Solution of quadratic equations and Nature of roots Questions in English

(D) Given equation: $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ $(i)$
For the square roots to be defined,we must have $x + 1 \ge 0$,$x - 1 \ge 0$,and $4x - 1 \ge 0$,which implies $x \ge 1$.
Squaring both sides of $(i)$:
$(x + 1) + (x - 1) - 2\sqrt{(x + 1)(x - 1)} = 4x - 1$
$2x - 2\sqrt{x^2 - 1} = 4x - 1$
$-2\sqrt{x^2 - 1} = 2x - 1$
Since $x \ge 1$,the left side $-2\sqrt{x^2 - 1}$ is $\le 0$,while the right side $2x - 1$ is $\ge 2(1) - 1 = 1$.
Since a non-positive number cannot equal a positive number,there is no real value of $x$ that satisfies the equation.
Thus,the equation has no solution.

(C) Given equation: $5^{x - 1} + 5 \cdot (0.2)^{x - 2} = 26$
Since $0.2 = \frac{1}{5} = 5^{-1}$,we have:
$5^{x - 1} + 5 \cdot (5^{-1})^{x - 2} = 26$
$5^{x - 1} + 5 \cdot 5^{-(x - 2)} = 26$
$5^{x - 1} + 5 \cdot 5^{2 - x} = 26$
$5^{x - 1} + 5^{3 - x} = 26$
$5^{x - 1} + \frac{5^3}{5^x} = 26$
$5^{x - 1} + \frac{125}{5^x} = 26$
Let $y = 5^{x - 1}$. Then $5^x = 5y$.
$y + \frac{125}{5y} = 26$
$y + \frac{25}{y} = 26$
$y^2 - 26y + 25 = 0$
$(y - 25)(y - 1) = 0$
So,$y = 25$ or $y = 1$.
If $5^{x - 1} = 25 = 5^2$,then $x - 1 = 2$,so $x = 3$.
If $5^{x - 1} = 1 = 5^0$,then $x - 1 = 0$,so $x = 1$.
There are $2$ values of $x$ ($1$ and $3$) that satisfy the equation.

(B) Given $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$.
Since $xy = 4$,we have $y = \frac{4}{x} = \frac{4}{\sqrt{7} + \sqrt{3}}$.
Rationalizing the denominator: $y = \frac{4(\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{4(\sqrt{7} - \sqrt{3})}{4} = \sqrt{7} - \sqrt{3}$.
Now,$x + y = (\sqrt{7} + \sqrt{3}) + (\sqrt{7} - \sqrt{3}) = 2\sqrt{7}$.
We know that $x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2$.
First,calculate $x^2 + y^2 = (x + y)^2 - 2xy = (2\sqrt{7})^2 - 2(4) = 28 - 8 = 20$.
Then,$x^4 + y^4 = (20)^2 - 2(xy)^2 = 400 - 2(4)^2 = 400 - 2(16) = 400 - 32 = 368$.

(B) Given equation: $\sqrt{x + 10} + \sqrt{x - 2} = 6$
Rearranging the terms: $\sqrt{x + 10} = 6 - \sqrt{x - 2}$
Squaring both sides: $x + 10 = (6 - \sqrt{x - 2})^2$
$x + 10 = 36 + (x - 2) - 12\sqrt{x - 2}$
$x + 10 = 34 + x - 12\sqrt{x - 2}$
Subtracting $x$ from both sides: $10 = 34 - 12\sqrt{x - 2}$
$-24 = -12\sqrt{x - 2}$
Dividing by $-12$: $2 = \sqrt{x - 2}$
Squaring both sides again: $4 = x - 2$
$x = 6$
Verification: $\sqrt{6 + 10} + \sqrt{6 - 2} = \sqrt{16} + \sqrt{4} = 4 + 2 = 6$. This satisfies the equation.

(A) Let the remainder be $Q(x) = ax^2 + bx + c$. Since the divisor is a cubic polynomial,the remainder must be at most a quadratic polynomial.
By the Remainder Theorem,we have:
$f(-1) = 6 \implies Q(-1) = a - b + c = 6$ $(i)$
$f(2) = 3 \implies Q(2) = 4a + 2b + c = 3$ (ii)
$f(-2) = 15 \implies Q(-2) = 4a - 2b + c = 15$ (iii)
Subtracting (iii) from (ii): $(4a + 2b + c) - (4a - 2b + c) = 3 - 15 \implies 4b = -12 \implies b = -3$.
Substituting $b = -3$ into (ii) and (iii):
$4a - 6 + c = 3 \implies 4a + c = 9$ (iv)
$4a + 6 + c = 15 \implies 4a + c = 9$ (consistent).
Substituting $b = -3$ into $(i)$: $a - (-3) + c = 6 \implies a + c = 3$ $(v)$.
Subtracting $(v)$ from (iv): $(4a + c) - (a + c) = 9 - 3 \implies 3a = 6 \implies a = 2$.
Substituting $a = 2$ into $(v)$: $2 + c = 3 \implies c = 1$.
Thus,the remainder is $Q(x) = 2x^2 - 3x + 1$.

