Solution of quadratic equations and Nature of roots Questions in English

(B) For the roots of a quadratic equation $ax^2 + bx + c = 0$ to have opposite signs,the product of the roots must be negative.
Product of roots $\alpha \beta = \frac{c}{a} < 0$.
Here,$a = 3$,$b = 2$,and $c = p(p - 1)$.
Substituting these values,we get $\frac{p(p - 1)}{3} < 0$.
Since $3 > 0$,we have $p(p - 1) < 0$.
Solving the inequality $p(p - 1) < 0$,we find that $p$ must lie between the roots $0$ and $1$.
Therefore,$p \in (0, 1)$.

(A) The given quadratic equation is $a^2x^2 + (a + b)x - b^2 = 0$.
Comparing this with $Ax^2 + Bx + C = 0$,we have $A = a^2$,$B = (a + b)$,and $C = -b^2$.
The discriminant $D$ is given by $D = B^2 - 4AC$.
$D = (a + b)^2 - 4(a^2)(-b^2) = (a + b)^2 + 4a^2b^2$.
Since $(a + b)^2 \ge 0$ and $4a^2b^2 \ge 0$,the sum $D = (a + b)^2 + 4a^2b^2$ must be greater than $0$ (assuming $a, b \neq 0$).
Since $D > 0$,the roots are real and distinct.

(C) Given the quadratic equation $(b - c)x^2 + (c - a)x + (a - b) = 0$.
Since the roots are equal,the discriminant $D = 0$.
Here,$A = (b - c)$,$B = (c - a)$,and $C = (a - b)$.
$D = B^2 - 4AC = (c - a)^2 - 4(b - c)(a - b) = 0$.
Expanding this: $(c^2 - 2ac + a^2) - 4(ab - b^2 - ac + bc) = 0$.
$c^2 - 2ac + a^2 - 4ab + 4b^2 + 4ac - 4bc = 0$.
$a^2 + 4b^2 + c^2 + 2ac - 4ab - 4bc = 0$.
This can be written as $(a - 2b + c)^2 = 0$.
Therefore,$a - 2b + c = 0$,which implies $a + c = 2b$.
This is the condition for $a, b, c$ to be in Arithmetic Progression.

(B) For the roots of a quadratic equation $Ax^2 + Bx + C = 0$ to have opposite signs,the product of the roots must be negative,i.e.,$\frac{C}{A} < 0$.
Given the equation $2x^2 - (a^3 + 8a - 1) x + (a^2 - 4a) = 0$,we have $A = 2$ and $C = a^2 - 4a$.
The condition for opposite signs is $\frac{a^2 - 4a}{2} < 0$.
This implies $a^2 - 4a < 0$.
Factoring the expression,we get $a(a - 4) < 0$.
This inequality holds when $0 < a < 4$.
Since the discriminant $D = B^2 - 4AC$ is always non-negative for these values of $a$ (as $AC < 0$ implies $D = B^2 - 4AC > 0$),the roots are guaranteed to be real and distinct.
Thus,the correct range is $0 < a < 4$.

(B) Given the quadratic equation $x^2 - (p - 4)x + 2e^{2 \ln p} - 4 = 0$.
Since $e^{2 \ln p} = e^{\ln p^2} = p^2$,the equation becomes $x^2 - (p - 4)x + 2p^2 - 4 = 0$.
For both roots to be negative,the following conditions must be satisfied:
$1.$ Discriminant $D \ge 0$: $(p - 4)^2 - 4(2p^2 - 4) \ge 0 \implies p^2 - 8p + 16 - 8p^2 + 16 \ge 0 \implies -7p^2 - 8p + 32 \ge 0 \implies 7p^2 + 8p - 32 \le 0$.
Solving $7p^2 + 8p - 32 = 0$,we get $p = \frac{-8 \pm \sqrt{64 - 4(7)(-32)}}{14} = \frac{-8 \pm \sqrt{64 + 896}}{14} = \frac{-8 \pm \sqrt{960}}{14} = \frac{-8 \pm 8\sqrt{15}}{14} = \frac{-4 \pm 4\sqrt{15}}{7}$.
$2.$ Sum of roots $< 0$: $p - 4 < 0 \implies p < 4$.
$3.$ Product of roots $> 0$: $2p^2 - 4 > 0 \implies p^2 > 2 \implies p > \sqrt{2}$ or $p < -\sqrt{2}$.
Also,for $\ln p$ to be defined,$p > 0$.
Combining $p > \sqrt{2}$ and $p < 4$,we get $p \in (\sqrt{2}, 4)$.

