Solution of quadratic equations and Nature of roots Questions in English

(D) We are given the equations:
$xy = \frac{1}{9}$
$x(y + 1) = \frac{7}{9}$
$y(x + 1) = \frac{5}{18}$
Expanding the expressions:
$xy + x = \frac{7}{9} \implies \frac{1}{9} + x = \frac{7}{9} \implies x = \frac{6}{9} = \frac{2}{3}$
$xy + y = \frac{5}{18} \implies \frac{1}{9} + y = \frac{5}{18} \implies y = \frac{5}{18} - \frac{2}{18} = \frac{3}{18} = \frac{1}{6}$
Now,calculate $(x + 1)(y + 1)$:
$(x + 1)(y + 1) = (\frac{2}{3} + 1)(\frac{1}{6} + 1)$
$= (\frac{5}{3})(\frac{7}{6})$
$= \frac{35}{18}$

(C) For a cubic polynomial $y = ax^3 + bx^2 + cx + d$ to be symmetric about a vertical line $x = k$,the coefficient of the cubic term must be zero,i.e.,$a = 0$.
Then the equation becomes $y = bx^2 + cx + d$,which is a parabola.
$A$ parabola $y = bx^2 + cx + d$ is symmetric about its axis of symmetry,given by $x = -\frac{c}{2b}$.
Given that the graph is symmetric about $x = k$,we have $k = -\frac{c}{2b}$.
This implies $2bk = -c$,or $2bk + c = 0$.
Since $a = 0$,we can write $a + \frac{c}{2b} + k = 0 + \frac{c}{2b} + (-\frac{c}{2b}) = 0$.
Thus,the condition is $a + \frac{c}{2b} + k = 0$.

(C) Let $u = x - 1$. Then $x = u + 1$. The equation becomes:
$\sqrt{u + 1 + 3 - 4\sqrt{u}} + \sqrt{u + 1 + 8 - 6\sqrt{u}} = 1$
$\sqrt{u + 4 - 4\sqrt{u}} + \sqrt{u + 9 - 6\sqrt{u}} = 1$
$\sqrt{(\sqrt{u} - 2)^2} + \sqrt{(\sqrt{u} - 3)^2} = 1$
$|\sqrt{u} - 2| + |\sqrt{u} - 3| = 1$
Let $y = \sqrt{u}$. Since $u \geq 0$,$y \geq 0$. The equation is $|y - 2| + |y - 3| = 1$.
This is of the form $|y - a| + |y - b| = |a - b|$,which holds for $y$ between $a$ and $b$ inclusive.
Here $a = 2$ and $b = 3$,so $2 \leq y \leq 3$.
Substituting $y = \sqrt{u}$,we get $2 \leq \sqrt{u} \leq 3$.
Squaring gives $4 \leq u \leq 9$.
Since $u = x - 1$,we have $4 \leq x - 1 \leq 9$,which implies $5 \leq x \leq 10$.
Thus,$x \in [5, 10]$.

(B) Given the quadratic equation $x^2 - \alpha x + \alpha + 1 = 0$.
For the roots to be integral,the discriminant $D$ must be a perfect square,say $\beta^2$,where $\beta \ge 0$.
$D = (-\alpha)^2 - 4(1)(\alpha + 1) = \alpha^2 - 4\alpha - 4 = \beta^2$.
Completing the square: $(\alpha - 2)^2 - 8 = \beta^2$.
Rearranging gives: $(\alpha - 2)^2 - \beta^2 = 8$,which factors as $(\alpha - 2 - \beta)(\alpha - 2 + \beta) = 8$.
Since $\alpha$ and $\beta$ are integers,we look for factor pairs of $8$ (where the product is $8$ and both factors have the same parity):
Case $1$: $(\alpha - 2 - \beta) = 2$ and $(\alpha - 2 + \beta) = 4$. Adding these gives $2(\alpha - 2) = 6$ $\Rightarrow \alpha - 2 = 3$ $\Rightarrow \alpha = 5$.
Case $2$: $(\alpha - 2 - \beta) = -4$ and $(\alpha - 2 + \beta) = -2$. Adding these gives $2(\alpha - 2) = -6$ $\Rightarrow \alpha - 2 = -3$ $\Rightarrow \alpha = -1$.
Case $3$: $(\alpha - 2 - \beta) = -2$ and $(\alpha - 2 + \beta) = -4$. Adding these gives $2(\alpha - 2) = -6 \Rightarrow \alpha = -1$.
Case $4$: $(\alpha - 2 - \beta) = 4$ and $(\alpha - 2 + \beta) = 2$. Adding these gives $2(\alpha - 2) = 6 \Rightarrow \alpha = 5$.
The distinct integral values for $\alpha$ are $-1$ and $5$.
The sum of these values is $-1 + 5 = 4$.

(C) From the graph,the parabola opens upwards,so $a > 0$.
The $y$-intercept is above the $x$-axis,so $c > 0$.
The vertex has a positive $x$-coordinate,$-b/(2a) > 0$. Since $a > 0$,we must have $-b > 0$,which implies $b < 0$.
Now,let us evaluate the options:
$A) ab^2c^3 = (+)(+)(+) = (+) > 0$ (Holds good)
$B) ab^3c^2 = (+)(-)(+) = (-) < 0$ (Holds good)
$C) ab^3c^5 = (+)(-)(+) = (-) < 0$. Thus,$ab^3c^5 > 0$ does $NOT$ hold good.
$D)$ Since the graph intersects the $x$-axis at two distinct points, the discriminant $D = b^2 - 4ac > 0$, so $b^2 > 4ac$ (Holds good).