(D) Given equation: $4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0$
Let $y = 2^{(x^2 + 2)}$. Since $4^{(x^2 + 2)} = (2^2)^{(x^2 + 2)} = (2^{(x^2 + 2)})^2$,the equation becomes:
$y^2 - 9y + 8 = 0$
Factoring the quadratic equation:
$(y - 8)(y - 1) = 0$
So,$y = 8$ or $y = 1$.
Case $1$: $y = 8$
$2^{(x^2 + 2)} = 8 = 2^3$
$x^2 + 2 = 3$
$x^2 = 1$
$x = \pm 1$
Case $2$: $y = 1$
$2^{(x^2 + 2)} = 1 = 2^0$
$x^2 + 2 = 0$
$x^2 = -2$
This has no real solution.
Thus,the solutions are $x = 1$ and $x = -1$.

(A) Given $x = 2 + \sqrt{3}$ and $xy = 1$,we have $y = \frac{1}{x} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$.
Let $S = \frac{x}{\sqrt{2} + \sqrt{x}} + \frac{y}{\sqrt{2} - \sqrt{y}}$.
Rationalizing the denominators:
$S = \frac{x(\sqrt{2} - \sqrt{x})}{2 - x} + \frac{y(\sqrt{2} + \sqrt{y})}{2 - y}$.
Since $x - 2 = \sqrt{3}$ and $2 - y = \sqrt{3}$,we have:
$S = \frac{x(\sqrt{2} - \sqrt{x})}{-\sqrt{3}} + \frac{y(\sqrt{2} + \sqrt{y})}{\sqrt{3}} = \frac{1}{\sqrt{3}} [y\sqrt{2} + y\sqrt{y} - x\sqrt{2} + x\sqrt{x}]$.
$S = \frac{1}{\sqrt{3}} [\sqrt{2}(y - x) + (x\sqrt{x} + y\sqrt{y})]$.
Since $y - x = (2 - \sqrt{3}) - (2 + \sqrt{3}) = -2\sqrt{3}$,and $x\sqrt{x} + y\sqrt{y} = (2 + \sqrt{3})^{3/2} + (2 - \sqrt{3})^{3/2} = \frac{(\sqrt{3}+1)^3 + (\sqrt{3}-1)^3}{2\sqrt{2}} = \frac{6\sqrt{3} + 6\sqrt{3}}{2\sqrt{2}} = \frac{12\sqrt{3}}{2\sqrt{2}} = 3\sqrt{6}$.
$S = \frac{1}{\sqrt{3}} [\sqrt{2}(-2\sqrt{3}) + 3\sqrt{6}] = \frac{1}{\sqrt{3}} [-2\sqrt{6} + 3\sqrt{6}] = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$.

(C) Given the equation: $x^4 - x^3 + x - 1 = 0$
Factor by grouping:
$x^3(x - 1) + 1(x - 1) = 0$
$(x^3 + 1)(x - 1) = 0$
This gives two cases:
$1) x - 1 = 0 \implies x = 1$
$2) x^3 + 1 = 0 \implies x^3 = -1$
The roots of $x^3 = -1$ are $x = -1$ and the complex roots $\frac{1 \pm \sqrt{3}i}{2}$.
Thus,the roots of the equation are $1, -1, \frac{1 + \sqrt{3}i}{2}, \text{ and } \frac{1 - \sqrt{3}i}{2}$.
The real roots are $1$ and $-1$.

(C) Given equation is $a^2 - 2a\sin x + 1 = 0$.
For $a$ to be a real value,the discriminant $D$ of this quadratic equation in $a$ must be greater than or equal to $0$.
$D = (-2\sin x)^2 - 4(1)(1) = 4\sin^2 x - 4 = 4(\sin^2 x - 1) = -4\cos^2 x$.
Since $D \ge 0$,we must have $-4\cos^2 x \ge 0$,which implies $\cos^2 x \le 0$.
Since $\cos^2 x$ cannot be negative,the only possibility is $\cos^2 x = 0$,which means $\cos x = 0$.
If $\cos x = 0$,then $\sin x = 1$ or $\sin x = -1$.
Case $1$: If $\sin x = 1$,the equation becomes $a^2 - 2a + 1 = 0$,which is $(a - 1)^2 = 0$,giving $a = 1$.
Case $2$: If $\sin x = -1$,the equation becomes $a^2 + 2a + 1 = 0$,which is $(a + 1)^2 = 0$,giving $a = -1$.
Thus,the real values of $a$ are $1$ and $-1$.
There are $2$ such real values.

(D) Let the two numbers be $a$ and $b$.
Since one number is the reciprocal of the other,we have $b = \frac{1}{a}$.
The arithmetic mean of the two numbers is given by $\frac{a + b}{2} = \frac{13}{12}$.
Substituting $b = \frac{1}{a}$,we get $\frac{a + \frac{1}{a}}{2} = \frac{13}{12}$.
Multiplying by $2$,we get $a + \frac{1}{a} = \frac{13}{6}$.
Multiplying the entire equation by $6a$,we get $6a^2 + 6 = 13a$,which simplifies to the quadratic equation $6a^2 - 13a + 6 = 0$.
Factoring the quadratic equation: $6a^2 - 9a - 4a + 6 = 0 \Rightarrow 3a(2a - 3) - 2(2a - 3) = 0$.
This gives $(3a - 2)(2a - 3) = 0$.
Thus,$a = \frac{2}{3}$ or $a = \frac{3}{2}$.
If $a = \frac{3}{2}$,then $b = \frac{2}{3}$. If $a = \frac{2}{3}$,then $b = \frac{3}{2}$.
Therefore,the numbers are $\frac{3}{2}$ and $\frac{2}{3}$.