(C) The given equation is $x^2 - 2mx + m^2 - 1 = 0$.
This can be rewritten as $(x - m)^2 - 1 = 0$.
$(x - m)^2 = 1$.
Taking the square root on both sides,we get $x - m = \pm 1$.
Thus,the roots are $x_1 = m - 1$ and $x_2 = m + 1$.
We are given that both roots are greater than $-2$ and less than $4$,so $-2 < m - 1$ and $m + 1 < 4$.
From $-2 < m - 1$,we get $m > -1$.
From $m + 1 < 4$,we get $m < 3$.
Combining these,we get $-1 < m < 3$.
Therefore,the interval is $(-1, 3)$.

(C) Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + ax + 1 = 0$.
Then,$\alpha + \beta = -a$ and $\alpha \beta = 1$.
The difference between the roots is given by $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}$.
Substituting the values,we get $|\alpha - \beta| = \sqrt{a^2 - 4}$.
According to the given condition,$|\alpha - \beta| < \sqrt{5}$.
Therefore,$\sqrt{a^2 - 4} < \sqrt{5}$.
Squaring both sides,we get $a^2 - 4 < 5$,which implies $a^2 < 9$.
This inequality holds when $|a| < 3$,which means $a \in (-3, 3)$.

(C) Given that the equation $bx^2 + cx + a = 0$ has imaginary roots,the discriminant $D < 0$.
Therefore,$c^2 - 4ab < 0$,which implies $c^2 < 4ab$.
Multiplying by $-1$,we get $-c^2 > -4ab$.
Now,consider the expression $E = 3b^2x^2 + 6bcx + 2c^2$.
$E = 3(b^2x^2 + 2bcx + c^2) - c^2$.
$E = 3(bx + c)^2 - c^2$.
Since $(bx + c)^2 \geq 0$,it follows that $3(bx + c)^2 \geq 0$.
Thus,$E \geq -c^2$.
Since $-c^2 > -4ab$,we have $E > -4ab$.

(B) The given equation is $x^2 - x + 1 = 0$.
Multiplying by $(x + 1)$,we get $(x + 1)(x^2 - x + 1) = 0$,which implies $x^3 + 1 = 0$,so $x^3 = -1$.
Since $\alpha$ and $\beta$ are roots of the equation,they satisfy $x^3 = -1$,so $\alpha^3 = -1$ and $\beta^3 = -1$.
We need to find $\alpha^{2009} + \beta^{2009}$.
$\alpha^{2009} = (\alpha^3)^{669} \cdot \alpha^2 = (-1)^{669} \cdot \alpha^2 = -\alpha^2$.
Similarly,$\beta^{2009} = -\beta^2$.
Thus,$\alpha^{2009} + \beta^{2009} = -(\alpha^2 + \beta^2)$.
From the equation $x^2 - x + 1 = 0$,we have $\alpha + \beta = 1$ and $\alpha\beta = 1$.
We know $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (1)^2 - 2(1) = 1 - 2 = -1$.
Therefore,$\alpha^{2009} + \beta^{2009} = -(-1) = 1$.