(A) The given quadratic equation is $(x - a)(x - 10) + 1 = 0$.
Expanding this,we get $x^{2} - (10 + a)x + 10a + 1 = 0$.
For the roots to be integers,the discriminant $D$ must be a perfect square.
$D = (10 + a)^{2} - 4(10a + 1) = 100 + 20a + a^{2} - 40a - 4 = a^{2} - 20a + 96$.
We can rewrite this as $D = (a - 10)^{2} - 4$.
Let $D = k^{2}$ for some non-negative integer $k$.
Then $(a - 10)^{2} - k^{2} = 4$,which implies $(a - 10 - k)(a - 10 + k) = 4$.
Since $a$ and $k$ are integers,the factors must be pairs of integers whose product is $4$.
The possible cases are:
$1) (a - 10 - k) = 2$ and $(a - 10 + k) = 2 \Rightarrow a - 10 = 2 \Rightarrow a = 12$.
$2) (a - 10 - k) = -2$ and $(a - 10 + k) = -2 \Rightarrow a - 10 = -2 \Rightarrow a = 8$.
$3) (a - 10 - k) = 1$ and $(a - 10 + k) = 4$ (No integer solution for $a$).
Thus,the integral values of $a$ are $8$ and $12$.

(B) Given the quadratic equation $ax^2 + bx + c = 0$ with $a > 0, b < 0, c < 0$.
Sum of roots $(\alpha + \beta) = -\frac{b}{a}$. Since $b < 0$ and $a > 0$,$-\frac{b}{a} > 0$. Thus,$\alpha + \beta > 0$.
Product of roots $(\alpha \beta) = \frac{c}{a}$. Since $c < 0$ and $a > 0$,$\frac{c}{a} < 0$. Thus,$\alpha \beta < 0$.
Since the product of the roots is negative,one root must be positive and the other must be negative.
Let $\alpha < 0$ and $\beta > 0$. Since $\alpha + \beta > 0$,the positive root must have a larger absolute value than the negative root.
Therefore,$|\beta| > |\alpha|$,which implies $0 < |\alpha| < \beta$.

(A) For the quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.
Here,$a = 1$,$b = -\sqrt{13}$,and $c = 1$.
$D = (-\sqrt{13})^2 - 4(1)(1) = 13 - 4 = 9$.
Since $D > 0$ and $D$ is a perfect square,the roots are real,distinct,and rational.
However,since the coefficient $b = -\sqrt{13}$ is irrational,the roots are real and distinct.

(A) The given equation is $x^2+x-n=0$.
For the roots to be integers,the discriminant $D = b^2-4ac = 1^2 - 4(1)(-n) = 1+4n$ must be a perfect square of an odd integer.
Let $1+4n = (2k+1)^2$ for some integer $k$.
$1+4n = 4k^2 + 4k + 1
$ $\Rightarrow 4n = 4k(k+1)
$ $\Rightarrow n = k(k+1)$.
Given $n \in [5, 100]$,we have $5 \le k(k+1) \le 100$.
For $k=2, n=2(3)=6$.
For $k=3, n=3(4)=12$.
For $k=4, n=4(5)=20$.
For $k=5, n=5(6)=30$.
For $k=6, n=6(7)=42$.
For $k=7, n=7(8)=56$.
For $k=8, n=8(9)=72$.
For $k=9, n=9(10)=90$.
For $k=10, n=10(11)=110 > 100$.
Thus,the possible values of $n$ are $6, 12, 20, 30, 42, 56, 72, 90$.
The total number of such values is $8$.

(D) Let $f(x) = -x^2 + 5x - 10$ and $g(x) = (\frac{3}{2})^x$.
For the quadratic function $f(x) = -x^2 + 5x - 10$,the discriminant $D = b^2 - 4ac = 5^2 - 4(-1)(-10) = 25 - 40 = -15$.
The maximum value of $f(x)$ is given by $\frac{-D}{4a} = \frac{-(-15)}{4(-1)} = -\frac{15}{4} = -3.75$.
Since $g(x) = (\frac{3}{2})^x$ is an exponential function,$g(x) > 0$ for all real $x$.
Because the maximum value of $f(x)$ is $-3.75$ (which is negative) and the minimum value of $g(x)$ is always positive,there is no real value of $x$ for which $f(x) = g(x)$.
Therefore,the equation has no real solutions.

(D) Let the quadratic expression be $f(x) = k(x - r_1)(x - r_2)$.
Given that $-1$ is a root,let $r_1 = -1$. Let the other root be $a$.
So,$f(x) = k(x + 1)(x - a) = k(x^2 + (1 - a)x - a)$.
We are given $f(1) + f(2) = 0$.
$f(1) = k(1 + 1)(1 - a) = 2k(1 - a) = 2k - 2ka$.
$f(2) = k(2 + 1)(2 - a) = 3k(2 - a) = 6k - 3ka$.
Summing these: $f(1) + f(2) = (2k - 2ka) + (6k - 3ka) = 8k - 5ka$.
Setting the sum to zero: $8k - 5ka = 0$.
Since $f(x)$ is a quadratic expression,$k \neq 0$,so we can divide by $k$.
$8 - 5a = 0 \Rightarrow 5a = 8 \Rightarrow a = \frac{8}{5}$.
Thus,the other root is $\frac{8}{5}$.