(C) The given quadratic equation is $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$.
Since the roots are equal,the discriminant $D = 0$.
Here,$A = a(b - c)$,$B = b(c - a)$,and $C = c(a - b)$.
$B^2 - 4AC = 0 \Rightarrow [b(c - a)]^2 - 4[a(b - c)][c(a - b)] = 0$.
$b^2(c - a)^2 - 4ac(b - c)(a - b) = 0$.
$b^2(c^2 - 2ac + a^2) - 4ac(ab - b^2 - ac + bc) = 0$.
$b^2c^2 - 2ab^2c + a^2b^2 - 4a^2bc + 4ab^2c + 4a^2c^2 - 4abc^2 = 0$.
$b^2c^2 + 2ab^2c + a^2b^2 - 4a^2bc - 4abc^2 + 4a^2c^2 = 0$.
$(bc + ab - 2ac)^2 = 0$.
$bc + ab = 2ac$.
Dividing by $abc$,we get $\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$.
This implies that $a, b, c$ are in $H.P.$

(B) Given the quadratic equation: ${x^2} - 2Ax + {G^2} = 0$ $(i)$
Let $a$ and $b$ be two positive numbers such that their arithmetic mean is $A$ and geometric mean is $G$.
Then,$A = \frac{a + b}{2}$ and $G^2 = ab$.
Substituting these into equation $(i)$,we get: ${x^2} - (a + b)x + ab = 0$.
Factoring the quadratic equation: ${x^2} - ax - bx + ab = 0$ $\Rightarrow x(x - a) - b(x - a) = 0$ $\Rightarrow (x - a)(x - b) = 0$.
Thus,the roots of the equation are $a$ and $b$.
For real roots $a$ and $b$ to exist,the discriminant $D$ of the quadratic equation must be non-negative: $D = (-2A)^2 - 4(1)(G^2) \ge 0$.
$4A^2 - 4G^2 \ge 0 \Rightarrow A^2 \ge G^2$.
Since $A$ and $G$ are means of positive numbers,$A, G > 0$,so $A \ge G$.
Specifically,$A - G = \frac{a + b}{2} - \sqrt{ab} = \frac{1}{2}(\sqrt{a} - \sqrt{b})^2 \ge 0$.
Therefore,$A \ge G$.

(A) Given $p, q, r$ are in $A.P.$ and are positive.
Therefore,$q = \frac{p + r}{2}$ ......$(i)$
For the roots of $px^2 + qx + r = 0$ to be real,the discriminant $D \ge 0$.
$D = q^2 - 4pr \ge 0$
Substituting $(i)$ into the inequality:
$\left( \frac{p + r}{2} \right)^2 - 4pr \ge 0$
$p^2 + r^2 + 2pr - 16pr \ge 0$
$p^2 + r^2 - 14pr \ge 0$
Dividing by $p^2$ (since $p > 0$):
$1 + \left( \frac{r}{p} \right)^2 - 14\left( \frac{r}{p} \right) \ge 0$
$\left( \frac{r}{p} \right)^2 - 14\left( \frac{r}{p} \right) + 1 \ge 0$
Completing the square:
$\left( \frac{r}{p} - 7 \right)^2 - 49 + 1 \ge 0$
$\left( \frac{r}{p} - 7 \right)^2 \ge 48$
$\left( \frac{r}{p} - 7 \right)^2 \ge (4\sqrt{3})^2$
Taking the square root on both sides:
$\left| \frac{r}{p} - 7 \right| \ge 4\sqrt{3}$.

(A) Given $a + b + c + d = 2$. Let $x = a + b$ and $y = c + d$. Then $x + y = 2$ and $M = xy$.
By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$,we have $\frac{x + y}{2} \ge \sqrt{xy}$.
Substituting the values,$\frac{2}{2} \ge \sqrt{M}$,which implies $1 \ge \sqrt{M}$,or $M \le 1$.
Since $a, b, c, d$ are positive real numbers,$x > 0$ and $y > 0$,so $M = xy > 0$.
Thus,the relation is $0 < M \le 1$.

(A) Given equation: $a(x^2 + 1) - (a^2 + 1)x = 0$
Expanding the terms: $ax^2 + a - a^2x - x = 0$
Rearranging the terms: $ax^2 - a^2x - x + a = 0$
Factoring by grouping: $ax(x - a) - 1(x - a) = 0$
$(ax - 1)(x - a) = 0$
Setting each factor to zero: $ax - 1 = 0$ or $x - a = 0$
Thus,$x = \frac{1}{a}$ or $x = a$.

