QUADRATIC EQUATION Questions in English

(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: $ax^2 - a^2x - x + a = 0$
Grouping the terms: $ax(x - a) - 1(x - a) = 0$
Factoring out $(x - a)$: $(ax - 1)(x - a) = 0$
Setting each factor to zero:
$ax - 1 = 0 \implies x = \frac{1}{a}$
$x - a = 0 \implies x = a$
Therefore,the roots are $a$ and $\frac{1}{a}$.

(B) Given equation: ${x^4} - 8{x^2} - 9 = 0$
Let ${x^2} = y$. 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 back ${x^2} = y$:
Case $1$: ${x^2} = 9 \Rightarrow x = \pm 3$.
Case $2$: ${x^2} = -1 \Rightarrow x = \pm \sqrt{-1} = \pm i$.
Thus,the roots are $\pm 3, \pm i$.

(C) Given the quadratic equation: $ix^2 - 4x - 4i = 0$
Divide the entire equation by $i$ (or multiply by $-i$):
$x^2 - (4/i)x - 4 = 0$
Since $1/i = -i$,we have:
$x^2 + 4ix - 4 = 0$
This can be written as:
$x^2 + 2ix + 2ix + (2i)^2 = 0$
$(x + 2i)^2 = 0$
Therefore,the roots are $x = -2i, -2i$.

(C) Given equation: $x^{2/3} + x^{1/3} - 2 = 0$
Let $a = x^{1/3}$. Then the equation becomes a quadratic equation in terms of $a$:
$a^2 + a - 2 = 0$
Factorizing the quadratic equation:
$a^2 + 2a - a - 2 = 0$
$a(a + 2) - 1(a + 2) = 0$
$(a - 1)(a + 2) = 0$
So,$a = 1$ or $a = -2$.
Now,substitute back $a = x^{1/3}$:
Case $1$: $x^{1/3} = 1 \implies x = 1^3 = 1$
Case $2$: $x^{1/3} = -2 \implies x = (-2)^3 = -8$
Thus,the roots are $1, -8$.

(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 use the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.
$(x - 2)^3 = (2^{2/3})^3 + (2^{1/3})^3 + 3(2^{2/3})(2^{1/3})(2^{2/3} + 2^{1/3})$.
$(x - 2)^3 = 2^2 + 2^1 + 3(2^1)(x - 2)$.
$x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = 4 + 2 + 6(x - 2)$.
$x^3 - 6x^2 + 12x - 8 = 6 + 6x - 12$.
$x^3 - 6x^2 + 12x - 8 = 6x - 6$.
Rearranging the terms to find $x^3 - 6x^2 + 6x$,we get:
$x^3 - 6x^2 + 6x = 8 - 12 + 6 = 2$.
Thus,the correct option is $B$.

(D) Given the quadratic equation $8\sec^2 \theta - 6\sec \theta + 1 = 0$.
Let $x = \sec \theta$. The equation becomes $8x^2 - 6x + 1 = 0$.
Factoring the quadratic equation: $8x^2 - 4x - 2x + 1 = 0$.
$4x(2x - 1) - 1(2x - 1) = 0$.
$(4x - 1)(2x - 1) = 0$.
This gives $x = \frac{1}{4}$ or $x = \frac{1}{2}$.
Thus,$\sec \theta = \frac{1}{4}$ or $\sec \theta = \frac{1}{2}$.
However,for any real value of $\theta$,the range of $\sec \theta$ is $(-\infty, -1] \cup [1, \infty)$.
Since both $\frac{1}{4}$ and $\frac{1}{2}$ lie in the interval $(-1, 1)$,there is no real value of $\theta$ that satisfies these equations.
Therefore,the number of roots is $0$.