(B) Given the equation: $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$
Simplify the left side: $\frac{x + q + x + p}{(x + p)(x + q)} = \frac{1}{r}$
Cross-multiply: $r(2x + p + q) = x^2 + (p + q)x + pq$
Rearrange into standard quadratic form: $x^2 + (p + q - 2r)x + (pq - pr - qr) = 0$
Let the roots be $\alpha$ and $\beta$. Since the roots are equal in magnitude but opposite in sign,$\alpha = -\beta$,which implies $\alpha + \beta = 0$.
From the sum of roots formula for a quadratic equation $ax^2 + bx + c = 0$,the sum of roots is $-b/a$. Thus,$-(p + q - 2r) = 0$,which means $p + q = 2r$.
The sum of squares of the roots is $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
Since $\alpha + \beta = 0$,this simplifies to $\alpha^2 + \beta^2 = -2\alpha\beta$.
From the product of roots formula,$\alpha\beta = pq - pr - qr$.
Substituting $\alpha\beta$: $\alpha^2 + \beta^2 = -2(pq - pr - qr) = -2pq + 2pr + 2qr$.
Since $2r = p + q$,substitute $2pr + 2qr = 2r(p + q) = (p + q)(p + q) = p^2 + 2pq + q^2$.
Therefore,$\alpha^2 + \beta^2 = -2pq + (p^2 + 2pq + q^2) = p^2 + q^2$.

(C) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$.
Given $p(0) = 1$,we have $c = 1$.
By the Remainder Theorem,$p(1) = 4$ and $p(-1) = 6$.
Substituting these into $p(x) = ax^2 + bx + 1$:
$p(1) = a(1)^2 + b(1) + 1 = 4 \Rightarrow a + b = 3$.
$p(-1) = a(-1)^2 + b(-1) + 1 = 6 \Rightarrow a - b = 5$.
Adding the two equations: $2a = 8 \Rightarrow a = 4$.
Subtracting the two equations: $2b = -2 \Rightarrow b = -1$.
Thus,$p(x) = 4x^2 - x + 1$.
Now,evaluating $p(-2)$:
$p(-2) = 4(-2)^2 - (-2) + 1 = 4(4) + 2 + 1 = 16 + 3 = 19$.
Therefore,$p(-2) = 19$ is the correct statement.

(C) Given the equation $2^{(x - 1)(x^2 + 5x - 50)} = 1$.
Since $2^0 = 1$,we equate the exponent to $0$:
$(x - 1)(x^2 + 5x - 50) = 0$.
Factoring the quadratic expression $x^2 + 5x - 50$:
$(x^2 + 10x - 5x - 50) = x(x + 10) - 5(x + 10) = (x - 5)(x + 10)$.
So,the equation becomes $(x - 1)(x - 5)(x + 10) = 0$.
The real values of $x$ are $x = 1, 5, -10$.
The sum of these values is $1 + 5 + (-10) = 6 - 10 = -4$.

(A) Given equation: $\sqrt{2x + 1} - \sqrt{2x - 1} = 1$
Squaring both sides: $(\sqrt{2x + 1} - \sqrt{2x - 1})^2 = 1^2$
$(2x + 1) + (2x - 1) - 2\sqrt{(2x + 1)(2x - 1)} = 1$
$4x - 2\sqrt{4x^2 - 1} = 1$
$4x - 1 = 2\sqrt{4x^2 - 1}$
Squaring again: $(4x - 1)^2 = 4(4x^2 - 1)$
$16x^2 - 8x + 1 = 16x^2 - 4$
$-8x = -5 \implies x = \frac{5}{8}$
Now,substitute $x = \frac{5}{8}$ into $\sqrt{4x^2 - 1}$:
$\sqrt{4(\frac{5}{8})^2 - 1} = \sqrt{4(\frac{25}{64}) - 1} = \sqrt{\frac{25}{16} - 1} = \sqrt{\frac{9}{16}} = \frac{3}{4}$

(B) Given equation: $(a-1)(x^4+x^2+1) + (a+1)(x^2+x+1)^2 = 0$
We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.
Substituting this,we get: $(a-1)(x^2+x+1)(x^2-x+1) + (a+1)(x^2+x+1)^2 = 0$
Factoring out $(x^2+x+1)$: $(x^2+x+1)[(a-1)(x^2-x+1) + (a+1)(x^2+x+1)] = 0$
Simplifying the expression inside the bracket: $(x^2+x+1)[ax^2-ax+a-x^2+x-1 + ax^2+ax+a+x^2+x+1] = 0$
$(x^2+x+1)(2ax^2+2x+2a) = 0$
$2(x^2+x+1)(ax^2+x+a) = 0$
The quadratic $x^2+x+1$ has a discriminant $D = 1^2 - 4(1)(1) = -3 < 0$,so it has no real roots.
For the equation to have real and distinct roots,the quadratic $ax^2+x+a = 0$ must have two distinct real roots.
This requires $a \neq 0$ and the discriminant $D > 0$.
$D = 1^2 - 4(a)(a) = 1 - 4a^2 > 0$
$4a^2 < 1$ $\Rightarrow a^2 < 1/4$ $\Rightarrow |a| < 1/2$.
Since $a \neq 0$,the set of values is $a \in (-1/2, 0) \cup (0, 1/2)$.

(C) Given the equation: $x^2 + |2x - 3| - 4 = 0$.
Case $1$: If $x \ge \frac{3}{2}$,then $|2x - 3| = 2x - 3$.
$x^2 + 2x - 3 - 4 = 0 \implies x^2 + 2x - 7 = 0$.
Using the quadratic formula: $x = \frac{-2 \pm \sqrt{4 - 4(1)(-7)}}{2} = \frac{-2 \pm \sqrt{32}}{2} = -1 \pm 2\sqrt{2}$.
Since $x \ge \frac{3}{2}$,we check the values: $-1 + 2\sqrt{2} \approx -1 + 2(1.414) = 1.828 > 1.5$ (Valid).
$-1 - 2\sqrt{2} < 0 < 1.5$ (Invalid).
So,$x_1 = 2\sqrt{2} - 1$.
Case $2$: If $x < \frac{3}{2}$,then $|2x - 3| = -(2x - 3) = -2x + 3$.
$x^2 - 2x + 3 - 4 = 0 \implies x^2 - 2x - 1 = 0$.
Using the quadratic formula: $x = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.
Since $x < \frac{3}{2}$,we check the values: $1 + \sqrt{2} \approx 2.414 > 1.5$ (Invalid).
$1 - \sqrt{2} \approx -0.414 < 1.5$ (Valid).
So,$x_2 = 1 - \sqrt{2}$.
Sum of the roots: $x_1 + x_2 = (2\sqrt{2} - 1) + (1 - \sqrt{2}) = \sqrt{2}$.