(B) Given equation: $x^4 - 8x^2 - 9 = 0$
Let $y = x^2$,then the equation becomes $y^2 - 8y - 9 = 0$.
Factoring the quadratic equation: $y^2 - 9y + y - 9 = 0$
$y(y - 9) + 1(y - 9) = 0$
$(y + 1)(y - 9) = 0$
So,$y = -1$ or $y = 9$.
Substituting $y = x^2$ back:
Case $1$: $x^2 = -1 \Rightarrow x = \pm i$
Case $2$: $x^2 = 9 \Rightarrow x = \pm 3$
Thus,the roots are $x = \pm 3, \pm i$.

(C) Given equation: $x^{2/3} + x^{1/3} - 2 = 0$
Let $a = x^{1/3}$. Substituting this into the equation,we get:
$a^2 + a - 2 = 0$
Factoring the quadratic equation:
$(a + 2)(a - 1) = 0$
So,$a = 1$ or $a = -2$.
Case $1$: If $a = 1$,then $x^{1/3} = 1$,which implies $x = 1^3 = 1$.
Case $2$: If $a = -2$,then $x^{1/3} = -2$,which implies $x = (-2)^3 = -8$.
Thus,the roots are $x = 1, -8$.

(B) Given $x = 2 + 2^{2/3} + 2^{1/3}$.
Subtracting $2$ from both sides,we get $x - 2 = 2^{2/3} + 2^{1/3}$.
Cubing both sides,we have $(x - 2)^3 = (2^{2/3} + 2^{1/3})^3$.
Using the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$,we get:
$x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = (2^{2/3})^3 + (2^{1/3})^3 + 3(2^{2/3})(2^{1/3})(2^{2/3} + 2^{1/3})$.
$x^3 - 6x^2 + 12x - 8 = 4 + 2 + 3(2^1)(x - 2)$.
$x^3 - 6x^2 + 12x - 8 = 6 + 6x - 12$.
$x^3 - 6x^2 + 12x - 8 = 6x - 6$.
Rearranging the terms,$x^3 - 6x^2 + 6x = 8 - 6 = 2$.

(D) Given equation is $\sqrt{3x + 1} + 1 = \sqrt{x}$.
Subtracting $1$ from both sides,we get $\sqrt{3x + 1} = \sqrt{x} - 1$.
Squaring both sides,we get $3x + 1 = x + 1 - 2\sqrt{x}$.
Simplifying,we get $2x = -2\sqrt{x}$,which implies $x = -\sqrt{x}$.
Squaring again,$x^2 = x$,so $x^2 - x = 0$,which gives $x(x - 1) = 0$,so $x = 0$ or $x = 1$.
Checking $x = 0$: $\sqrt{3(0) + 1} + 1 = 1 + 1 = 2$,while $\sqrt{0} = 0$. Since $2 \neq 0$,$x = 0$ is not a solution.
Checking $x = 1$: $\sqrt{3(1) + 1} + 1 = 2 + 1 = 3$,while $\sqrt{1} = 1$. Since $3 \neq 1$,$x = 1$ is not a solution.
Therefore,there are no real roots for the given equation.

(B) Let the required number be $x$.
According to the problem,$x = \sqrt{x} + 12$.
Rearranging the equation,we get $x - 12 = \sqrt{x}$.
Squaring both sides,we get $(x - 12)^2 = x$.
$x^2 - 24x + 144 = x$.
$x^2 - 25x + 144 = 0$.
Factoring the quadratic equation: $(x - 16)(x - 9) = 0$.
So,$x = 16$ or $x = 9$.
Check for $x = 16$: $16 - \sqrt{16} = 16 - 4 = 12$. This satisfies the condition.
Check for $x = 9$: $9 - \sqrt{9} = 9 - 3 = 6 \neq 12$. This does not satisfy the condition.
Therefore,the correct number is $16$.

(B) The given equation is $3^{2x} - 10 \cdot 3^x + 9 = 0$,which can be written as $(3^x)^2 - 10(3^x) + 9 = 0$.
Let $a = 3^x$. Then the equation reduces to $a^2 - 10a + 9 = 0$.
Factoring the quadratic equation,we get $(a - 9)(a - 1) = 0$,which gives $a = 9$ or $a = 1$.
Case $1$: If $a = 9$,then $3^x = 9 = 3^2$,so $x = 2$.
Case $2$: If $a = 1$,then $3^x = 1 = 3^0$,so $x = 0$.
Thus,the roots of the equation are $0$ and $2$.

(C) Given: ${x^2} + {y^2} = 25$ and ${xy} = 12$.
From the second equation,$y = \frac{12}{x}$.
Substituting this into the first equation: ${x^2} + \left( \frac{12}{x} \right)^2 = 25$.
${x^2} + \frac{144}{x^2} = 25$.
Multiplying by ${x^2}$: ${x^4} + 144 = 25{x^2}$.
Rearranging: ${x^4} - 25{x^2} + 144 = 0$.
Let ${x^2} = t$,then ${t^2} - 25t + 144 = 0$.
Factoring: $(t - 16)(t - 9) = 0$.
So,${x^2} = 16$ or ${x^2} = 9$.
Therefore,$x = \pm 4$ or $x = \pm 3$.
The set of possible values for $x$ is $\{3, 4, -3, -4\}$.

