dividing the polynomial $x^4 - 2x^3 + x - 380$ by $(x - 5)(x + 4) = x^2 - x - 20$,we get the quotient $x^2 - x + 19$.
So,$(x - 5)(x + 4)(x^2 - x + 19) = 0$.
Setting $x^2 - x + 19 = 0$,we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
$x = \frac{1 \pm \sqrt{1 - 4(19)}}{2} = \frac{1 \pm \sqrt{-75}}{2} = \frac{1 \pm 5\sqrt{-3}}{2}$.
Thus,the roots are $5, -4, \frac{1 \pm 5\sqrt{-3}}{2}$.

(D) Given the quadratic equation $pqx^2 - (p + q)^2x + (p + q)^2 = 0$.
Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = pq$,$b = -(p + q)^2$,and $c = (p + q)^2$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
Discriminant $D = b^2 - 4ac = (-(p + q)^2)^2 - 4(pq)(p + q)^2 = (p + q)^4 - 4pq(p + q)^2$.
$D = (p + q)^2 [(p + q)^2 - 4pq] = (p + q)^2 [p^2 + 2pq + q^2 - 4pq] = (p + q)^2 (p - q)^2$.
So,$\sqrt{D} = (p + q)(p - q) = p^2 - q^2$.
$x = \frac{(p + q)^2 \pm (p^2 - q^2)}{2pq}$.
Case $1$: $x = \frac{p^2 + 2pq + q^2 + p^2 - q^2}{2pq} = \frac{2p^2 + 2pq}{2pq} = \frac{2p(p + q)}{2pq} = \frac{p + q}{q}$.
Case $2$: $x = \frac{p^2 + 2pq + q^2 - p^2 + q^2}{2pq} = \frac{2pq + 2q^2}{2pq} = \frac{2q(p + q)}{2pq} = \frac{p + q}{p}$.
Thus,the solution set is $\left\{ \frac{p + q}{p}, \frac{p + q}{q} \right\}$.

(A) Given $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots \infty}}}$.
Since the expression is infinite,we can write $x = \sqrt{1 + x}$.
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 $a = 1, b = -1, c = -1$:
$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$.
Since $x$ represents a square root,$x$ must be positive $(x > 0)$.
Therefore,we discard the negative root $\frac{1 - \sqrt{5}}{2}$.
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 $t = |x|$,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 discard $t = -3$.
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 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:
$\frac{2c}{-b - \sqrt{b^2 - 4ac}} \times \frac{-b + \sqrt{b^2 - 4ac}}{-b + \sqrt{b^2 - 4ac}} = \frac{2c(-b + \sqrt{b^2 - 4ac})}{b^2 - (b^2 - 4ac)} = \frac{2c(-b + \sqrt{b^2 - 4ac})}{4ac} = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$.
Similarly,$\frac{2c}{-b + \sqrt{b^2 - 4ac}} = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$.
Thus,the expression in option $(C)$ is equivalent to the standard quadratic formula.

(A) Given that the equation $x^2 + \lambda x + \mu = 0$ has equal roots,the discriminant must be zero: $\lambda^2 - 4\mu = 0$,which implies $\lambda^2 = 4\mu$.
For the second equation $x^2 + \lambda x - 12 = 0$,$x = 2$ is a root.
Substituting $x = 2$ into the equation: $2^2 + \lambda(2) - 12 = 0$.
$4 + 2\lambda - 12 = 0$.
$2\lambda = 8$,so $\lambda = 4$.
Now,substitute $\lambda = 4$ into the first relation $\lambda^2 = 4\mu$:
$4^2 = 4\mu$.
$16 = 4\mu$,so $\mu = 4$.
Thus,$(\lambda, \mu) = (4, 4)$.

(A) Let $f(x) = (x - a)(x - c) + 2(x - b)(x - d)$.
Since $f(x)$ is a quadratic polynomial with a positive leading coefficient,we evaluate $f(x)$ at $x = a, b, c, d$:
$f(a) = (a - a)(a - c) + 2(a - b)(a - d) = 2(a - b)(a - d)$. Since $a < b$ and $a < d$,$(a - b) < 0$ and $(a - d) < 0$,so $f(a) > 0$.
$f(b) = (b - a)(b - c) + 2(b - b)(b - d) = (b - a)(b - c)$. Since $b > a$ and $b < c$,$(b - a) > 0$ and $(b - c) < 0$,so $f(b) < 0$.
$f(c) = (c - a)(c - c) + 2(c - b)(c - d) = 2(c - b)(c - d)$. Since $c > b$ and $c < d$,$(c - b) > 0$ and $(c - d) < 0$,so $f(c) < 0$.
$f(d) = (d - a)(d - c) + 2(d - b)(d - d) = (d - a)(d - c)$. Since $d > a$ and $d > c$,$(d - a) > 0$ and $(d - c) > 0$,so $f(d) > 0$.
Since $f(a) > 0$ and $f(b) < 0$,there exists a root in $(a, b)$.
Since $f(c) < 0$ and $f(d) > 0$,there exists a root in $(c, d)$.
Thus,the equation has two real and distinct roots.