(A) Given the equation $\sqrt{3x^2 + x + 5} = x - 3$.
For the square root to be defined and equal to a real value,we must have $x - 3 \geq 0$,which implies $x \geq 3$.
Squaring both sides:
$3x^2 + x + 5 = (x - 3)^2$
$3x^2 + x + 5 = x^2 - 6x + 9$
$2x^2 + 7x - 4 = 0$
Solving the quadratic equation $2x^2 + 7x - 4 = 0$ using the quadratic formula or factorization:
$2x^2 + 8x - x - 4 = 0$
$2x(x + 4) - 1(x + 4) = 0$
$(2x - 1)(x + 4) = 0$
So,$x = \frac{1}{2}$ or $x = -4$.
Checking these values against the condition $x \geq 3$:
For $x = \frac{1}{2}$,$\frac{1}{2} < 3$ (Invalid).
For $x = -4$,$-4 < 3$ (Invalid).
Since neither value satisfies the condition $x \geq 3$,the equation has no real solution.

(B) For the quadratic equation $(k-2)x^2 + 8x + k+4 = 0$ to have real and distinct roots,the discriminant $D > 0$:
$D = 8^2 - 4(k-2)(k+4) > 0$
$64 - 4(k^2 + 2k - 8) > 0$
$16 - (k^2 + 2k - 8) > 0$
$-k^2 - 2k + 24 > 0 \Rightarrow k^2 + 2k - 24 < 0$
$(k+6)(k-4) < 0 \Rightarrow -6 < k < 4$
For the roots to be negative,the sum of roots $\alpha + \beta = -\frac{8}{k-2} < 0$ and the product of roots $\alpha \beta = \frac{k+4}{k-2} > 0$.
From $\alpha + \beta < 0$: $\frac{8}{k-2} > 0 \Rightarrow k > 2$.
From $\alpha \beta > 0$: $\frac{k+4}{k-2} > 0 \Rightarrow k < -4$ or $k > 2$.
Combining all conditions: $(-6 < k < 4)$ $AND$ $(k > 2)$ $AND$ $(k < -4 \text{ or } k > 2)$.
The intersection is $2 < k < 4$.
Among the given options,only $k = 3$ satisfies $2 < k < 4$.

(C) The given quadratic equation is $px^2 + qx + r = 0$ with $p, q, r \in \mathbb{R}$ and $r > p > 0.$
Since the roots $\alpha$ and $\beta$ are complex,the discriminant $D = q^2 - 4pr < 0.$
This implies $q^2 < 4pr.$
Since the coefficients are real,the complex roots must be conjugates,i.e.,$\beta = \bar{\alpha}.$
Thus,$|\alpha| = |\beta| = \sqrt{\alpha \bar{\alpha}} = \sqrt{\alpha \beta}.$
From the properties of quadratic equations,the product of the roots is $\alpha \beta = \frac{r}{p}.$
Therefore,$|\alpha| = |\beta| = \sqrt{\frac{r}{p}}.$
Given $r > p > 0,$ we have $\frac{r}{p} > 1,$ which implies $\sqrt{\frac{r}{p}} > 1.$
Then,$|\alpha| + |\beta| = 2|\alpha| = 2\sqrt{\frac{r}{p}}.$
Since $\sqrt{\frac{r}{p}} > 1,$ it follows that $2\sqrt{\frac{r}{p}} > 2.$
Hence,$|\alpha| + |\beta| > 2.$

(C) For the roots of the quadratic equation $ax^2 + bx + c = 0$ to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square.
Here,$a = 6$,$b = -11$,and $c = \alpha$.
$D = (-11)^2 - 4(6)(\alpha) = 121 - 24\alpha$.
Since $\alpha$ is a positive integer,$121 - 24\alpha \ge 0$,which implies $24\alpha \le 121$,so $\alpha \le 5.04$. Thus,$\alpha \in \{1, 2, 3, 4, 5\}$.
We check each value:
If $\alpha = 1$,$D = 121 - 24(1) = 97$ (not a perfect square).
If $\alpha = 2$,$D = 121 - 24(2) = 121 - 48 = 73$ (not a perfect square).
If $\alpha = 3$,$D = 121 - 24(3) = 121 - 72 = 49 = 7^2$ (perfect square).
If $\alpha = 4$,$D = 121 - 24(4) = 121 - 96 = 25 = 5^2$ (perfect square).
If $\alpha = 5$,$D = 121 - 24(5) = 121 - 120 = 1 = 1^2$ (perfect square).
The possible values for $\alpha$ are $3, 4, 5$. Therefore,there are $3$ such values.

(D) The given quadratic equation is $x^2 + (3 - \lambda)x + (2 - \lambda) = 0$.
Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = - (3 - \lambda) = \lambda - 3$ and $\alpha \beta = 2 - \lambda$.
The sum of the squares of the roots is $S = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta$.
Substituting the values,we get $S = (\lambda - 3)^2 - 2(2 - \lambda)$.
$S = \lambda^2 - 6 \lambda + 9 - 4 + 2 \lambda$.
$S = \lambda^2 - 4 \lambda + 5$.
To find the least value,we complete the square: $S = (\lambda - 2)^2 + 1$.
The expression $S$ has the least value when $\lambda - 2 = 0$,which gives $\lambda = 2$.