(C) The equation $f(x)^{g(x)} = 1$ holds if:
$1)$ $f(x) = 1$
$x^2 - 5x + 5 = 1$ $\Rightarrow x^2 - 5x + 4 = 0$ $\Rightarrow (x-1)(x-4) = 0$ $\Rightarrow x = 1, 4$.
$2)$ $f(x) = -1$ and $g(x)$ is an even integer.
$x^2 - 5x + 5 = -1$ $\Rightarrow x^2 - 5x + 6 = 0$ $\Rightarrow (x-2)(x-3) = 0$ $\Rightarrow x = 2, 3$.
For $x = 2$,$g(2) = 2^2 + 4(2) - 60 = 4 + 8 - 60 = -48$ (even),so $x = 2$ is a solution.
For $x = 3$,$g(3) = 3^2 + 4(3) - 60 = 9 + 12 - 60 = -39$ (odd),so $x = 3$ is rejected.
$3)$ $f(x) = 0$ and $g(x) > 0$ (not applicable here as $f(x)$ is not $0$ for these roots) or $g(x) = 0$ and $f(x) \neq 0$.
$g(x) = x^2 + 4x - 60 = 0$ $\Rightarrow (x+10)(x-6) = 0$ $\Rightarrow x = -10, 6$.
For $x = -10$,$f(-10) = (-10)^2 - 5(-10) + 5 = 100 + 50 + 5 = 155 \neq 0$.
For $x = 6$,$f(6) = 6^2 - 5(6) + 5 = 36 - 30 + 5 = 11 \neq 0$.
Thus,the valid values of $x$ are $1, 4, 2, -10, 6$.
The sum is $1 + 4 + 2 - 10 + 6 = 3$.

(A) The given equation is $\sum_{r=1}^{n} (x + r - 1)(x + r) = 10n$.
Expanding the terms: $\sum_{r=1}^{n} (x^2 + (2r - 1)x + r^2 - r) = 10n$.
This simplifies to $nx^2 + x \sum_{r=1}^{n} (2r - 1) + \sum_{r=1}^{n} (r^2 - r) = 10n$.
Using summation formulas: $nx^2 + n^2x + \frac{(n-1)n(n+1)}{3} = 10n$.
Dividing by $n$ (since $n$ is a positive integer): $x^2 + nx + \frac{n^2 - 1}{3} = 10$.
$x^2 + nx + \frac{n^2 - 31}{3} = 0$.
Let the two consecutive integral solutions be $\alpha$ and $\alpha + 1$.
Sum of roots: $2\alpha + 1 = -n \Rightarrow \alpha = \frac{-(n+1)}{2}$.
Product of roots: $\alpha(\alpha + 1) = \frac{n^2 - 31}{3}$.
Substituting $\alpha$: $(\frac{-(n+1)}{2})(\frac{-(n+1)}{2} + 1) = \frac{n^2 - 31}{3}$.
$(\frac{-(n+1)}{2})(\frac{1-n}{2}) = \frac{n^2 - 31}{3}$.
$\frac{n^2 - 1}{4} = \frac{n^2 - 31}{3}$.
$3n^2 - 3 = 4n^2 - 124$.
$n^2 = 121 \Rightarrow n = 11$.

(D) Given $p(x) = f(x) - g(x) = (a - a_1)x^2 + (b - b_1)x + (c - c_1).$
Since $p(x) = 0$ has only one root at $x = -1,$ the quadratic $p(x)$ must be a perfect square of the form $p(x) = k(x + 1)^2$ for some constant $k = a - a_1 \ne 0.$
Given $p(-2) = 2,$
$k(-2 + 1)^2 = 2 \Rightarrow k(-1)^2 = 2 \Rightarrow k = 2.$
Thus,$p(x) = 2(x + 1)^2.$
Now,we need to find $p(2):$
$p(2) = 2(2 + 1)^2 = 2(3)^2 = 2 \times 9 = 18.$

(C) Given the linear congruence $8x \equiv 6 \pmod{14}$.
This is equivalent to $8x - 6 = 14k$ for some integer $k \in \mathbb{Z}$.
Dividing by $2$,we get $4x - 3 = 7k$,which implies $4x \equiv 3 \pmod{7}$.
Multiplying by the modular inverse of $4$ modulo $7$,which is $2$ (since $4 \times 2 = 8 \equiv 1 \pmod{7}$):
$2(4x) \equiv 2(3) \pmod{7} \implies 8x \equiv 6 \pmod{7} \implies x \equiv 6 \pmod{7}$.
Thus,$x$ can be written as $x = 7n + 6$ for $n \in \mathbb{Z}$.
For $n = 0, x = 6$. For $n = 1, x = 13$. For $n = 2, x = 20$,and so on.
The solutions are $x \in \{ \dots, -1, 6, 13, 20, 27, \dots \}$.
This set can be represented as the union of two equivalence classes modulo $14$: $[6] = \{ \dots, -8, 6, 20, 34, \dots \}$ and $[13] = \{ \dots, -1, 13, 27, 41, \dots \}$.
Therefore,the solution set is $[6] \cup [13]$.