(A) The given equation is $qx^2 + px + q = 0$. Since the roots are complex,the discriminant $D_1 = p^2 - 4q^2 < 0$,which implies $p^2 < 4q^2$.
Now,consider the equation $x^2 - 4qx + p^2 = 0$. The discriminant $D_2$ of this equation is given by $D_2 = (-4q)^2 - 4(1)(p^2) = 16q^2 - 4p^2$.
We can rewrite this as $D_2 = 4(4q^2 - p^2)$.
Since $p^2 < 4q^2$,it follows that $4q^2 - p^2 > 0$,and therefore $D_2 > 0$.
Since the discriminant $D_2$ is positive,the roots of the equation $x^2 - 4qx + p^2 = 0$ are real and unequal.

(D) For a quadratic expression $Ax^2 + Bx + C$ to be positive for all $x$,we must have $A > 0$ and the discriminant $D = B^2 - 4AC < 0$.
Here,$A = a^2 - 1$,$B = 2(a - 1)$,and $C = 2$.
Condition $1$: $A > 0 \implies a^2 - 1 > 0 \implies a^2 > 1 \implies a < -1$ or $a > 1$.
Condition $2$: $D < 0 \implies [2(a - 1)]^2 - 4(a^2 - 1)(2) < 0$.
$4(a - 1)^2 - 8(a - 1)(a + 1) < 0$.
Dividing by $4(a - 1)$ (assuming $a \neq 1$): $(a - 1) - 2(a + 1) < 0$.
$a - 1 - 2a - 2 < 0 \implies -a - 3 < 0 \implies a > -3$.
Combining the conditions: $(a < -1 \text{ or } a > 1)$ $AND$ $(a > -3)$.
This gives $a \in (-3, -1) \cup (1, \infty)$.
Note: If $a = 1$,the expression becomes $2$,which is positive for all $x$. If $a = -1$,the expression becomes $-4x + 2$,which is not positive for all $x$.
Re-evaluating the discriminant condition: $4(a-1)^2 - 8(a^2-1) < 0 \implies 4(a-1)^2 - 8(a-1)(a+1) < 0$.
$4(a-1)[(a-1) - 2(a+1)] < 0 \implies 4(a-1)(-a-3) < 0 \implies -4(a-1)(a+3) < 0 \implies (a-1)(a+3) > 0$.
This holds for $a < -3$ or $a > 1$.
Intersection with $a^2 - 1 > 0$ ($a < -1$ or $a > 1$) gives $a < -3$ or $a > 1$.

(A) Given the equation $\frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1}$.
Cross-multiplying,we get $(m + 1)(x^2 - bx) = (m - 1)(ax - c)$.
Expanding this,we have $(m + 1)x^2 - (m + 1)bx = (m - 1)ax - (m - 1)c$.
Rearranging the terms into the standard quadratic form $Ax^2 + Bx + C = 0$:
$(m + 1)x^2 - [b(m + 1) + a(m - 1)]x + c(m - 1) = 0$.
Since the roots are equal in magnitude but opposite in sign,their sum must be zero.
For a quadratic equation $Ax^2 + Bx + C = 0$,the sum of roots is $-\frac{B}{A}$.
Thus,$-\frac{B}{A} = 0$,which implies $B = 0$.
Therefore,$b(m + 1) + a(m - 1) = 0$.
$bm + b + am - a = 0$.
$m(a + b) = a - b$.
$m = \frac{a - b}{a + b}$.

(B) The given equation is ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$.
Since ${e^{2\log k}} = {e^{\log {k^2}}} = {k^2}$,the equation becomes ${x^2} - 3kx + 2{k^2} - 1 = 0$.
The product of the roots of a quadratic equation $ax^2 + bx + c = 0$ is $\frac{c}{a}$.
Here,the product of roots is $2{k^2} - 1$.
Given that the product of roots is $7$,we have $2{k^2} - 1 = 7$,which implies $2{k^2} = 8$,so ${k^2} = 4$,giving $k = \pm 2$.
Since the term $\log k$ exists in the original equation,$k$ must be positive,so $k = 2$.
For the roots to be real,the discriminant $D = b^2 - 4ac$ must be $\ge 0$.
$D = (-3k)^2 - 4(1)(2{k^2} - 1) = 9{k^2} - 8{k^2} + 4 = {k^2} + 4$.
Since ${k^2} + 4 > 0$ for all real $k$,the roots are always real for any real $k$.
However,given the options and the condition $k=2$,the roots are real when $k=2$.