(C) The given quadratic equation is $x^2 \sin \theta - x(\sin \theta \cos \theta + 1) + \cos \theta = 0$.
Using the quadratic formula,$x = \frac{(\sin \theta \cos \theta + 1) \pm \sqrt{(\sin \theta \cos \theta + 1)^2 - 4 \sin \theta \cos \theta}}{2 \sin \theta}$.
Simplifying the discriminant: $(\sin \theta \cos \theta + 1)^2 - 4 \sin \theta \cos \theta = (\sin \theta \cos \theta - 1)^2$.
Thus,$x = \frac{\sin \theta \cos \theta + 1 \pm (\sin \theta \cos \theta - 1)}{2 \sin \theta}$.
This gives $x_1 = \frac{2 \sin \theta \cos \theta}{2 \sin \theta} = \cos \theta$ and $x_2 = \frac{2}{2 \sin \theta} = \csc \theta$.
Since $0 < \theta < 45^\circ$,$\cos \theta > \sin \theta$,so $\cos \theta < \csc \theta$. Thus $\alpha = \cos \theta$ and $\beta = \csc \theta$.
The sum is $S = \sum_{n=0}^\infty \alpha^n + \sum_{n=0}^\infty (-\frac{1}{\beta})^n = \frac{1}{1 - \alpha} + \frac{1}{1 + 1/\beta} = \frac{1}{1 - \cos \theta} + \frac{1}{1 + \sin \theta}$.

(C) Let $t = \sqrt{x}$,where $t > 0$.
The equation becomes $|t - 2| + t(t - 4) + 2 = 0$.
$|t - 2| + t^2 - 4t + 2 = 0$.
We can rewrite $t^2 - 4t + 2$ as $(t^2 - 4t + 4) - 2 = (t - 2)^2 - 2$.
So,$|t - 2| + (t - 2)^2 - 2 = 0$.
Let $u = |t - 2|$,then $u^2 + u - 2 = 0$.
$(u + 2)(u - 1) = 0$.
Since $u = |t - 2| \ge 0$,we must have $u = 1$.
$|t - 2| = 1 \implies t - 2 = 1$ or $t - 2 = -1$.
$t = 3$ or $t = 1$.
Since $t = \sqrt{x}$,we have $\sqrt{x} = 3 \implies x = 9$ and $\sqrt{x} = 1 \implies x = 1$.
The sum of the solutions is $9 + 1 = 10$.

(A) For the quadratic equation $ax^2 + bx + c = 0$ to have no real roots,the discriminant $D$ must be less than $0$.
Here,$a = (1 + m^2)$,$b = -2(1 + 3m)$,and $c = (1 + 8m)$.
$D = b^2 - 4ac = [-2(1 + 3m)]^2 - 4(1 + m^2)(1 + 8m) < 0$.
$D = 4(1 + 9m^2 + 6m) - 4(1 + 8m + m^2 + 8m^3) < 0$.
$D = 4(1 + 9m^2 + 6m - 1 - 8m - m^2 - 8m^3) < 0$.
$D = 4(-8m^3 + 8m^2 - 2m) < 0$.
$D = -8m(4m^2 - 4m + 1) < 0$.
$D = -8m(2m - 1)^2 < 0$.
Since $(2m - 1)^2 \ge 0$,for $D < 0$,we must have $-8m < 0$ and $2m - 1 \neq 0$.
This implies $m > 0$ and $m \neq \frac{1}{2}$.
Since there are infinitely many integers $m > 0$ (excluding $m = 0.5$,which is not an integer),there are infinitely many integral values of $m$.

(B) Given that $p, q \in \mathbb{Q}$ and $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$.
Since the coefficients are rational,the irrational roots must occur in conjugate pairs.
Therefore,the other root is $2 + \sqrt{3}$.
Sum of roots $= (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$.
From the equation $x^2 + px + q = 0$,the sum of roots $= -p$.
So,$-p = 4 \implies p = -4$.
Product of roots $= (2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.
From the equation,the product of roots $= q$.
So,$q = 1$.
Now,check the options:
For option $(B)$,$p^2 - 4q - 12 = (-4)^2 - 4(1) - 12 = 16 - 4 - 12 = 0$.
Thus,option $(B)$ is correct.

(D) Let $2^x = t$. Since $2^x > 0$,we must have $t > 0$.
The equation becomes $5 + |t - 1| = t(t - 2) = t^2 - 2t$.
Rearranging,we get $|t - 1| = t^2 - 2t - 5$.
Let $g(t) = |t - 1|$ and $f(t) = t^2 - 2t - 5$.
We look for solutions where $t > 0$.
Case $1$: $t \ge 1$.
$t - 1 = t^2 - 2t - 5$ $\Rightarrow t^2 - 3t - 4 = 0$ $\Rightarrow (t - 4)(t + 1) = 0$.
Since $t \ge 1$,we have $t = 4$. Thus $2^x = 4 \Rightarrow x = 2$.
Case $2$: $0 < t < 1$.
$-(t - 1) = t^2 - 2t - 5$ $\Rightarrow -t + 1 = t^2 - 2t - 5$ $\Rightarrow t^2 - t - 6 = 0$ $\Rightarrow (t - 3)(t + 2) = 0$.
Neither $t = 3$ nor $t = -2$ satisfy $0 < t < 1$.
Thus,there is only $1$ real root.