(A) Given that $a, b, c$ are integers,the coefficients of the quadratic equation $ax^2 + bx + c = 0$ are rational.
If one root of a quadratic equation with rational coefficients is of the form $\alpha + \sqrt{\beta}$ (where $\sqrt{\beta}$ is an irrational surd),then the other root must be its conjugate,$\alpha - \sqrt{\beta}$.
Here,one root is $3 + \sqrt{5}$.
Therefore,the other root is $3 - \sqrt{5}$.

(D) Given the equation $|x|^2 - 3|x| + 2 = 0$.
Let $|x| = t$,where $t \ge 0$.
The equation becomes $t^2 - 3t + 2 = 0$.
Factoring the quadratic equation: $(t - 1)(t - 2) = 0$.
This gives $t = 1$ or $t = 2$.
Since $t = |x|$,we have $|x| = 1$ or $|x| = 2$.
For $|x| = 1$,the solutions are $x = 1$ and $x = -1$.
For $|x| = 2$,the solutions are $x = 2$ and $x = -2$.
Thus,the real solutions are $x \in \{1, -1, 2, -2\}$.
The total number of real solutions is $4$.

(B) The given equation is $|x^2 + 4x + 3| + 2x + 5 = 0$.
Case $I$: $x^2 + 4x + 3 \ge 0$,which implies $(x+1)(x+3) \ge 0$,so $x \in (-\infty, -3] \cup [-1, \infty)$.
The equation becomes $x^2 + 4x + 3 + 2x + 5 = 0$,which simplifies to $x^2 + 6x + 8 = 0$.
Factoring gives $(x+2)(x+4) = 0$,so $x = -2$ or $x = -4$.
Checking the condition: $x = -4$ satisfies $x \in (-\infty, -3]$,but $x = -2$ does not satisfy $x \in [-1, \infty)$. Thus,$x = -4$ is a solution.
Case $II$: $x^2 + 4x + 3 < 0$,which implies $x \in (-3, -1)$.
The equation becomes $-(x^2 + 4x + 3) + 2x + 5 = 0$,which simplifies to $-x^2 - 2x + 2 = 0$,or $x^2 + 2x - 2 = 0$.
Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{-2 \pm \sqrt{12}}{2} = -1 \pm \sqrt{3}$.
Checking the condition: $\sqrt{3} \approx 1.732$,so $x_1 = -1 + 1.732 = 0.732$ (not in $(-3, -1)$) and $x_2 = -1 - 1.732 = -2.732$ (in $(-3, -1)$).
Thus,$x = -1 - \sqrt{3}$ is a solution.
There are exactly $2$ real solutions.

(C) Given equation is $(p - q){x^2} + (q - r)x + (r - p) = 0$.
Notice that the sum of the coefficients is $(p - q) + (q - r) + (r - p) = 0$.
If the sum of the coefficients of a quadratic equation $ax^2 + bx + c = 0$ is $0$,then $x = 1$ is always one of the roots.
Let the roots be $\alpha$ and $\beta$. We know that the product of the roots $\alpha \beta = \frac{c}{a}$.
Here,$\alpha = 1$,so $1 \times \beta = \frac{r - p}{p - q}$.
Therefore,the roots are $1$ and $\frac{r - p}{p - q}$.

(C) Given that $4$ is a root of the equation $x^2 + px + 12 = 0$.
Substituting $x = 4$ into the equation: $4^2 + p(4) + 12 = 0$.
$16 + 4p + 12 = 0$ $\Rightarrow 4p = -28$ $\Rightarrow p = -7$.
Now,the second equation is $x^2 + px + q = 0$,which becomes $x^2 - 7x + q = 0$.
Since the roots of this equation are equal,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
$(-7)^2 - 4(1)(q) = 0$.
$49 - 4q = 0$ $\Rightarrow 4q = 49$ $\Rightarrow q = \frac{49}{4}$.

(D) Given the equation: $x - \frac{2}{x - 1} = 1 - \frac{2}{x - 1}$.
For the expression to be defined,the denominator must not be zero,so $x - 1 \neq 0$,which implies $x \neq 1$.
Adding $\frac{2}{x - 1}$ to both sides of the equation,we get $x = 1$.
However,we have the condition $x \neq 1$ from the original equation.
Since the only potential solution $x = 1$ is excluded by the domain of the equation,there are no roots.

(D) Given the equation: $x + \frac{1}{x} = 2$ (where $x \neq 0$).
Multiply the entire equation by $x$:
$x^2 + 1 = 2x$
Rearrange the terms to form a quadratic equation:
$x^2 - 2x + 1 = 0$
This is a perfect square trinomial:
$(x - 1)^2 = 0$
Solving for $x$:
$x = 1$.
Since $1$ is not among the options $A, B, C$,the correct choice is $D$.