(D) Given equation is $\sqrt{3x + 1} + 1 = \sqrt{x}$.
Rearranging the terms,we get $\sqrt{3x + 1} = \sqrt{x} - 1$.
Squaring both sides:
$(\sqrt{3x + 1})^2 = (\sqrt{x} - 1)^2$
$3x + 1 = x + 1 - 2\sqrt{x}$.
Simplifying the equation:
$2x = -2\sqrt{x}$
$x = -\sqrt{x}$.
Squaring again:
$x^2 = x$
$x^2 - x = 0$
$x(x - 1) = 0$,which gives $x = 0$ 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 $x = 1$: $\sqrt{3(1) + 1} + 1 = \sqrt{4} + 1 = 2 + 1 = 3$,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,the number exceeds its positive square root by $12$,so we have the equation:
$x = \sqrt{x} + 12$
Rearranging the terms,we get:
$x - 12 = \sqrt{x}$
Squaring both sides:
$(x - 12)^2 = x$
$x^2 - 24x + 144 = x$
$x^2 - 25x + 144 = 0$
Factoring the quadratic equation:
$(x - 16)(x - 9) = 0$
This gives $x = 16$ or $x = 9$.
Checking the values:
If $x = 16$,$\sqrt{16} = 4$. $16 - 4 = 12$. (This satisfies the condition).
If $x = 9$,$\sqrt{9} = 3$. $9 - 3 = 6 \neq 12$. (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 rewritten as $(3^x)^2 - 10(3^x) + 9 = 0$.
Let $a = 3^x$. Substituting this into the equation, we get the quadratic equation:
$a^2 - 10a + 9 = 0$
Factoring the quadratic equation:
$(a - 9)(a - 1) = 0$
This gives the roots $a = 9$ and $a = 1$.
Now, substitute back $a = 3^x$:
For $a = 9$: $3^x = 9 \Rightarrow 3^x = 3^2 \Rightarrow x = 2$.
For $a = 1$: $3^x = 1 \Rightarrow 3^x = 3^0 \Rightarrow x = 0$.
Thus, the roots of the equation are $x = 0$ and $x = 2$.

(D) Given the equation: ${x^{2/3}} - 7{x^{1/3}} + 10 = 0$.
This can be rewritten as: ${({x^{1/3}})^2} - 7({x^{1/3}}) + 10 = 0$.
Let $a = {x^{1/3}}$. Substituting $a$ into the equation,we get the quadratic equation: ${a^2} - 7a + 10 = 0$.
Factoring the quadratic equation: $(a - 5)(a - 2) = 0$.
This gives the roots: $a = 5$ or $a = 2$.
Since $a = {x^{1/3}}$,we have $x = {a^3}$.
For $a = 5$,$x = {5^3} = 125$.
For $a = 2$,$x = {2^3} = 8$.
Therefore,the solution set is ${125, 8}$.

(C) Given equations are ${x^2} + {y^2} = 25$ and $xy = 12$.
From the second equation,we have $y = \frac{12}{x}$.
Substitute this into the first equation: ${x^2} + {\left( \frac{12}{x} \right)^2} = 25$.
This simplifies to ${x^2} + \frac{144}{x^2} = 25$.
Multiplying by ${x^2}$,we get ${x^4} + 144 = 25{x^2}$,which rearranges to ${x^4} - 25{x^2} + 144 = 0$.
Let $u = {x^2}$,then ${u^2} - 25u + 144 = 0$.
Factoring the quadratic equation: $(u - 16)(u - 9) = 0$.
So,${x^2} = 16$ or ${x^2} = 9$.
Taking the square root,we get $x = \pm 4$ or $x = \pm 3$.
Thus,the possible values for $x$ are $\{3, 4, -3, -4\}$.

(B) Given the equation: $x^{\log_x(1 - x)^2} = 9$.
Using the logarithmic property $a^{\log_a(M)} = M$,the equation simplifies to:
$(1 - x)^2 = 9$.
Expanding the square:
$1 - 2x + x^2 = 9$.
Rearranging the terms to form a quadratic equation:
$x^2 - 2x - 8 = 0$.
Factoring the quadratic equation:
$(x - 4)(x + 2) = 0$.
This gives the potential solutions $x = 4$ and $x = -2$.
Now,we must check the validity of these solutions in the original logarithmic expression $\log_x(1 - x)^2$. The base of a logarithm $x$ must satisfy $x > 0$ and $x \neq 1$.
For $x = 4$: The base is $4$,which is valid ($4 > 0$ and $4 \neq 1$).
For $x = -2$: The base is $-2$,which is invalid as the base of a logarithm must be positive.
Therefore,the only valid solution is $x = 4$.