(B) Let $x = \sqrt {a + \sqrt {a + \sqrt {a + .....\infty } } } $
Since the expression is infinite,we can write:
$x = \sqrt {a + x}$
Squaring both sides:
$x^2 = a + x$
Rearranging the terms:
$x^2 - x - a = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-a)}}{2(1)}$
$x = \frac{1 \pm \sqrt{1 + 4a}}{2}$
Since $a > 0$,the value of $x$ must be positive. Therefore,we discard the negative root:
$x = \frac{1 + \sqrt{4a + 1}}{2}$

(B) Given equation: $|{x^2} - x - 6| = x + 2$
Case $I$: ${x^2} - x - 6 < 0$
$(x - 3)(x + 2) < 0 \Rightarrow -2 < x < 3$
In this case,the equation becomes: $-(x^2 - x - 6) = x + 2$
$-x^2 + x + 6 = x + 2$ $\Rightarrow x^2 = 4$ $\Rightarrow x = \pm 2$
Since the domain is $-2 < x < 3$,only $x = 2$ is a valid solution.
Case $II$: ${x^2} - x - 6 \ge 0$
$(x - 3)(x + 2) \ge 0 \Rightarrow x \le -2$ or $x \ge 3$
In this case,the equation becomes: $x^2 - x - 6 = x + 2$
$x^2 - 2x - 8 = 0$ $\Rightarrow (x - 4)(x + 2) = 0$ $\Rightarrow x = 4$ or $x = -2$
Both values satisfy the domain $x \le -2$ or $x \ge 3$.
Combining both cases,the solutions are $x = -2, 2, 4$.

(C) For a quadratic equation $Ax^2 + Bx + C = 0$ to have roots with opposite signs,the product of the roots must be negative,i.e.,$\frac{C}{A} < 0$.
Here,$A = 3$ and $C = a^2 - 3a + 2$.
Thus,$\frac{a^2 - 3a + 2}{3} < 0$,which implies $a^2 - 3a + 2 < 0$.
Factoring the quadratic expression,we get $(a - 1)(a - 2) < 0$.
This inequality holds when $1 < a < 2$.
Since the discriminant $D = B^2 - 4AC = [2(a^2 + 1)]^2 - 4(3)(a^2 - 3a + 2) = 4(a^4 + 2a^2 + 1) - 12(a^2 - 3a + 2) = 4(a^4 - a^2 + 9a - 5)$ is always positive for $a \in (1, 2)$,the roots are guaranteed to be real.
Therefore,the value of $a$ lies in $(1, 2)$.

(C) Given the roots of $x^2 + px + q = 0$ are $\alpha$ and $\beta$,we have $\alpha + \beta = -p$ and $\alpha\beta = q$.
Given the roots of $x^2 - xr + s = 0$ are $\alpha^4$ and $\beta^4$,we have $\alpha^4 + \beta^4 = r$ and $\alpha^4\beta^4 = s$.
Consider the equation $x^2 - 4qx + 2q^2 - r = 0$. The discriminant $D$ is given by:
$D = (-4q)^2 - 4(1)(2q^2 - r) = 16q^2 - 8q^2 + 4r = 8q^2 + 4r$.
Substituting $q = \alpha\beta$ and $r = \alpha^4 + \beta^4$:
$D = 8(\alpha\beta)^2 + 4(\alpha^4 + \beta^4) = 4(2\alpha^2\beta^2 + \alpha^4 + \beta^4) = 4(\alpha^2 + \beta^2)^2$.
Since $(\alpha^2 + \beta^2)^2 \ge 0$,it follows that $D \ge 0$.
Therefore,the roots of the equation $x^2 - 4qx + 2q^2 - r = 0$ are always real.