(D) Let $3^{x} = t$,where $t > 0$.
The equation becomes $t(t-1) + 2 = |t-1| + |t-2|$.
$t^{2} - t + 2 = |t-1| + |t-2|$.
Case-$I$: $0 < t < 1$.
$t^{2} - t + 2 = (1 - t) + (2 - t) = 3 - 2t$.
$t^{2} + t - 1 = 0$.
$t = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Since $t > 0$,$t = \frac{\sqrt{5}-1}{2} \approx 0.618$,which satisfies $0 < t < 1$.
Case-$II$: $1 \leq t < 2$.
$t^{2} - t + 2 = (t - 1) + (2 - t) = 1$.
$t^{2} - t + 1 = 0$.
The discriminant $D = (-1)^{2} - 4(1)(1) = -3 < 0$,so no real solution.
Case-$III$: $t \geq 2$.
$t^{2} - t + 2 = (t - 1) + (t - 2) = 2t - 3$.
$t^{2} - 3t + 5 = 0$.
The discriminant $D = (-3)^{2} - 4(1)(5) = 9 - 20 = -11 < 0$,so no real solution.
Thus,the only valid value is $t = \frac{\sqrt{5}-1}{2}$.
Since $3^{x} = \frac{\sqrt{5}-1}{2}$,there is exactly one real value for $x$,which is $x = \log_{3}\left(\frac{\sqrt{5}-1}{2}\right)$.
Therefore,$S$ is a singleton set.

(C) The given equation is $2x^{2} + (a-10)x + (\frac{33}{2} - 2a) = 0$.
For real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = (a-10)^{2} - 4(2)(\frac{33}{2} - 2a) \geq 0$.
$D = a^{2} - 20a + 100 - 8(\frac{33-4a}{2}) \geq 0$.
$D = a^{2} - 20a + 100 - 4(33-4a) \geq 0$.
$D = a^{2} - 20a + 100 - 132 + 16a \geq 0$.
$D = a^{2} - 4a - 32 \geq 0$.
$(a-8)(a+4) \geq 0$.
This inequality holds for $a \in (-\infty, -4] \cup [8, \infty)$.
Since we are looking for the least positive value of $a$,we consider the interval $[8, \infty)$.
The least positive value is $8$.

(B) For the equation $a x^{2}-2 b x+5=0$,the roots are $\alpha, \alpha$.
Thus,the sum of roots $2\alpha = \frac{2b}{a} \Rightarrow \alpha = \frac{b}{a}$ and the product of roots $\alpha^{2} = \frac{5}{a}$.
From these,$b = a\alpha$ and $a = \frac{5}{\alpha^{2}}$. Substituting $a$,we get $b = \frac{5}{\alpha}$.
Since $\alpha$ is also a root of $x^{2}-2 b x-10=0$,we have $\alpha^{2}-2 b \alpha-10=0$.
Substituting $b = \frac{5}{\alpha}$ into this equation: $\alpha^{2}-2(\frac{5}{\alpha})\alpha-10=0$ $\Rightarrow \alpha^{2}-10-10=0$ $\Rightarrow \alpha^{2}=20$.
Now,for the equation $x^{2}-2 b x-10=0$,the product of roots $\alpha \beta = -10$.
Since $\alpha^{2} = 20$,we have $\alpha = \pm \sqrt{20}$.
Then $\beta = \frac{-10}{\alpha}$.
Thus,$\beta^{2} = \frac{100}{\alpha^{2}} = \frac{100}{20} = 5$.
Therefore,$\alpha^{2}+\beta^{2} = 20 + 5 = 25$.

(D) Given equation: $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$.
Divide the entire equation by $e^{2x}$ (since $e^{2x} \neq 0$):
$e^{2x} + e^x - 4 + \frac{1}{e^x} + \frac{1}{e^{2x}} = 0$.
Rearrange the terms:
$(e^{2x} + \frac{1}{e^{2x}}) + (e^x + \frac{1}{e^x}) - 4 = 0$.
Using the identity $a^2 + \frac{1}{a^2} = (a + \frac{1}{a})^2 - 2$,we get:
$(e^x + \frac{1}{e^x})^2 - 2 + (e^x + \frac{1}{e^x}) - 4 = 0$.
Let $t = e^x + \frac{1}{e^x}$. Since $e^x > 0$,by $AM$-$GM$ inequality,$t = e^x + \frac{1}{e^x} \geq 2$.
The equation becomes $t^2 + t - 6 = 0$.
Factoring the quadratic: $(t + 3)(t - 2) = 0$.
This gives $t = -3$ or $t = 2$.
Since $t \geq 2$,we must have $t = 2$.
$e^x + \frac{1}{e^x} = 2$ $\Rightarrow e^{2x} - 2e^x + 1 = 0$ $\Rightarrow (e^x - 1)^2 = 0$.
$e^x = 1 \Rightarrow x = 0$.
Thus,there is only $1$ real root.

(D) The given function is $f(x) = 9x^{2} + 12x + 2$.
We can rewrite this by completing the square:
$f(x) = (3x)^{2} + 2(3x)(2) + 2^{2} - 2^{2} + 2$
$f(x) = (3x + 2)^{2} - 4 + 2$
$f(x) = (3x + 2)^{2} - 2$.
Since $(3x + 2)^{2} \geq 0$ for all $x \in \mathbb{R}$,the minimum value of $(3x + 2)^{2}$ is $0$.
Therefore,the minimum value of $f(x)$ is $0 - 2 = -2$,which occurs when $3x + 2 = 0$,i.e.,$x = -\frac{2}{3}$.
As $x \to \infty$,$f(x) \to \infty$,so the function does not have a maximum value.