(B) Given $5\cos A + 3 = 0$,so $\cos A = -\frac{3}{5}$.
Since $A$ is an angle in a triangle and $\cos A < 0$,$A$ is an obtuse angle,so $\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - (-\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Then $\tan A = \frac{\sin A}{\cos A} = \frac{4/5}{-3/5} = -\frac{4}{3}$.
Let the roots be $\alpha = \sin A = \frac{4}{5}$ and $\beta = \tan A = -\frac{4}{3}$.
Sum of roots: $\alpha + \beta = \frac{4}{5} - \frac{4}{3} = \frac{12 - 20}{15} = -\frac{8}{15}$.
Product of roots: $\alpha \cdot \beta = (\frac{4}{5})(-\frac{4}{3}) = -\frac{16}{15}$.
The quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - (-\frac{8}{15})x + (-\frac{16}{15}) = 0$.
$x^2 + \frac{8}{15}x - \frac{16}{15} = 0$.
Multiplying by $15$,we get $15x^2 + 8x - 16 = 0$.

(A) Given equation: $x^2 - |x| - 6 = 0$
Case $1$: If $x \ge 0$,then $|x| = x$. The equation becomes $x^2 - x - 6 = 0$.
$(x - 3)(x + 2) = 0$.
This gives $x = 3$ or $x = -2$. Since $x \ge 0$,we have $x = 3$.
Case $2$: If $x < 0$,then $|x| = -x$. The equation becomes $x^2 - (-x) - 6 = 0$,which is $x^2 + x - 6 = 0$.
$(x + 3)(x - 2) = 0$.
This gives $x = -3$ or $x = 2$. Since $x < 0$,we have $x = -3$.
The real roots are $3$ and $-3$.
The product of all real roots is $3 \times (-3) = -9$.

(C) For the equation ${x^2} + px + q = 0$,the sum of roots is $\alpha + \beta = -p$ and the product of roots is $\alpha \beta = q$.
For the equation ${x^2} + rx + s = 0$,the roots are $\alpha + h$ and $\beta + h$. Thus,the sum of roots is $(\alpha + h) + (\beta + h) = -r$,which simplifies to $(\alpha + \beta) + 2h = -r$. Substituting $\alpha + \beta = -p$,we get $-p + 2h = -r$,so $h = \frac{p - r}{2}$.
The product of roots for the second equation is $(\alpha + h)(\beta + h) = s$,which expands to $\alpha \beta + h(\alpha + \beta) + {h^2} = s$. Substituting the known values,we get $q + h(-p) + {h^2} = s$. Substituting $h = \frac{p - r}{2}$,we have $q - p\left( \frac{p - r}{2} \right) + \left( \frac{p - r}{2} \right)^2 = s$. Multiplying by $4$,we get $4q - 2p(p - r) + (p - r)^2 = 4s$,which simplifies to $4q - 2p^2 + 2pr + p^2 + r^2 - 2pr = 4s$. This results in $4q - p^2 + r^2 = 4s$,or ${p^2} - 4q = {r^2} - 4s$.

(A) Let $f(x) = {x^4} - (p - 3){x^3} - (3p - 5){x^2} + (2p - 7)x + 6$.
Since $(x + 1)$ is a factor of $f(x)$,by the Factor Theorem,$f(-1) = 0$.
Substituting $x = -1$ into the expression:
$(-1)^4 - (p - 3)(-1)^3 - (3p - 5)(-1)^2 + (2p - 7)(-1) + 6 = 0$
$1 - (p - 3)(-1) - (3p - 5)(1) - (2p - 7) + 6 = 0$
$1 + (p - 3) - (3p - 5) - (2p - 7) + 6 = 0$
$1 + p - 3 - 3p + 5 - 2p + 7 + 6 = 0$
Combining like terms:
$(p - 3p - 2p) + (1 - 3 + 5 + 7 + 6) = 0$
$-4p + 16 = 0$
$-4p = -16$
$p = 4$.