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

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

(D) Given equation: ${e^{\sin x}} - {e^{-\sin x}} - 4 = 0$.
Let ${e^{\sin x}} = y$. Since ${e^{\sin x}} > 0$,we have $y > 0$.
The equation becomes $y - \frac{1}{y} - 4 = 0$.
Multiplying by $y$,we get ${y^2} - 4y - 1 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$.
Since $y > 0$ and $\sqrt{5} \approx 2.236$,we must have $y = 2 + \sqrt{5}$.
Thus,${e^{\sin x}} = 2 + \sqrt{5}$.
Taking the natural logarithm on both sides: $\sin x = \ln(2 + \sqrt{5})$.
Since $\sqrt{5} > 1$,$2 + \sqrt{5} > 3$. Therefore,$\ln(2 + \sqrt{5}) > \ln(3) > 1$.
However,the range of $\sin x$ is $[-1, 1]$.
Since $\ln(2 + \sqrt{5}) > 1$,there is no real value of $x$ that satisfies this equation.
Therefore,the number of real roots is $0$ (None).

(B) The given equation is $|x^2 + 4x + 3| + 2x + 5 = 0$.
Case $I$: $x^2 + 4x + 3 \ge 0$
This 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 valid solution.
Case $II$: $x^2 + 4x + 3 < 0$
This implies $x \in (-3, -1)$.
The equation becomes $-(x^2 + 4x + 3) + 2x + 5 = 0$,which simplifies to $-x^2 - 4x - 3 + 2x + 5 = 0$,or $-x^2 - 2x + 2 = 0$.
Multiplying by $-1$,we get $x^2 + 2x - 2 = 0$.
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: $\sqrt{3} \approx 1.732$.
$x_1 = -1 + 1.732 = 0.732$ (does not satisfy $x \in (-3, -1)$).
$x_2 = -1 - 1.732 = -2.732$ (satisfies $x \in (-3, -1)$).
Thus,$x = -1 - \sqrt{3}$ is a valid solution.
There are two real solutions: $x = -4$ and $x = -1 - \sqrt{3}$.

(C) Given equation is $(p - q)x^2 + (q - r)x + (r - p) = 0$.
Notice that the sum of the coefficients is $(p - q) + (q - r) + (r - p) = p - q + q - r + r - p = 0$.
If the sum of the coefficients of a quadratic equation $ax^2 + bx + c = 0$ is zero,then $x = 1$ is always one of the roots.
Let the roots be $\alpha$ and $\beta$. We know that the product of the roots $\alpha \cdot \beta = \frac{c}{a}$.
Here,$\alpha = 1$,so $1 \cdot \beta = \frac{r - p}{p - q}$.
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: $(4)^2 + p(4) + 12 = 0$.
$16 + 4p + 12 = 0$.
$4p + 28 = 0$.
$4p = -28$,which gives $p = -7$.
Now,consider the second equation $x^2 + px + q = 0$. Substituting $p = -7$,we get $x^2 - 7x + q = 0$.
Since the roots of this equation are equal,the discriminant $D$ must be zero.
$D = b^2 - 4ac = 0$.
$(-7)^2 - 4(1)(q) = 0$.
$49 - 4q = 0$.
$4q = 49$.
Therefore,$q = 49/4$.

(D) Given equation: $x - \frac{2}{x - 1} = 1 - \frac{2}{x - 1}$.
For the equation 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 for this equation.

(D) Given the equation: $x + \frac{1}{x} = 2$ (where $x \neq 0$).
Multiply the entire equation by $x$ to clear the denominator:
$x^2 + 1 = 2x$
Rearrange the terms to form a standard quadratic equation:
$x^2 - 2x + 1 = 0$
This is a perfect square trinomial,which can be written as:
$(x - 1)^2 = 0$
Taking the square root of both sides:
$x - 1 = 0$
$x = 1$
Since the value $x = 1$ is not among the options $A, B,$ or $C$,the correct choice is $D$.

(A) For a quadratic equation of the form $ax^2 + bx + c = 0$, the product of the roots is given by $\frac{c}{a}$.
Comparing the given equation $2x^2 + 6x + (\alpha^2 + 1) = 0$ with the standard form, we have $a = 2$ and $c = \alpha^2 + 1$.
The product of the roots is $\frac{\alpha^2 + 1}{2}$.
According to the problem, the product of the roots is $-\alpha$.
Therefore, $\frac{\alpha^2 + 1}{2} = -\alpha$.
Multiplying both sides by $2$, we get $\alpha^2 + 1 = -2\alpha$.
Rearranging the terms, we get $\alpha^2 + 2\alpha + 1 = 0$.
This is a perfect square trinomial: $(\alpha + 1)^2 = 0$.
Solving for $\alpha$, we get $\alpha = -1$.