(B) The given quadratic equation is $x^{2}+x+1=0$.
Comparing this with $ax^{2}+bx+c=0$,we get $a=1, b=1, c=1$.
The discriminant is $D = b^{2}-4ac = 1^{2}-4(1)(1) = 1-4 = -3$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$,we get:
$x = \frac{-1 \pm \sqrt{-3}}{2(1)} = \frac{-1 \pm \sqrt{3}i}{2}$.

(A) Given the quadratic equation $\sqrt{5} x^{2} + x + \sqrt{5} = 0$.
Comparing this with $ax^{2} + bx + c = 0$,we have $a = \sqrt{5}$,$b = 1$,and $c = \sqrt{5}$.
The discriminant $D = b^{2} - 4ac$ is calculated as:
$D = (1)^{2} - 4(\sqrt{5})(\sqrt{5}) = 1 - 4(5) = 1 - 20 = -19$.
The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.
Substituting the values,we get $x = \frac{-1 \pm \sqrt{-19}}{2\sqrt{5}}$.
Since $\sqrt{-1} = i$,the solutions are $x = \frac{-1 \pm \sqrt{19} i}{2\sqrt{5}}$.

(A) The given quadratic equation is $2x^{2}+x+1=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=2, b=1, c=1$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = 1^{2}-4(2)(1) = 1-8 = -7$.
The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.
Substituting the values,we get $x = \frac{-1 \pm \sqrt{-7}}{2(2)} = \frac{-1 \pm \sqrt{7}i}{4}$ (since $\sqrt{-1} = i$).

(A) The given quadratic equation is $x^{2}+3x+9=0$.
On comparing the given equation with $ax^{2}+bx+c=0$,we obtain $a=1, b=3,$ and $c=9$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = 3^{2}-4(1)(9) = 9-36 = -27$.
The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.
$x = \frac{-3 \pm \sqrt{-27}}{2(1)} = \frac{-3 \pm \sqrt{27}i}{2} = \frac{-3 \pm 3\sqrt{3}i}{2}$.

(D) The given quadratic equation is $-x^{2}+x-2=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=-1, b=1, c=-2$.
The discriminant $D$ is given by $D = b^{2}-4ac = (1)^{2}-4(-1)(-2) = 1-8 = -7$.
The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.
Substituting the values,$x = \frac{-1 \pm \sqrt{-7}}{2(-1)} = \frac{-1 \pm \sqrt{7}i}{-2}$.
This can be simplified as $x = \frac{1 \pm \sqrt{7}i}{2}$.

(A) The given quadratic equation is $x^{2}+3x+5=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=3, c=5$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = 3^{2}-4(1)(5) = 9-20 = -11$.
The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.
$x = \frac{-3 \pm \sqrt{-11}}{2(1)} = \frac{-3 \pm \sqrt{11}i}{2}$ (since $\sqrt{-1} = i$).

(A) The given quadratic equation is $x^{2}-x+2=0$.
On comparing the given equation with $ax^{2}+bx+c=0$,we obtain $a=1, b=-1,$ and $c=2$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = (-1)^{2}-4(1)(2) = 1-8 = -7$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$,we get:
$x = \frac{-(-1) \pm \sqrt{-7}}{2(1)} = \frac{1 \pm \sqrt{7}i}{2}$.

(A) The given quadratic equation is $\sqrt{2} x^{2} + x + \sqrt{2} = 0$.
On comparing the given equation with $ax^{2} + bx + c = 0$,we obtain $a = \sqrt{2}$,$b = 1$,and $c = \sqrt{2}$.
The discriminant $D$ is given by $D = b^{2} - 4ac$.
$D = (1)^{2} - 4(\sqrt{2})(\sqrt{2}) = 1 - 8 = -7$.
The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.
$x = \frac{-1 \pm \sqrt{-7}}{2(\sqrt{2})} = \frac{-1 \pm \sqrt{7}i}{2\sqrt{2}}$.

(A) The given quadratic equation is $\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0$.
Comparing this with $a x^{2}+b x+c=0$,we get $a=\sqrt{3}, b=-\sqrt{2}, c=3 \sqrt{3}$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = (-\sqrt{2})^{2} - 4(\sqrt{3})(3 \sqrt{3}) = 2 - 4(3)(3) = 2 - 36 = -34$.
The roots are given by $x = \frac{-b \pm \sqrt{D}}{2a}$.
$x = \frac{-(-\sqrt{2}) \pm \sqrt{-34}}{2 \sqrt{3}} = \frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}}$ (since $\sqrt{-1} = i$).

(A) The given quadratic equation is $x^{2}+x+\frac{1}{\sqrt{2}}=0$.
Multiplying by $\sqrt{2}$,we get $\sqrt{2}x^{2}+\sqrt{2}x+1=0$.
Comparing with $ax^{2}+bx+c=0$,we have $a=\sqrt{2}$,$b=\sqrt{2}$,and $c=1$.
The discriminant $D = b^{2}-4ac = (\sqrt{2})^{2}-4(\sqrt{2})(1) = 2-4\sqrt{2}$.
Since $D < 0$,the roots are given by $x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-\sqrt{2} \pm \sqrt{2-4\sqrt{2}}}{2\sqrt{2}}$.
$x = \frac{-\sqrt{2} \pm i\sqrt{4\sqrt{2}-2}}{2\sqrt{2}} = \frac{-\sqrt{2} \pm i\sqrt{2(2\sqrt{2}-1)}}{2\sqrt{2}}$.
$x = \frac{-\sqrt{2} \pm i\sqrt{2}\sqrt{2\sqrt{2}-1}}{2\sqrt{2}} = \frac{-1 \pm i\sqrt{2\sqrt{2}-1}}{2}$.