(C) Given equation: $\sqrt{3x^2 - 7x - 30} + \sqrt{2x^2 - 7x - 5} = x + 5$
Rearranging the terms: $\sqrt{3x^2 - 7x - 30} = (x + 5) - \sqrt{2x^2 - 7x - 5}$
Squaring both sides:
$3x^2 - 7x - 30 = (x + 5)^2 + (2x^2 - 7x - 5) - 2(x + 5)\sqrt{2x^2 - 7x - 5}$
$3x^2 - 7x - 30 = x^2 + 10x + 25 + 2x^2 - 7x - 5 - 2(x + 5)\sqrt{2x^2 - 7x - 5}$
$3x^2 - 7x - 30 = 3x^2 + 3x + 20 - 2(x + 5)\sqrt{2x^2 - 7x - 5}$
$-10x - 50 = -2(x + 5)\sqrt{2x^2 - 7x - 5}$
$5(x + 5) = (x + 5)\sqrt{2x^2 - 7x - 5}$
This implies either $x + 5 = 0$ or $5 = \sqrt{2x^2 - 7x - 5}$.
If $x = -5$,the original equation becomes $\sqrt{75 + 35 - 30} + \sqrt{50 + 35 - 5} = 0$,which is $\sqrt{80} + \sqrt{80} = 0$,which is false.
If $5 = \sqrt{2x^2 - 7x - 5}$,then $25 = 2x^2 - 7x - 5$,which gives $2x^2 - 7x - 30 = 0$.
Factoring the quadratic: $(2x + 5)(x - 6) = 0$.
Thus,$x = 6$ or $x = -2.5$. Checking $x = 6$ in the original equation: $\sqrt{108 - 42 - 30} + \sqrt{72 - 42 - 5} = \sqrt{36} + \sqrt{25} = 6 + 5 = 11$,which matches $x + 5 = 11$. Therefore,$x = 6$ is the solution.

(B) Let $x = 2 + \frac{1}{2 + \frac{1}{2 + \dots \infty}}$.
Since the expression is infinite,we can write $x = 2 + \frac{1}{x}$.
Multiplying by $x$,we get $x^2 = 2x + 1$,which simplifies to $x^2 - 2x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$.
$x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$.
Since the expression $2 + \frac{1}{2 + \dots}$ must be positive and greater than $2$,we discard $1 - \sqrt{2} \approx -0.414$.
Thus,the value is $1 + \sqrt{2}$.

(D) quadratic equation $ax^2 + bx + c = 0$ can have at most $2$ roots unless it is an identity.
Since the equation has $3$ distinct roots,it must be an identity.
Therefore,all coefficients must be zero,i.e.,$a = 0, b = 0, c = 0$.

(D) Given the equation $|x|^2 - 7|x| + 12 = 0$.
Let $|x| = t$,where $t \ge 0$.
The equation becomes $t^2 - 7t + 12 = 0$.
Factoring the quadratic equation: $(t - 4)(t - 3) = 0$.
This gives $t = 4$ or $t = 3$.
Substituting back $|x| = t$:
Case $1$: $|x| = 4 \implies x = 4, -4$.
Case $2$: $|x| = 3 \implies x = 3, -3$.
Thus,the roots are $4, -4, 3, -3$.
The total number of roots is $4$.

(C) Given $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}$
Squaring both sides,we get $x^2 = 2 + \sqrt{2 + \sqrt{2 + \dots}}$
Since the expression inside the square root is $x$,we can write $x^2 = 2 + x$
Rearranging the terms,we get $x^2 - x - 2 = 0$
Factoring the quadratic equation: $(x - 2)(x + 1) = 0$
This gives $x = 2$ or $x = -1$
Since the square root of a positive number must be positive,$x$ cannot be $-1$.
Therefore,$x = 2$.

(A) Given equation: $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$.
Squaring both sides: $(x + 1) + (x - 1) - 2\sqrt{(x + 1)(x - 1)} = 4x - 1$.
$2x - 2\sqrt{x^2 - 1} = 4x - 1$.
$-2\sqrt{x^2 - 1} = 2x - 1$.
Squaring both sides again: $4(x^2 - 1) = (2x - 1)^2$.
$4x^2 - 4 = 4x^2 - 4x + 1$.
$-4 = -4x + 1$.
$4x = 5 \implies x = 5/4$.
Checking $x = 5/4$ in the original equation: $\sqrt{5/4 + 1} - \sqrt{5/4 - 1} = \sqrt{9/4} - \sqrt{1/4} = 3/2 - 1/2 = 1$.
However,the $RHS$ is $\sqrt{4(5/4) - 1} = \sqrt{5 - 1} = \sqrt{4} = 2$.
Since $1 \neq 2$,the value $x = 5/4$ is an extraneous solution.
Therefore,the equation has no solution.

(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 the quadratic equation $x^2 - x - 6 = 0$.
Factoring the quadratic equation: $(x - 3)(x + 2) = 0$.
This gives $x = 3$ or $x = -2$.
Since $x$ is a square root,$x$ must be positive,so $x = 3$.

(D) Given equation: $x^2 + 5|x| + 4 = 0$.
Since $x^2 = |x|^2$,the equation becomes $|x|^2 + 5|x| + 4 = 0$.
Let $t = |x|$,where $t \ge 0$. The equation is $t^2 + 5t + 4 = 0$.
Factoring the quadratic: $(t + 1)(t + 4) = 0$.
This gives $t = -1$ or $t = -4$.
Since $|x| = t$ and the absolute value must be non-negative $(|x| \ge 0)$,there are no real values of $x$ that satisfy $|x| = -1$ or $|x| = -4$.
Therefore,the equation has no real roots.