(C) Given the expression $f(x) = mx - 1 + \frac{1}{x} \ge 0$ for $x > 0$.
Multiplying by $x$ (since $x > 0$),we get $mx^2 - x + 1 \ge 0$.
For a quadratic expression $ax^2 + bx + c \ge 0$ to be non-negative for all $x > 0$,we must have $a > 0$ and the discriminant $D \le 0$.
Here,$a = m$,$b = -1$,and $c = 1$.
Condition $1$: $m > 0$.
Condition $2$: $D = b^2 - 4ac = (-1)^2 - 4(m)(1) = 1 - 4m \le 0$.
Solving $1 - 4m \le 0$ gives $4m \ge 1$,or $m \ge \frac{1}{4}$.
Since $m \ge \frac{1}{4}$ satisfies $m > 0$,the minimum value of $m$ is $\frac{1}{4}$.

(C) Given the quadratic equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$.
Since the sum of the coefficients is $a(b - c) + b(c - a) + c(a - b) = ab - ac + bc - ab + ac - bc = 0$,one root of the equation must be $1$.
Since the roots are equal,both roots must be $1$.
For a quadratic equation $Ax^2 + Bx + C = 0$,the product of roots is $\frac{C}{A}$.
Thus,$1 \times 1 = \frac{c(a - b)}{a(b - c)}$.
$a(b - c) = c(a - b)$
$ab - ac = ac - bc$
$ab + bc = 2ac$
Dividing both sides by $abc$,we get $\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$.
This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$,which means $a, b, c$ are in $H.P.$

(C) The given cubic equation is $x^3 + 3Hx + G = 0$,where $G$ and $H$ are real constants.
For a cubic equation of the form $x^3 + px + q = 0$,the discriminant is given by $\Delta = - (4p^3 + 27q^2)$.
In our case,$p = 3H$ and $q = G$.
Substituting these,we get $\Delta = - (4(3H)^3 + 27G^2) = - (108H^3 + 27G^2) = -27(4H^3 + G^2)$.
Given that $G^2 + 4H^3 > 0$,it follows that $\Delta = -27(G^2 + 4H^3) < 0$.
For a cubic equation with real coefficients,if the discriminant $\Delta < 0$,the equation has one real root and two complex conjugate (imaginary) roots.

(B) Given the quadratic equation $x^2 + px + (1 - p) = 0$.
Since $(1 - p)$ is a root,it must satisfy the equation:
$(1 - p)^2 + p(1 - p) + (1 - p) = 0$
$(1 - p) [ (1 - p) + p + 1 ] = 0$
$(1 - p) [ 2 ] = 0$
This implies $1 - p = 0$,so $p = 1$.
Substituting $p = 1$ into the original equation:
$x^2 + 1x + (1 - 1) = 0$
$x^2 + x = 0$
$x(x + 1) = 0$
Thus,the roots are $x = 0$ and $x = -1$.

(D) The domain of definition of the function $y = \sqrt{x(x - 3)}$ is $x(x - 3) \ge 0$,which implies $x \le 0$ or $x \ge 3$. $(i)$
The given equation is $(3|x| - 3)^2 = |x| + 7$.
Expanding the left side: $9|x|^2 - 18|x| + 9 = |x| + 7$.
Rearranging the terms: $9|x|^2 - 19|x| + 2 = 0$.
Let $t = |x|$,then $9t^2 - 19t + 2 = 0$.
Factoring the quadratic: $(9t - 1)(t - 2) = 0$.
Thus,$|x| = 1/9$ or $|x| = 2$.
This gives $x = \pm 1/9$ or $x = \pm 2$.
Checking these values against the domain condition $(i)$ ($x \le 0$ or $x \ge 3$):
For $x = 2$,$2$ is not in the domain.
For $x = -2$,$-2 \le 0$ (Valid).
For $x = 1/9$,$1/9$ is not in the domain.
For $x = -1/9$,$-1/9 \le 0$ (Valid).
Therefore,the solutions are $-2$ and $-1/9$.