(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:
$3x^2 - 7x - 30 = (x + 5)^2 + (2x^2 - 7x - 5) - 2(x + 5)\sqrt {2x^2 - 7x - 5}$
$3x^2 - 7x - 30 = x^2 + 10x + 25 + 2x^2 - 7x - 5 - 2(x + 5)\sqrt {2x^2 - 7x - 5}$
$3x^2 - 7x - 30 = 3x^2 + 3x + 20 - 2(x + 5)\sqrt {2x^2 - 7x - 5}$
$-10x - 50 = -2(x + 5)\sqrt {2x^2 - 7x - 5}$
$5(x + 5) = (x + 5)\sqrt {2x^2 - 7x - 5}$
Case $1$: $x + 5 = 0 \Rightarrow x = -5$. Checking in the original equation: $\sqrt{75+35-30} + \sqrt{50+35-5} = 0 \Rightarrow \sqrt{80} + \sqrt{80} = 0$, which is false.
Case $2$: $5 = \sqrt {2x^2 - 7x - 5}$
Squaring both sides: $25 = 2x^2 - 7x - 5$
$2x^2 - 7x - 30 = 0$
$(2x + 5)(x - 6) = 0$
$x = 6$ or $x = -2.5$. Checking $x = 6$ in the original equation: $\sqrt{108-42-30} + \sqrt{72-42-5} = 6+5 \Rightarrow \sqrt{36} + \sqrt{25} = 11 \Rightarrow 6+5 = 11$. This is true.
Thus, $x = 6$.

(B) Let $x = 2 + \frac{1}{2 + \frac{1}{2 + \dots \infty}}$.
Since the expression is infinite,we can substitute the repeating part with $x$:
$x = 2 + \frac{1}{x}$.
Multiplying both sides by $x$,we get $x^2 = 2x + 1$,which simplifies to the quadratic equation $x^2 - 2x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 1, b = -2, c = -1$:
$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$.
Since the expression $2 + \frac{1}{2 + \dots}$ must be positive and greater than $2$,we reject $1 - \sqrt{2}$ (which is negative).
Therefore,the value is $1 + \sqrt{2}$.

(D) Given equation: ${2^{x + 2}} \cdot {3^{3x/(x - 1)}} = {3^2}$
Taking $\log$ on both sides:
$(x + 2)\log 2 + \frac{{3x}}{{x - 1}}\log 3 = 2\log 3$
$(x + 2)\log 2 = 2\log 3 - \frac{{3x}}{{x - 1}}\log 3$
$(x + 2)\log 2 = \log 3 \left( {2 - \frac{{3x}}{{x - 1}}} \right)$
$(x + 2)\log 2 = \log 3 \left( {\frac{{2x - 2 - 3x}}{{x - 1}}} \right)$
$(x + 2)\log 2 = \log 3 \left( {\frac{{ - x - 2}}{{x - 1}}} \right)$
$(x + 2)\log 2 = - \log 3 \left( {\frac{{x + 2}}{{x - 1}}} \right)$
$(x + 2) \left[ {\log 2 + \frac{{\log 3}}{{x - 1}}} \right] = 0$
Case $1$: $x + 2 = 0 \implies x = -2$
Case $2$: $\log 2 + \frac{{\log 3}}{{x - 1}} = 0$
$\frac{{\log 3}}{{x - 1}} = - \log 2$
$x - 1 = - \frac{{\log 3}}{{\log 2}}$
$x = 1 - \frac{{\log 3}}{{\log 2}}$
Thus,the roots are $-2$ and $1 - \frac{{\log 3}}{{\log 2}}$.

(D) The given equation is $x^2 + x + 1 = 0$.
The roots of this equation are the complex cube roots of unity,$\omega$ and $\omega^2$.
Let $\alpha = \omega$ and $\beta = \omega^2$.
We need to find the equation whose roots are $\alpha^{19}$ and $\beta^7$.
Since $\omega^3 = 1$,we have $\alpha^{19} = \omega^{19} = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$.
Similarly,$\beta^7 = (\omega^2)^7 = \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.
The roots of the new equation are $\omega$ and $\omega^2$.
Therefore,the required equation is the same as the original equation: $x^2 + x + 1 = 0$.