(D) Given that $\alpha$ is a root of $a^2x^2 + bx + c = 0$,we have $a^2\alpha^2 + b\alpha + c = 0$,which implies $b\alpha + c = -a^2\alpha^2$.
Given that $\beta$ is a root of $a^2x^2 - bx - c = 0$,we have $a^2\beta^2 - b\beta - c = 0$,which implies $b\beta + c = a^2\beta^2$.
Let $f(x) = a^2x^2 + 2bx + 2c$.
Evaluating $f(x)$ at $\alpha$: $f(\alpha) = a^2\alpha^2 + 2(b\alpha + c) = a^2\alpha^2 + 2(-a^2\alpha^2) = -a^2\alpha^2$. Since $a \ne 0$ and $\alpha > 0$,$f(\alpha) < 0$.
Evaluating $f(x)$ at $\beta$: $f(\beta) = a^2\beta^2 + 2(b\beta + c) = a^2\beta^2 + 2(a^2\beta^2) = 3a^2\beta^2$. Since $a \ne 0$ and $\beta > 0$,$f(\beta) > 0$.
Since $f(\alpha) < 0$ and $f(\beta) > 0$,by the Intermediate Value Theorem,there exists a root $\gamma$ of $f(x) = 0$ such that $\alpha < \gamma < \beta$.

(A) The given expression is in the form of the exponential series expansion $e^x = 1 + x + \frac{x^2}{2!} + \dots \infty$.
Numerator $N = e^{m \ln 3} \times e^{n \ln 3} = e^{(m+n) \ln 3} = 3^{m+n}$.
Denominator $D = e^{mn \ln 3} = 3^{mn}$.
For the equation $x^2 - x - 1 = 0$,the sum of roots $m+n = 1$ and the product of roots $mn = -1$.
Substituting these values,we get $\frac{N}{D} = \frac{3^{m+n}}{3^{mn}} = \frac{3^1}{3^{-1}} = 3^{1 - (-1)} = 3^2 = 9$.

(B) For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given equations: $(k + 1)x + 8y = 4k$ and $kx + (k + 3)y = 3k - 1$.
Applying the condition: $\frac{k + 1}{k} = \frac{8}{k + 3} = \frac{4k}{3k - 1}$.
First,solve $\frac{k + 1}{k} = \frac{8}{k + 3}$:
$(k + 1)(k + 3) = 8k \Rightarrow k^2 + 4k + 3 = 8k \Rightarrow k^2 - 4k + 3 = 0$.
Factoring gives $(k - 1)(k - 3) = 0$,so $k = 1$ or $k = 3$.
Next,check these values in the second equality $\frac{8}{k + 3} = \frac{4k}{3k - 1}$:
If $k = 1$: $\frac{8}{1 + 3} = \frac{8}{4} = 2$ and $\frac{4(1)}{3(1) - 1} = \frac{4}{2} = 2$. Since $2 = 2$,$k = 1$ is a solution.
If $k = 3$: $\frac{8}{3 + 3} = \frac{8}{6} = \frac{4}{3}$ and $\frac{4(3)}{3(3) - 1} = \frac{12}{8} = \frac{3}{2}$. Since $\frac{4}{3} \neq \frac{3}{2}$,$k = 3$ is not a solution.
Thus,there is only $1$ value of $k$ for which the system has infinitely many solutions.

(D) Given the equation $\cos \theta = x + \frac{1}{x}$.
Multiplying by $x$,we get $x^2 - x \cos \theta + 1 = 0$.
Since $x$ is a real number,the discriminant $D$ of this quadratic equation must be greater than or equal to $0$.
$D = b^2 - 4ac = (-\cos \theta)^2 - 4(1)(1) \ge 0$.
$\cos^2 \theta - 4 \ge 0$.
$\cos^2 \theta \ge 4$.
However,the range of $\cos \theta$ is $[-1, 1]$,so $\cos^2 \theta$ must be in the interval $[0, 1]$.
Since $\cos^2 \theta \ge 4$ is impossible,there is no real value of $x$ for which this equation holds.
Therefore,no value of $\theta$ is possible.

(D) The given point $(t^2 + 2t + 5, 2t^2 + t - 2)$ lies on the line $x + y = 2$ if the coordinates satisfy the equation:
$(t^2 + 2t + 5) + (2t^2 + t - 2) = 2$
$3t^2 + 3t + 3 = 2$
$3t^2 + 3t + 1 = 0$
For this quadratic equation,the discriminant $D = b^2 - 4ac = (3)^2 - 4(3)(1) = 9 - 12 = -3$.
Since the discriminant is negative $(D < 0)$,there are no real values of $t$ that satisfy this equation.
Therefore,the point cannot lie on the line for any real value of $t$.