(C) Given that ${x^2} + 2ax + {a^2} = {(x + a)^2}$ is a factor of $f(x) = {x^3} - 3px + 2q$.
Since $(x+a)^2$ is a factor,$x = -a$ must be a root of $f(x) = 0$ and $f'(x) = 0$.
First,$f(-a) = {(-a)^3} - 3p(-a) + 2q = 0$.
$-{a^3} + 3pa + 2q = 0$ ...$(i)$
Next,$f'(x) = 3{x^2} - 3p$.
Setting $f'(-a) = 0$,we get $3{(-a)^2} - 3p = 0$.
$3{a^2} = 3p \Rightarrow p = {a^2}$ ...$(ii)$
Substitute $p = {a^2}$ into equation $(i)$:
$-{a^3} + 3({a^2})a + 2q = 0$
$-{a^3} + 3{a^3} + 2q = 0$
$2{a^3} + 2q = 0 \Rightarrow {a^3} = -q$.
Squaring both sides,we get ${a^6} = {q^2}$.
Since $p = {a^2}$,then ${p^3} = {({a^2})^3} = {a^6}$.
Therefore,${p^3} = {q^2}$.

(A) The congruence $8x \equiv 6 \pmod{14}$ can be written as $8x - 6 = 14k$ for some integer $k \in \mathbb{Z}$.
Dividing by $2$,we get $4x - 3 = 7k$,which implies $4x \equiv 3 \pmod{7}$.
To solve $4x \equiv 3 \pmod{7}$,we multiply by the modular inverse of $4$ modulo $7$. Since $4 \times 2 = 8 \equiv 1 \pmod{7}$,the inverse is $2$.
Multiplying both sides by $2$: $8x \equiv 6 \pmod{7}$,which simplifies to $x \equiv 6 \pmod{7}$.
This means $x$ can be written in the form $x = 7n + 6$ for $n \in \mathbb{Z}$.
For $n = 0, x = 6$. For $n = 1, x = 13$. For $n = 2, x = 20$,and so on.
The solution set consists of integers congruent to $6 \pmod{7}$ or $13 \pmod{7}$,which can be represented as the union of equivalence classes $[6]$ and $[13]$ modulo $14$.

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

(A) Given $x = {(\sqrt 2 + 1)^{1/3}} - {(\sqrt 2 - 1)^{1/3}}$.
Using the identity $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$,we have:
${x^3} = (\sqrt 2 + 1) - (\sqrt 2 - 1) - 3{(\sqrt 2 + 1)^{1/3}}{(\sqrt 2 - 1)^{1/3}} \left[ {(\sqrt 2 + 1)^{1/3}} - {(\sqrt 2 - 1)^{1/3}} \right]$.
Since ${(\sqrt 2 + 1)^{1/3}}{(\sqrt 2 - 1)^{1/3}} = {((\sqrt 2 + 1)(\sqrt 2 - 1))^{1/3}} = {(2 - 1)^{1/3}} = 1$,the equation becomes:
${x^3} = 2 - 3(1)x$.
Therefore,${x^3} + 3x = 2$.

(C) Given $x = 2^{1/3} - 2^{-1/3}$.
Cube both sides: $x^3 = (2^{1/3} - 2^{-1/3})^3$.
Using the identity $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$:
$x^3 = (2^{1/3})^3 - (2^{-1/3})^3 - 3(2^{1/3})(2^{-1/3})(2^{1/3} - 2^{-1/3})$.
$x^3 = 2 - 2^{-1} - 3(1)(x)$.
$x^3 = 2 - 1/2 - 3x$.
$x^3 = 3/2 - 3x$.
Multiply by $2$: $2x^3 = 3 - 6x$.
Rearranging gives: $2x^3 + 6x = 3$.

(B) Let $x = \sqrt{a + \sqrt{a + \sqrt{a + \dots \infty}}}$.
Squaring both sides,we get $x^2 = a + \sqrt{a + \sqrt{a + \dots \infty}}$.
Since the expression inside the square root is the same as $x$,we can write $x^2 = a + x$.
Rearranging the terms,we get the quadratic equation $x^2 - x - a = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a=1, b=-1, c=-a$,we get $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-a)}}{2(1)}$.
This simplifies to $x = \frac{1 \pm \sqrt{1 + 4a}}{2}$.
Since $x$ must be positive (as it is a square root),we take the positive root: $x = \frac{1 + \sqrt{4a + 1}}{2}$.