(C) The critical points for the given equation are $x = 2$ and $x = 3$.
Case $1$: If $x < 2$,then $|x - 2| = -(x - 2)$ and $|x - 3| = -(x - 3)$.
The equation becomes $-(x - 2) - (x - 3) = 7$,which simplifies to $-x + 2 - x + 3 = 7$,or $-2x + 5 = 7$.
$-2x = 2$,so $x = -1$. Since $-1 < 2$,this is a valid solution.
Case $2$: If $2 \le x < 3$,then $|x - 2| = x - 2$ and $|x - 3| = -(x - 3)$.
The equation becomes $(x - 2) - (x - 3) = 7$,which simplifies to $x - 2 - x + 3 = 7$,or $1 = 7$.
This is a contradiction,so there is no solution in this interval.
Case $3$: If $x \ge 3$,then $|x - 2| = x - 2$ and $|x - 3| = x - 3$.
The equation becomes $(x - 2) + (x - 3) = 7$,which simplifies to $2x - 5 = 7$,or $2x = 12$.
Thus,$x = 6$. Since $6 \ge 3$,this is a valid solution.
Therefore,the solutions are $x = 6$ or $x = -1$.

(D) quadratic equation of the form $ax^2 + bx + c = 0$ can have at most two distinct roots unless it is an identity.
If an equation has more than two distinct roots,it must satisfy the condition that all its coefficients are zero.
Therefore,for the equation to have three distinct roots ${x_1}, {x_2}, {x_3}$,it must be an identity,which implies $a = 0, b = 0,$ and $c = 0$.

(D) Given 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 two possible values for $t$: $t = 4$ or $t = 3$.
Substituting back $|x| = t$:
Case $1$: $|x| = 4 \implies x = 4$ or $x = -4$.
Case $2$: $|x| = 3 \implies x = 3$ or $x = -3$.
Thus,the roots are $x \in \{4, -4, 3, -3\}$.
The total number of roots is $4$.

(D) Given equation: $\frac{\log 5 + \log (x^2 + 1)}{\log (x - 2)} = 2$.
For the logarithmic terms to be defined,we must have $x - 2 > 0$ and $x - 2 \neq 1$,which implies $x > 2$ and $x \neq 3$.
Using the property $\log a + \log b = \log (ab)$,the equation becomes $\log (5(x^2 + 1)) = 2 \log (x - 2)$.
Using the property $n \log a = \log (a^n)$,we get $\log (5x^2 + 5) = \log ((x - 2)^2)$.
Equating the arguments: $5x^2 + 5 = x^2 - 4x + 4$.
Rearranging terms: $4x^2 + 4x + 1 = 0$.
This is a perfect square: $(2x + 1)^2 = 0$,which gives $x = -\frac{1}{2}$.
However,the domain condition requires $x > 2$. Since $x = -\frac{1}{2}$ does not satisfy $x > 2$,there is no valid solution.
Thus,the number of solutions is $0$.

(A) Given $x = \sqrt{7 + 4\sqrt{3}}.$
We can simplify the expression inside the square root: $7 + 4\sqrt{3} = 4 + 3 + 2(2)(\sqrt{3}) = (2 + \sqrt{3})^2.$
Therefore,$x = \sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3}.$
Now,find $\frac{1}{x}:$
$\frac{1}{x} = \frac{1}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.$
Finally,calculate $x + \frac{1}{x}:$
$x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4.$

(D) Let $t = \log_2 x$. Then the equation becomes $t + \frac{1}{t} = \frac{10}{3}$.
Multiplying by $3t$,we get $3t^2 - 10t + 3 = 0$.
Factoring the quadratic equation: $3t^2 - 9t - t + 3 = 0 \Rightarrow 3t(t - 3) - 1(t - 3) = 0$.
This gives $(3t - 1)(t - 3) = 0$,so $t = 3$ or $t = 1/3$.
Since $x \neq y$,we assign $\log_2 x = 3$ and $\log_2 y = 1/3$ (or vice versa).
Thus,$x = 2^3 = 8$ and $y = 2^{1/3} = \sqrt[3]{2}$.
Therefore,$x + y = 8 + \sqrt[3]{2}$.