(C) The formula for the resultant $R$ of two forces $P$ and $Q$ acting at an angle $\theta$ is given by $R^2 = P^2 + Q^2 + 2PQ \cos \theta$.
Given $R = \sqrt{7}Q$ and $\theta = 60^\circ$.
Substituting these values: $(\sqrt{7}Q)^2 = P^2 + Q^2 + 2PQ \cos 60^\circ$.
$7Q^2 = P^2 + Q^2 + 2PQ(1/2)$.
$7Q^2 = P^2 + Q^2 + PQ$.
Rearranging the terms: $P^2 + PQ - 6Q^2 = 0$.
Factoring the quadratic equation: $P^2 + 3PQ - 2PQ - 6Q^2 = 0$.
$P(P + 3Q) - 2Q(P + 3Q) = 0$.
$(P - 2Q)(P + 3Q) = 0$.
Since $P$ and $Q$ are magnitudes of forces,they must be positive,so $P + 3Q \neq 0$.
Therefore,$P - 2Q = 0$,which implies $P = 2Q$.
Thus,$P/Q = 2$.

(D) The given function is a quadratic in $x$ of the form $f(x) = Ax^2 + Bx + C$,where $A = 1 + b^2$,$B = 2b$,and $C = 1$.
Since $A = 1 + b^2 > 0$ for all real $b$,the function has a minimum value at $x = -\frac{B}{2A}$.
The minimum value $m(b)$ is given by $f(-\frac{B}{2A}) = C - \frac{B^2}{4A}$.
Substituting the values: $m(b) = 1 - \frac{(2b)^2}{4(1 + b^2)} = 1 - \frac{4b^2}{4(1 + b^2)} = 1 - \frac{b^2}{1 + b^2}$.
Simplifying this,we get $m(b) = \frac{1 + b^2 - b^2}{1 + b^2} = \frac{1}{1 + b^2}$.
Since $b^2 \ge 0$,we have $1 + b^2 \ge 1$,which implies $0 < \frac{1}{1 + b^2} \le 1$.
Thus,the range of $m(b)$ is $(0, 1]$.

(B) Let the two numbers be $a$ and $b$. The arithmetic mean is $A = \frac{a + b}{2}$ and the geometric mean is $G = \sqrt{ab}$.
Thus,$G^2 = ab$.
The given equation is $x^2 - 2Ax + G^2 = 0$.
Substituting $2A = a + b$ and $G^2 = ab$,we get $x^2 - (a + b)x + ab = 0$.
Factoring the quadratic equation: $(x - a)(x - b) = 0$.
This implies that $a$ and $b$ are the roots of the equation.
For any two positive real numbers $a$ and $b$,the arithmetic mean is always greater than or equal to the geometric mean $(A \ge G)$.
Specifically,$A - G = \frac{a + b}{2} - \sqrt{ab} = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0$.
Since $a$ and $b$ are roots of a quadratic equation with real coefficients,$A \ge G$ holds true.

(C) quadratic equation is of the form $ax^2 + bx + c = 0$,where $a \neq 0$.
Since one root is $0$,the constant term $c$ must be $0$.
Thus,the equation is $ax^2 + bx = 0$,where $a, b \in \{0, 1, 2, \dots, 9\}$ and $a \neq 0$.
The number of choices for $a$ is $9$ (as $a \in \{1, 2, \dots, 9\}$).
The number of choices for $b$ is $10$ (as $b \in \{0, 1, 2, \dots, 9\}$).
Therefore,the total number of such quadratic equations is $N = 9 \times 10 = 90$.

(A) Given that $4$ is a root of the equation $x^2 + px + 12 = 0$.
Substituting $x = 4$ into the equation: $(4)^2 + p(4) + 12 = 0$.
$16 + 4p + 12 = 0$.
$4p + 28 = 0$.
$4p = -28$.
$p = -7$.
Now,the second equation is $x^2 + px + q = 0$,which becomes $x^2 - 7x + q = 0$.
Since this equation has equal roots,its discriminant $D = b^2 - 4ac$ must be $0$.
Here,$a = 1$,$b = -7$,and $c = q$.
$(-7)^2 - 4(1)(q) = 0$.
$49 - 4q = 0$.
$4q = 49$.
$q = 49/4$.

(A) Given that the roots of $bx^2 + cx + a = 0$ are imaginary,the discriminant $D < 0$.
$D = c^2 - 4ba < 0$,which implies $c^2 < 4ab$.
Let $f(x) = 3b^2x^2 + 6bcx + 2c^2$.
We can rewrite $f(x)$ by completing the square:
$f(x) = 3(b^2x^2 + 2bcx) + 2c^2$
$f(x) = 3(bx + c)^2 - 3c^2 + 2c^2$
$f(x) = 3(bx + c)^2 - c^2$.
Since $(bx + c)^2 \ge 0$,the minimum value of $f(x)$ is $-c^2$.
From $c^2 < 4ab$,we have $-c^2 > -4ab$.
Therefore,$f(x) = 3(bx + c)^2 - c^2 \ge -c^2 > -4ab$.
Thus,$f(x) > -4ab$ for all real values of $x$.