(D) Given the equation $|x - 2| = x^2$.
Case $1$: If $x - 2 \ge 0$ (i.e.,$x \ge 2$),then $x - 2 = x^2 \Rightarrow x^2 - x + 2 = 0$. The discriminant $D = (-1)^2 - 4(1)(2) = 1 - 8 = -7 < 0$,so there are no real roots in this case.
Case $2$: If $x - 2 < 0$ (i.e.,$x < 2$),then $-(x - 2) = x^2$ $\Rightarrow 2 - x = x^2$ $\Rightarrow x^2 + x - 2 = 0$.
Factoring the quadratic equation: $(x + 2)(x - 1) = 0$.
This gives $x = -2$ or $x = 1$.
Both values satisfy the condition $x < 2$.
Thus,the solution set is $\{ 1, -2 \}$.

(A) Let $y = |x - 2|$. The equation becomes $y^2 + y - 6 = 0$.
Factoring the quadratic: $(y + 3)(y - 2) = 0$.
This gives $y = -3$ or $y = 2$.
Since $y = |x - 2|$,$y$ must be non-negative,so we discard $y = -3$.
Thus,$|x - 2| = 2$.
This implies $x - 2 = 2$ or $x - 2 = -2$.
Solving these,we get $x = 4$ or $x = 0$.
Therefore,the roots are $0, 4$.

(A) Given equation: $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$
Add $3$ to both sides of the equation:
$(\frac{p + q - x}{r} + 1) + (\frac{q + r - x}{p} + 1) + (\frac{r + p - x}{q} + 1) + \frac{3x}{p + q + r} = 3$
$\frac{p + q + r - x}{r} + \frac{q + r + p - x}{p} + \frac{r + p + q - x}{q} + \frac{3x}{p + q + r} = 3$
$(p + q + r - x) (\frac{1}{r} + \frac{1}{p} + \frac{1}{q}) = 3 - \frac{3x}{p + q + r}$
$(p + q + r - x) (\frac{1}{p} + \frac{1}{q} + \frac{1}{r}) = 3 (\frac{p + q + r - x}{p + q + r})$
$(p + q + r - x) [\frac{1}{p} + \frac{1}{q} + \frac{1}{r} - \frac{3}{p + q + r}] = 0$
This implies $p + q + r - x = 0$ or the bracket term is zero.
Therefore,$x = p + q + r$.

(C) For a quadratic equation $Ax^2 + Bx + C = 0$,the roots are equal if the discriminant $D = B^2 - 4AC = 0$.
Here,$A = (m^2 + 1)$,$B = 2am$,and $C = a^2 - b^2$.
Substituting these values into the condition $B^2 - 4AC = 0$:
$(2am)^2 - 4(m^2 + 1)(a^2 - b^2) = 0$
$4a^2m^2 - 4(m^2a^2 - m^2b^2 + a^2 - b^2) = 0$
Dividing by $4$:
$a^2m^2 - (m^2a^2 - m^2b^2 + a^2 - b^2) = 0$
$a^2m^2 - m^2a^2 + m^2b^2 - a^2 + b^2 = 0$
$m^2b^2 - a^2 + b^2 = 0$
$b^2(m^2 + 1) - a^2 = 0$
Therefore,$a^2 - b^2(m^2 + 1) = 0$.

(B) Let the discriminants of $P(x) = 0$ and $Q(x) = 0$ be $D_1$ and $D_2$ respectively.
$D_1 = b^2 - 4ac$ and $D_2 = d^2 - 4(-a)c = d^2 + 4ac$.
Consider the sum of the discriminants: $D_1 + D_2 = (b^2 - 4ac) + (d^2 + 4ac) = b^2 + d^2$.
Since $b^2 \ge 0$ and $d^2 \ge 0$,we have $D_1 + D_2 \ge 0$.
If both $D_1 < 0$ and $D_2 < 0$,then $D_1 + D_2 < 0$,which contradicts $D_1 + D_2 \ge 0$.
Thus,at least one of $D_1$ or $D_2$ must be greater than or equal to $0$.
If $D_1 \ge 0$,$P(x) = 0$ has at least two real roots.
If $D_2 \ge 0$,$Q(x) = 0$ has at least two real roots.
Therefore,the equation $P(x) \cdot Q(x) = 0$ must have at least two real roots.

(C) The given equation is $(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$.
Expanding the terms,we get:
$(x^2 - (a+b)x + ab) + (x^2 - (b+c)x + bc) + (x^2 - (c+a)x + ca) = 0$.
Combining like terms,we obtain the quadratic equation:
$3x^2 - 2(a + b + c)x + (ab + bc + ca) = 0$.
For a quadratic equation $Ax^2 + Bx + C = 0$,the discriminant is $D = B^2 - 4AC$.
Here,$A = 3$,$B = -2(a + b + c)$,and $C = (ab + bc + ca)$.
$D = [-2(a + b + c)]^2 - 4(3)(ab + bc + ca)$
$D = 4(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) - 12(ab + bc + ca)$
$D = 4(a^2 + b^2 + c^2 - ab - bc - ca)$
$D = 2[2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca]$
$D = 2[(a-b)^2 + (b-c)^2 + (c-a)^2]$.
Since the sum of squares of real numbers is always non-negative,$D \ge 0$.
Therefore,the roots are always real.