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

(B) Given equation: $4^x - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2x - 1}$
Rearranging the terms:
$2^{2x} + 2^{2x - 1} = 3^{x + 1/2} + 3^{x - 1/2}$
$2^{2x}(1 + 1/2) = 3^{x - 1/2}(3 + 1)$
$2^{2x} \cdot (3/2) = 3^{x - 1/2} \cdot 4$
$2^{2x} \cdot 3 \cdot 2^{-1} = 3^{x - 1/2} \cdot 2^2$
$2^{2x - 1} \cdot 3 = 3^{x - 1/2} \cdot 2^2$
$2^{2x - 3} = 3^{x - 1/2 - 1} = 3^{x - 3/2}$
Taking logarithm on both sides:
$(2x - 3) \log 2 = (x - 3/2) \log 3$
$2x \log 2 - 3 \log 2 = x \log 3 - (3/2) \log 3$
$x(2 \log 2 - \log 3) = 3 \log 2 - (3/2) \log 3$
$x \log(4/3) = \log(2^3) - \log(3^{3/2})$
$x \log(4/3) = \log(8 / (3\sqrt{3}))$
Since $8 / (3\sqrt{3}) = (4/3)^{3/2}$,we have:
$x \log(4/3) = (3/2) \log(4/3)$
Therefore,$x = 3/2$.

(A) Let $f(x) = e^x - x - 1$.
To find the roots,we analyze the function $f(x)$.
The derivative is $f'(x) = e^x - 1$.
Setting $f'(x) = 0$ gives $e^x = 1$,which implies $x = 0$.
At $x = 0$,$f(0) = e^0 - 0 - 1 = 1 - 1 = 0$.
Since $f'(x) < 0$ for $x < 0$ and $f'(x) > 0$ for $x > 0$,the function $f(x)$ has a global minimum at $x = 0$.
Since the minimum value is $0$,the graph of $f(x)$ touches the $x$-axis only at $x = 0$.
Therefore,the equation has exactly one real root,$x = 0$.

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

(B) Given the equation: $\log_e x + \log_e(1 + x) = 0$
Using the logarithmic property $\log_e a + \log_e b = \log_e(ab)$,we get:
$\log_e(x(1 + x)) = 0$
By the definition of the natural logarithm,if $\log_e y = 0$,then $y = e^0 = 1$.
Therefore,$x(1 + x) = 1$
$x^2 + x = 1$
$x^2 + x - 1 = 0$
Thus,the correct option is $(b)$.

(C) Given the expression $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots \infty}}}$.
Since the expression repeats infinitely,we can write it as $x = \sqrt{6 + x}$.
Squaring both sides,we get $x^2 = 6 + x$,where $x > 0$.
Rearranging the terms gives 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 $x$ must be positive (as it is a square root),we discard $x = -2$.
Therefore,$x = 3$.

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

(A) Given equation: $\log_{4}\{\log_{2}(\sqrt{x+8} - \sqrt{x})\} = 0$
Using the definition of logarithm,$\log_{b}(a) = c \Rightarrow a = b^{c}$:
$\log_{2}(\sqrt{x+8} - \sqrt{x}) = 4^{0} = 1$
Again,using the definition of logarithm:
$\sqrt{x+8} - \sqrt{x} = 2^{1} = 2$
Rearranging the terms:
$\sqrt{x+8} = 2 + \sqrt{x}$
Squaring both sides:
$x + 8 = (2 + \sqrt{x})^{2}$
$x + 8 = 4 + x + 4\sqrt{x}$
Subtracting $x$ and $4$ from both sides:
$4 = 4\sqrt{x}$
$1 = \sqrt{x}$
Squaring both sides again:
$x = 1$
Verification: For $x=1$,$\sqrt{1+8} - \sqrt{1} = 3 - 1 = 2$. Then $\log_{2}(2) = 1$,and $\log_{4}(1) = 0$. The root is valid.

(D) The given equation is $|x - 2| = x^2$.
Case $1$: If $x - 2 \ge 0$ (i.e.,$x \ge 2$),then $x - 2 = x^2$,which implies $x^2 - x + 2 = 0$. The discriminant $D = (-1)^2 - 4(1)(2) = 1 - 8 = -7 < 0$. Thus,there are no real roots in this case.
Case $2$: If $x - 2 < 0$ (i.e.,$x < 2$),then $-(x - 2) = x^2$,which implies $2 - x = x^2$,or $x^2 + x - 2 = 0$.
Factoring the quadratic equation: $x^2 + 2x - x - 2 = 0 \Rightarrow x(x + 2) - 1(x + 2) = 0 \Rightarrow (x - 1)(x + 2) = 0$.
This gives $x = 1$ or $x = -2$.
Since both $1 < 2$ and $-2 < 2$,both are valid solutions.
Therefore,the set of solutions is $\{1, -2\}$.