(B) For the quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.
Here,$a = 1$,$b = -2\sqrt{2}$,and $c = 1$.
$D = (-2\sqrt{2})^2 - 4(1)(1) = 8 - 4 = 4$.
Since $D > 0$,the roots are real and distinct.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{2\sqrt{2} \pm \sqrt{4}}{2} = \frac{2\sqrt{2} \pm 2}{2} = \sqrt{2} \pm 1$.
Since $\sqrt{2} \pm 1$ are irrational,the roots are real and distinct.

(C) For a quadratic equation $ax^2 + bx + c = 0$ to have equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Here,$a = 1$,$b = -(3k - 1)$,and $c = 2k^2 + 2k$.
Setting the discriminant to zero:
$D = [-(3k - 1)]^2 - 4(1)(2k^2 + 2k) = 0$
$(3k - 1)^2 - 4(2k^2 + 2k) = 0$
$9k^2 - 6k + 1 - 8k^2 - 8k = 0$
$k^2 - 14k + 1 = 0$
Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$k = \frac{14 \pm \sqrt{(-14)^2 - 4(1)(1)}}{2(1)}$
$k = \frac{14 \pm \sqrt{196 - 4}}{2} = \frac{14 \pm \sqrt{192}}{2} = \frac{14 \pm 8\sqrt{3}}{2} = 7 \pm 4\sqrt{3}$.

(B) For the quadratic equation $Ax^2 + Bx + C = 0$,if the roots are equal,the discriminant $D = B^2 - 4AC = 0$.
Here,$A = (m - n)$,$B = (n - l)$,and $C = (l - m)$.
Substituting these into $B^2 - 4AC = 0$:
$(n - l)^2 - 4(m - n)(l - m) = 0$
$n^2 + l^2 - 2nl - 4(ml - m^2 - nl + nm) = 0$
$n^2 + l^2 - 2nl - 4ml + 4m^2 + 4nl - 4nm = 0$
$l^2 + n^2 + 4m^2 + 2nl - 4ml - 4nm = 0$
This expression is the expansion of $(l + n - 2m)^2 = 0$.
Therefore,$l + n - 2m = 0$,which implies $2m = n + l$.

(B) Given the equation: $x(x + 2) = 3 - ax^2$
Rearranging the terms: $x^2 + 2x = 3 - ax^2$
$(1 + a)x^2 + 2x - 3 = 0$
For a quadratic equation $Ax^2 + Bx + C = 0$ to have one root approaching infinity,the coefficient of the highest degree term must approach zero.
Therefore,$1 + a \to 0$
$a \to -1$

(A) Given $a + b + c = 0$,we have $b + c = -a$,$c + a = -b$,and $a + b = -c$.
Substituting these into the equation $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$:
$(-a - a)x^2 + (-b - b)x + (-c - c) = 0$
$-2ax^2 - 2bx - 2c = 0$
$ax^2 + bx + c = 0$.
The discriminant $D = b^2 - 4ac$.
Since $c = -(a + b)$,$D = b^2 - 4a(-(a + b)) = b^2 + 4a^2 + 4ab = (2a + b)^2$.
Since $a, b, c$ are rational,$D$ is a perfect square of a rational number,so the roots are rational.
Thus,Statement-$I$ is true.
Statement-$II$ is also true because the discriminant is $(2a + b)^2$,which is a perfect square,and this explains why the roots are rational.

(C) For a quadratic equation $Ax^2 + Bx + C = 0$ to have roots of opposite signs,the product of the roots must be negative,i.e.,$\frac{C}{A} < 0$.
Here,$A = 3$ and $C = a^2 - 3a + 2$.
So,$\frac{a^2 - 3a + 2}{3} < 0$.
This implies $a^2 - 3a + 2 < 0$.
Factoring the quadratic expression,we get $(a - 1)(a - 2) < 0$.
The inequality holds when $a$ lies between the roots of the quadratic expression $(a - 1)(a - 2) = 0$.
Thus,$1 < a < 2$,which can be written as the interval $(1, 2)$.

(B) Let the two numbers be $a$ and $b$.
Given that the arithmetic mean is $\frac{a + b}{2} = 9$,so $a + b = 18$.
Given that the geometric mean is $\sqrt{ab} = 4$,so $ab = 16$.
The quadratic equation whose roots are $a$ and $b$ is given by $x^2 - (a + b)x + ab = 0$.
Substituting the values,we get $x^2 - 18x + 16 = 0$.