(B) The given quadratic equation is $({p^2} + {q^2}){x^2} - 2q(p + r)x + ({q^2} + {r^2}) = 0$.
Since the roots are real and equal,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Here,$a = ({p^2} + {q^2})$,$b = -2q(p + r)$,and $c = ({q^2} + {r^2})$.
So,$D = [-2q(p + r)]^2 - 4({p^2} + {q^2})({q^2} + {r^2}) = 0$.
$4q^2(p + r)^2 - 4({p^2}q^2 + p^2r^2 + q^4 + q^2r^2) = 0$.
Dividing by $4$,we get $q^2(p^2 + r^2 + 2pr) - (p^2q^2 + p^2r^2 + q^4 + q^2r^2) = 0$.
$p^2q^2 + q^2r^2 + 2pq^2r - p^2q^2 - p^2r^2 - q^4 - q^2r^2 = 0$.
Simplifying the expression,we get $2pq^2r - p^2r^2 - q^4 = 0$.
$-(q^4 - 2pq^2r + p^2r^2) = 0$.
$-(q^2 - pr)^2 = 0$.
This implies $q^2 - pr = 0$,or $q^2 = pr$.
Therefore,$p, q, r$ are in $G.P.$

(C) Given the quadratic equation $4ax^2 + 3bx + 2c = 0$.
The discriminant $D$ is given by $D = B^2 - 4AC = (3b)^2 - 4(4a)(2c) = 9b^2 - 32ac$.
Given $a + b + c = 0$,we have $b = -(a + c)$.
Substituting $b$ into the discriminant:
$D = 9(-(a + c))^2 - 32ac = 9(a^2 + 2ac + c^2) - 32ac = 9a^2 + 18ac + 9c^2 - 32ac = 9a^2 - 14ac + 9c^2$.
We can rewrite this as $D = 9a^2 - 14ac + 9c^2 = (3a - 3c)^2 + 4ac$.
Alternatively,consider the function $f(x) = 4ax^2 + 3bx + 2c$.
Since $a+b+c=0$,$b = -a-c$.
$f(1) = 4a + 3b + 2c = 4a + 3(-a-c) + 2c = a - c$.
$f(0) = 2c$.
$f(-1) = 4a - 3b + 2c = 4a - 3(-a-c) + 2c = 7a + 5c$.
Since the quadratic takes values of different signs for different $x$,the roots must be real.

(D) Given equation: $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$
Comparing with $Ax^2 + Bx + C = 0$,we have:
$A = 2(a^2 + b^2)$,$B = 2(a + b)$,$C = 1$
The discriminant $D = B^2 - 4AC$ is:
$D = [2(a + b)]^2 - 4 \times 2(a^2 + b^2) \times 1$
$D = 4(a^2 + b^2 + 2ab) - 8(a^2 + b^2)$
$D = 4a^2 + 4b^2 + 8ab - 8a^2 - 8b^2$
$D = -4a^2 - 4b^2 + 8ab$
$D = -4(a^2 + b^2 - 2ab)$
$D = -4(a - b)^2$
Since $(a - b)^2 \ge 0$,it follows that $D = -4(a - b)^2 \le 0$.
If $a \neq b$,then $D < 0$,which implies the roots are imaginary.

(D) Given that the roots of $ax^2 + x + b = 0$ are real,the discriminant $D_1 \ge 0$.
$D_1 = (1)^2 - 4ab \ge 0 \implies 4ab \le 1$.
Now,consider the second equation $x^2 - 4\sqrt{ab}x + 1 = 0$.
The discriminant $D_2$ of this equation is given by:
$D_2 = (-4\sqrt{ab})^2 - 4(1)(1) = 16ab - 4$.
Since $4ab \le 1$,we have $16ab \le 4$,which implies $16ab - 4 \le 0$.
Therefore,$D_2 \le 0$.
Since the discriminant is less than or equal to zero,the roots are imaginary.

(B) The roots of the quadratic equation $ax^2 + bx + c = 0$ are given by the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Case $1$: If the discriminant $D = b^2 - 4ac \ge 0$,the roots are real.
Since $a, b, c > 0$,we have $\sqrt{b^2 - 4ac} < \sqrt{b^2} = b$.
Thus,both roots $x_1 = \frac{-b + \sqrt{D}}{2a}$ and $x_2 = \frac{-b - \sqrt{D}}{2a}$ are negative because the numerator is negative and $a > 0$.
Case $2$: If $D = b^2 - 4ac < 0$,the roots are complex conjugates:
$x = \frac{-b}{2a} \pm i \frac{\sqrt{4ac - b^2}}{2a}$
The real part of these roots is $\text{Re}(x) = -\frac{b}{2a}$.
Since $a > 0$ and $b > 0$,the real part $-\frac{b}{2a}$ is always negative.
Conclusion: In all cases,the roots have negative real parts.