(B) Given equation: $\log_4(x - 1) = \log_2(x - 3)$
First,define the domain: $x - 1 > 0 \Rightarrow x > 1$ and $x - 3 > 0 \Rightarrow x > 3$. Thus,$x > 3$.
Using the change of base formula $\log_{a^n}(b) = \frac{1}{n}\log_a(b)$,we have $\log_4(x - 1) = \frac{1}{2}\log_2(x - 1)$.
So,$\frac{1}{2}\log_2(x - 1) = \log_2(x - 3)$.
Multiplying by $2$,we get $\log_2(x - 1) = 2\log_2(x - 3) = \log_2((x - 3)^2)$.
Equating the arguments: $x - 1 = (x - 3)^2$.
$x - 1 = x^2 - 6x + 9$.
$x^2 - 7x + 10 = 0$.
$(x - 5)(x - 2) = 0$.
This gives $x = 5$ or $x = 2$.
Since the domain requires $x > 3$,$x = 2$ is an extraneous solution.
Therefore,the only valid solution is $x = 5$.
The number of solutions is $1$.

(A) Let $y = |x - 2|$. Since $|x - 2| \ge 0$,we have $y \ge 0$.
The equation becomes $y^2 + y - 6 = 0$.
Factoring the quadratic equation: $(y + 3)(y - 2) = 0$.
This gives $y = -3$ or $y = 2$.
Since $y \ge 0$,we discard $y = -3$. Thus,$y = 2$.
Substituting back,$|x - 2| = 2$.
This implies $x - 2 = 2$ or $x - 2 = -2$.
Solving these,$x = 4$ or $x = 0$.
Therefore,the roots are $0$ and $4$.

(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 $1$ to each of the first three terms to adjust the numerators:
$(\frac{p + q - x}{r} + 1) + (\frac{q + r - x}{p} + 1) + (\frac{r + p - x}{q} + 1) + \frac{3x}{p + q + r} - 3 = 0$
$\frac{p + q + r - x}{r} + \frac{p + q + r - x}{p} + \frac{p + q + r - x}{q} + \frac{3x - 3(p + q + r)}{p + q + r} = 0$
$(p + q + r - x) (\frac{1}{r} + \frac{1}{p} + \frac{1}{q}) - \frac{3(p + q + r - x)}{p + q + r} = 0$
$(p + q + r - x) [\frac{1}{p} + \frac{1}{q} + \frac{1}{r} - \frac{3}{p + q + r}] = 0$
This implies either $(p + q + r - x) = 0$ or the bracketed term is zero.
Thus,$x = p + q + r$.

(A) Given equation: ${x^2} - 5|x| + 6 = 0$.
Since ${x^2} = |x|^2$,we can rewrite the equation as $|x|^2 - 5|x| + 6 = 0$.
Let $|x| = t$,where $t \ge 0$. The equation becomes $t^2 - 5t + 6 = 0$.
Factoring the quadratic equation: $(t - 2)(t - 3) = 0$.
This gives $t = 2$ or $t = 3$.
Substituting back $|x| = t$:
Case $1$: $|x| = 2 \implies x = 2, -2$.
Case $2$: $|x| = 3 \implies x = 3, -3$.
Thus,the solutions are $x \in \{2, -2, 3, -3\}$.
The total number of solutions is $4$.

(C) For a quadratic equation $Ax^2 + Bx + C = 0$,if the roots are equal,the discriminant $D = B^2 - 4AC$ must be equal to $0$.
Here,$A = ({m^2} + 1)$,$B = 2am$,and $C = ({a^2} - {b^2})$.
Setting $D = 0$:
$(2am)^2 - 4({m^2} + 1)({a^2} - {b^2}) = 0$
$4a^2m^2 - 4({m^2}a^2 - {m^2}b^2 + a^2 - b^2) = 0$
Dividing by $4$:
$a^2m^2 - (a^2m^2 - m^2b^2 + a^2 - b^2) = 0$
$a^2m^2 - a^2m^2 + m^2b^2 - a^2 + b^2 = 0$
$m^2b^2 + b^2 - a^2 = 0$
$b^2(m^2 + 1) - a^2 = 0$
Therefore,${a^2} - {b^2}({m^2} + 1) = 0$.