(D) For a quadratic equation $At^2 + Bt + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be $0$.
Here,$A = (a^2 + b^2)$,$B = -2(ac + bd)$,and $C = (c^2 + d^2)$.
Setting $B^2 - 4AC = 0$:
$[-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0$
$4(a^2c^2 + b^2d^2 + 2abcd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0$
Dividing by $4$:
$(a^2c^2 + b^2d^2 + 2abcd) - (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0$
$2abcd - a^2d^2 - b^2c^2 = 0$
$-(a^2d^2 + b^2c^2 - 2abcd) = 0$
$-(ad - bc)^2 = 0$
$(ad - bc)^2 = 0$
$ad = bc$
$\frac{a}{b} = \frac{c}{d}$

(A) Given equation is $x^2 - |x| - 6 = 0$.
Case $1$: If $x \ge 0$,then $|x| = x$. The equation becomes $x^2 - x - 6 = 0$.
$(x - 3)(x + 2) = 0$.
Since $x \ge 0$,we have $x = 3$.
Case $2$: If $x < 0$,then $|x| = -x$. The equation becomes $x^2 - (-x) - 6 = 0$,which is $x^2 + x - 6 = 0$.
$(x + 3)(x - 2) = 0$.
Since $x < 0$,we have $x = -3$.
The real roots are $3$ and $-3$.
The product of the real roots is $3 \times (-3) = -9$.

(D) Let $f(x) = x^2 + ax + b$.
Since $f(x)$ is an integer for every integer $x$,we have:
$f(0) = 0^2 + a(0) + b = b$.
Since $f(0)$ must be an integer,$b$ must be an integer.
Now,$f(1) = 1^2 + a(1) + b = 1 + a + b$.
Since $f(1)$ is an integer and $b$ is an integer,$1 + a + b$ is an integer,which implies $a$ must be an integer.
Therefore,both $a$ and $b$ must be integers.
Since the options provided do not state that both are integers,the correct choice is $D$.

(B) For the quadratic equation $x^2 - 4x + \log_{1/2} a = 0$ to have non-real roots,the discriminant $D$ must be less than $0$.
$D = b^2 - 4ac < 0$
Here,$a = 1$,$b = -4$,and $c = \log_{1/2} a$.
$(-4)^2 - 4(1)(\log_{1/2} a) < 0$
$16 - 4 \log_{1/2} a < 0$
$16 < 4 \log_{1/2} a$
$4 < \log_{1/2} a$
Since the base of the logarithm is $1/2$ (which is between $0$ and $1$),the inequality reverses when we remove the logarithm:
$a < (1/2)^4$
$a < 1/16$
Thus,the maximum value of $a$ is $1/16$ (as $a$ must be strictly less than $1/16$ for the roots to be non-real,the supremum is $1/16$).

(C) Given that $\alpha$ and $\beta$ are the roots of the equation $(x - a)(x - b) = c$,we can write:
$(x - a)(x - b) - c = (x - \alpha)(x - \beta)$
Rearranging the terms,we get:
$(x - \alpha)(x - \beta) + c = (x - a)(x - b)$
Therefore,the equation $(x - \alpha)(x - \beta) + c = 0$ is equivalent to $(x - a)(x - b) = 0$.
The roots of this equation are $x = a$ and $x = b$.

(C) Given $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots \infty}}}$.
Since the expression is infinite,we can write $x = \sqrt{6 + x}$.
Squaring both sides,we get $x^2 = 6 + x$,where $x > 0$.
Rearranging the terms,we get $x^2 - x - 6 = 0$.
Factoring the quadratic equation: $(x - 3)(x + 2) = 0$.
Since $x > 0$,we must have $x = 3$.

(C) Given equation: $\sqrt{3x^2 - 7x - 30} + \sqrt{2x^2 - 7x - 5} = x + 5$.
Let $f(x) = \sqrt{3x^2 - 7x - 30} + \sqrt{2x^2 - 7x - 5} - (x + 5) = 0$.
Testing the options:
For $x = 6$:
$\sqrt{3(6)^2 - 7(6) - 30} + \sqrt{2(6)^2 - 7(6) - 5} = \sqrt{108 - 42 - 30} + \sqrt{72 - 42 - 5} = \sqrt{36} + \sqrt{25} = 6 + 5 = 11$.
Right side: $x + 5 = 6 + 5 = 11$.
Since $11 = 11$,$x = 6$ is the correct solution.

(B) The given equation is $(a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0$.
Expanding the middle term: $(a^2 + b^2)x^2 - 2(ab + bc)x + (b^2 + c^2) = 0$.
Rearranging the terms: $(a^2x^2 - 2abx + b^2) + (b^2x^2 - 2bcx + c^2) = 0$.
This simplifies to: $(ax - b)^2 + (bx - c)^2 = 0$.
Since the sum of squares is zero,each term must be zero: $ax - b = 0$ and $bx - c = 0$.
Thus,$x = \frac{b}{a}$ and $x = \frac{c}{b}$.
Equating the values of $x$: $\frac{b}{a} = \frac{c}{b}$.
This implies $b^2 = ac$,which means $a, b, c$ are in Geometric Progression.