(A) The given quadratic equation is $x^{2}+\frac{x}{\sqrt{2}}+1=0$.
Multiply the entire equation by $\sqrt{2}$ to simplify: $\sqrt{2}x^{2}+x+\sqrt{2}=0$.
Comparing this with the standard quadratic form $ax^{2}+bx+c=0$,we get $a=\sqrt{2}$,$b=1$,and $c=\sqrt{2}$.
The discriminant $D$ is given by $D = b^{2}-4ac = (1)^{2}-4(\sqrt{2})(\sqrt{2}) = 1-8 = -7$.
The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.
Substituting the values,we get $x = \frac{-1 \pm \sqrt{-7}}{2\sqrt{2}} = \frac{-1 \pm \sqrt{7}i}{2\sqrt{2}}$ (since $\sqrt{-1}=i$).

(A) The given quadratic equation is $3x^{2} - 4x + \frac{20}{3} = 0$.
Multiplying the entire equation by $3$,we get $9x^{2} - 12x + 20 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we have $a = 9$,$b = -12$,and $c = 20$.
The discriminant $D$ is given by $D = b^{2} - 4ac = (-12)^{2} - 4(9)(20) = 144 - 720 = -576$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$,we get:
$x = \frac{-(-12) \pm \sqrt{-576}}{2(9)} = \frac{12 \pm \sqrt{576}i}{18} = \frac{12 \pm 24i}{18}$.
Simplifying the expression,we get $x = \frac{12}{18} \pm \frac{24}{18}i = \frac{2}{3} \pm \frac{4}{3}i$.

(A) The given quadratic equation is $x^{2}-2x+\frac{3}{2}=0$.
Multiplying by $2$,we get $2x^{2}-4x+3=0$.
Comparing this with $ax^{2}+bx+c=0$,we have $a=2, b=-4, c=3$.
The discriminant $D = b^{2}-4ac = (-4)^{2}-4(2)(3) = 16-24 = -8$.
The roots are given by $x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-4) \pm \sqrt{-8}}{2(2)}$.
$x = \frac{4 \pm 2\sqrt{2}i}{4} = \frac{4}{4} \pm \frac{2\sqrt{2}i}{4} = 1 \pm \frac{\sqrt{2}}{2}i$.

(A) The given quadratic equation is $27 x^{2}-10 x+1=0$.
On comparing this equation with $a x^{2}+b x+c=0$,we obtain $a=27, b=-10$,and $c=1$.
Therefore,the discriminant $D$ is given by $D=b^{2}-4 a c = (-10)^{2}-4(27)(1) = 100-108 = -8$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2 a}$,we get:
$x = \frac{-(-10) \pm \sqrt{-8}}{2(27)} = \frac{10 \pm 2 \sqrt{2} i}{54}$.
Dividing the numerator and denominator by $2$,we obtain $x = \frac{5 \pm \sqrt{2} i}{27}$.

(A) The given quadratic equation is $21x^{2} - 28x + 10 = 0$.
Comparing this with $ax^{2} + bx + c = 0$,we get $a = 21$,$b = -28$,and $c = 10$.
The discriminant $D$ is given by $D = b^{2} - 4ac$.
$D = (-28)^{2} - 4(21)(10) = 784 - 840 = -56$.
The roots are given by $x = \frac{-b \pm \sqrt{D}}{2a}$.
$x = \frac{-(-28) \pm \sqrt{-56}}{2(21)} = \frac{28 \pm \sqrt{56}i}{42}$.
Since $\sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}$,we have:
$x = \frac{28 \pm 2\sqrt{14}i}{42} = \frac{28}{42} \pm \frac{2\sqrt{14}}{42}i = \frac{2}{3} \pm \frac{\sqrt{14}}{21}i$.

(C) Let the quadratic polynomial be $f(x) = a(x - 3)(x - \alpha)$,where $\alpha$ is the other root.
Given $f(2) = a(2 - 3)(2 - \alpha) = a(-1)(2 - \alpha) = a(\alpha - 2)$.
Given $f(-1) = a(-1 - 3)(-1 - \alpha) = a(-4)(-1 - \alpha) = 4a(1 + \alpha)$.
Since $f(-1) + f(2) = 0$,we have $4a(1 + \alpha) + a(\alpha - 2) = 0$.
Since $a \neq 0$,we can divide by $a$: $4 + 4\alpha + \alpha - 2 = 0$.
$5\alpha + 2 = 0$ $\Rightarrow 5\alpha = -2$ $\Rightarrow \alpha = -\frac{2}{5} = -0.4$.
Thus,the other root $\alpha = -0.4$ lies in the interval $(-1, 0)$.

(B) Given equation: $9x^{2}-18|x|+5=0$
Since $x^{2} = |x|^{2}$,the equation becomes $9|x|^{2}-18|x|+5=0$.
Let $t = |x|$,then $9t^{2}-18t+5=0$.
Factoring the quadratic: $9t^{2}-15t-3t+5=0$.
$3t(3t-5)-1(3t-5)=0$.
$(3t-1)(3t-5)=0$.
So,$|x| = \frac{1}{3}$ or $|x| = \frac{5}{3}$.
The roots are $x = \frac{1}{3}, -\frac{1}{3}, \frac{5}{3}, -\frac{5}{3}$.
The product of the roots is $(\frac{1}{3}) \times (-\frac{1}{3}) \times (\frac{5}{3}) \times (-\frac{5}{3}) = \frac{25}{81}$.

(B) $1$. The graph is a parabola opening downwards. This indicates that the coefficient of $x^2$ is negative,so $a < 0$.
$2$. The graph intersects the $x$-axis at two distinct points. This indicates that the quadratic equation has two distinct real roots,which means the discriminant $D = b^2 - 4ac > 0$.
$3$. Combining these two observations,we get $a < 0$ and $D > 0$.