(B) Let the roots of $P(x) = 0$ and $Q(x) = 0$ be considered.
For $P(x) = ax^2 + bx + c = 0$,the discriminant is $D_1 = b^2 - 4ac$.
For $Q(x) = -ax^2 + dx + c = 0$,the discriminant is $D_2 = d^2 - 4(-a)(c) = d^2 + 4ac$.
If all four roots were imaginary,then both $D_1 < 0$ and $D_2 < 0$ must hold.
Adding these inequalities: $(b^2 - 4ac) + (d^2 + 4ac) < 0$,which simplifies to $b^2 + d^2 < 0$.
Since $b^2$ and $d^2$ are non-negative,$b^2 + d^2 < 0$ is impossible unless $b=0$ and $d=0$.
If $b=0$ and $d=0$,then $P(x) = ax^2 + c = 0$ and $Q(x) = -ax^2 + c = 0$.
This gives $x^2 = -c/a$ and $x^2 = c/a$.
Since $ac \neq 0$,one of $c/a$ or $-c/a$ must be positive,ensuring at least two real roots.
Thus,in all cases,$P(x) \cdot Q(x) = 0$ has 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 get $3x^2 - 2(a+b+c)x + (ab + bc + ca) = 0$.
For a quadratic equation $Ax^2 + Bx + C = 0$,the discriminant is $D = B^2 - 4AC$.
Here,$D = [-2(a+b+c)]^2 - 4(3)(ab + bc + ca) = 4(a+b+c)^2 - 12(ab + bc + ca)$.
$D = 4[a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca] = 4[a^2 + b^2 + c^2 - ab - bc - ca]$.
This can be written as $D = 2[(a-b)^2 + (b-c)^2 + (c-a)^2]$.
Since the sum of squares is always non-negative,$D \ge 0$.
Therefore,the roots are always real.

(B) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $-\alpha$ because they are equal in magnitude but opposite in sign.
For any quadratic equation,the sum of the roots is given by $-\frac{b}{a}$.
Here,$a = 2$,$b = 3(\lambda - 2)$,and $c = \lambda + 4$.
Therefore,the sum of the roots is $\alpha + (-\alpha) = -\frac{3(\lambda - 2)}{2}$.
$0 = -\frac{3(\lambda - 2)}{2}$.
Multiplying both sides by $-2/3$,we get $0 = \lambda - 2$.
Thus,$\lambda = 2$.

(B) Given the quadratic equation $({p^2} + {q^2}){x^2} - 2q(p + r)x + ({q^2} + {r^2}) = 0$.
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 $c = ({q^2} + {r^2})$.
Substituting these into $D = 0$:
$[-2q(p + r)]^2 - 4({p^2} + {q^2})({q^2} + {r^2}) = 0$
$4q^2(p + r)^2 - 4({p^2}q^2 + p^2r^2 + q^4 + q^2r^2) = 0$
Dividing by $4$:
$q^2(p^2 + r^2 + 2pr) - (p^2q^2 + p^2r^2 + q^4 + q^2r^2) = 0$
$p^2q^2 + q^2r^2 + 2pq^2r - p^2q^2 - p^2r^2 - q^4 - q^2r^2 = 0$
Simplifying the expression:
$2pq^2r - p^2r^2 - q^4 = 0$
$-(q^4 - 2pq^2r + p^2r^2) = 0$
$-(q^2 - pr)^2 = 0$
$(q^2 - pr)^2 = 0$
This implies $q^2 = pr$.
Therefore,$p, q, r$ are in $G.P.$ (Geometric Progression).

(C) Given the quadratic equation $4ax^2 + 3bx + 2c = 0$.
For the roots to be real,the discriminant $D = B^2 - 4AC$ must be $\ge 0$.
Here,$A = 4a$,$B = 3b$,and $C = 2c$.
$D = (3b)^2 - 4(4a)(2c) = 9b^2 - 32ac$.
Given $a + b + c = 0$,we have $b = -(a + c)$.
Substituting $b$ into the discriminant expression:
$D = 9(-(a + c))^2 - 32ac = 9(a^2 + 2ac + c^2) - 32ac$.
$D = 9a^2 + 18ac + 9c^2 - 32ac = 9a^2 - 14ac + 9c^2$.
We can rewrite this as $D = 9(a^2 - \frac{14}{9}ac + c^2) = 9((a - \frac{7}{9}c)^2 + c^2(1 - \frac{49}{81})) = 9((a - \frac{7}{9}c)^2 + \frac{32}{81}c^2)$.
Since $(a - \frac{7}{9}c)^2 \ge 0$ and $\frac{32}{81}c^2 \ge 0$,the discriminant $D$ is always $\ge 0$.
Therefore,the roots are real.