(A) Given the equation: $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$
Multiply both sides by $r(x + p)(x + q)$ to clear the denominators:
$r(x + q) + r(x + p) = (x + p)(x + q)$
$rx + rq + rx + rp = x^2 + px + qx + pq$
$2rx + r(p + q) = x^2 + (p + q)x + pq$
Rearranging into standard quadratic form $ax^2 + bx + c = 0$:
$x^2 + (p + q - 2r)x + (pq - r(p + q)) = 0$
Let the roots be $\alpha$ and $-\alpha$.
For a quadratic equation $x^2 + Bx + C = 0$,the sum of roots is $-B$.
Here,$\alpha + (-\alpha) = 0$,so the coefficient of $x$ must be $0$.
$p + q - 2r = 0$
$2r = p + q$
$r = \frac{p + q}{2}$

(A) Given the equation: $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{4x}{p + q + r} = 0$.
Add $1$ to each of the first three terms to simplify:
$(\frac{p + q - x}{r} + 1) + (\frac{q + r - x}{p} + 1) + (\frac{r + p - x}{q} + 1) + \frac{4x}{p + q + r} - 3 = 0$
$\frac{p + q + r - x}{r} + \frac{q + r + p - x}{p} + \frac{r + p + q - x}{q} + \frac{4x - 3(p + q + r)}{p + q + r} = 0$
$(p + q + r - x) (\frac{1}{r} + \frac{1}{p} + \frac{1}{q}) + \frac{4x - 3(p + q + r)}{p + q + r} = 0$
Alternatively,rewrite the equation as:
$(\frac{p + q - x}{r} + 1) + (\frac{q + r - x}{p} + 1) + (\frac{r + p - x}{q} + 1) + \frac{4x}{p + q + r} - 3 = 0$
$(p + q + r - x) (\frac{1}{p} + \frac{1}{q} + \frac{1}{r}) = 3 - \frac{4x}{p + q + r}$
$(p + q + r - x) (\frac{1}{p} + \frac{1}{q} + \frac{1}{r}) = \frac{3(p + q + r) - 4x}{p + q + r}$
Setting $(p + q + r - x) = 0$ gives $x = p + q + r$.

(A) Given that $\sin A, \sin B, \cos A$ are in $G.P.$,we have $\sin^2 B = \sin A \cos A$.
Multiplying by $2$,we get $2 \sin^2 B = 2 \sin A \cos A = \sin 2A$.
Since $\sin 2A \leq 1$,it follows that $2 \sin^2 B \leq 1$,or $\sin^2 B \leq \frac{1}{2}$.
Now,for the quadratic equation $x^2 + 2x \cot B + 1 = 0$,the discriminant $D$ is given by:
$D = (2 \cot B)^2 - 4(1)(1) = 4 \cot^2 B - 4 = 4(\cot^2 B - 1)$.
Since $\sin^2 B \leq \frac{1}{2}$,we have $\csc^2 B \geq 2$,which implies $\cot^2 B = \csc^2 B - 1 \geq 2 - 1 = 1$.
Therefore,$\cot^2 B - 1 \geq 0$,which means $D \geq 0$.
Since the discriminant is non-negative,the roots are always real.

(D) Given the equation: $x^2 + 5|x| + 4 = 0$.
Since $x^2 = |x|^2$,we can rewrite the equation as: $|x|^2 + 5|x| + 4 = 0$.
Factoring the quadratic expression in terms of $|x|$: $(|x| + 1)(|x| + 4) = 0$.
This gives two possibilities: $|x| = -1$ or $|x| = -4$.
Since the absolute value $|x|$ must always be non-negative $(|x| \ge 0)$,neither $|x| = -1$ nor $|x| = -4$ is possible.
Therefore,there are no real roots for the given equation.

(A) For the equation $ax^2 + 2bx + c = 0$,the roots are real,so the discriminant $D_1 \ge 0$.
$D_1 = (2b)^2 - 4ac = 4b^2 - 4ac \ge 0 \implies b^2 \ge ac$.
For the equation $bx^2 - 2\sqrt{ac}x + b = 0$,the roots are real,so the discriminant $D_2 \ge 0$.
$D_2 = (-2\sqrt{ac})^2 - 4(b)(b) = 4ac - 4b^2 \ge 0 \implies ac \ge b^2$.
Since $b^2 \ge ac$ and $ac \ge b^2$,it must be that $b^2 = ac$.