(B) Given $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$.
First,find $y$: $y = \frac{4}{x} = \frac{4}{\sqrt{7} + \sqrt{3}}$.
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 need to find $x^4 + y^4$.
Note that $x^2 + y^2 = (x + y)^2 - 2xy = (2\sqrt{7})^2 - 2(4) = 28 - 8 = 20$.
Then,$x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2 = (20)^2 - 2(4)^2 = 400 - 2(16) = 400 - 32 = 368$.

(B) Given equation: $\sqrt{x + 10} + \sqrt{x - 2} = 6$
Let $u = \sqrt{x + 10}$ and $v = \sqrt{x - 2}$.
Then $u^2 = x + 10$ and $v^2 = x - 2$.
Subtracting the two: $u^2 - v^2 = (x + 10) - (x - 2) = 12$.
We know $(u - v)(u + v) = 12$.
Since $u + v = 6$,we have $(u - v)(6) = 12$,which implies $u - v = 2$.
Now we have a system:
$u + v = 6$
$u - v = 2$
Adding these: $2u = 8 \implies u = 4$.
Since $u = \sqrt{x + 10} = 4$,squaring both sides gives $x + 10 = 16$,so $x = 6$.
Checking the value: $\sqrt{6 + 10} + \sqrt{6 - 2} = \sqrt{16} + \sqrt{4} = 4 + 2 = 6$.
The solution is $x = 6$.

(A) Let the remainder be $R(x) = ax^2 + bx + c$ since the divisor is a cubic polynomial.
By the Remainder Theorem,we have:
$f(-1) = 6 \implies a(-1)^2 + b(-1) + c = 6 \implies a - b + c = 6$ $(i)$
$f(2) = 3 \implies a(2)^2 + b(2) + c = 3 \implies 4a + 2b + c = 3$ (ii)
$f(-2) = 15 \implies a(-2)^2 + b(-2) + c = 15 \implies 4a - 2b + c = 15$ (iii)
Subtracting (iii) from (ii): $(4a + 2b + c) - (4a - 2b + c) = 3 - 15 \implies 4b = -12 \implies b = -3$.
Substituting $b = -3$ into $(i)$: $a - (-3) + c = 6 \implies a + c = 3 \implies c = 3 - a$.
Substituting $b = -3$ and $c = 3 - a$ into (ii): $4a + 2(-3) + (3 - a) = 3 \implies 3a - 3 = 3 \implies 3a = 6 \implies a = 2$.
Then $c = 3 - 2 = 1$.
Thus,the remainder is $R(x) = 2x^2 - 3x + 1$.

(D) For the $x$-coordinates: $x^2 - 3|x| + 2 = 0$.
Let $|x| = t$,then $t^2 - 3t + 2 = 0$,which gives $(t-1)(t-2) = 0$.
So,$|x| = 1$ or $|x| = 2$,which implies $x \in \{1, -1, 2, -2\}$.
For the $y$-coordinates: $y^2 - 3y + 2 = 0$.
This gives $(y-1)(y-2) = 0$,so $y \in \{1, 2\}$.
The square has unit area,meaning the side length is $1$.
Possible sets of vertices forming a square of side $1$ are:
$S_1 = \{(1, 1), (2, 1), (2, 2), (1, 2)\}$
$S_2 = \{(-1, 1), (-2, 1), (-2, 2), (-1, 2)\}$
Both sets represent squares of unit area.
Thus,both $(A)$ and $(B)$ are correct.

(C) Given the polynomial $P(x) = 2x^3 + x^2 + 3x - 2$.
First,we check the derivative $P'(x) = 6x^2 + 2x + 3$.
The discriminant of $P'(x)$ is $D = (2)^2 - 4(6)(3) = 4 - 72 = -68 < 0$.
Since the leading coefficient is positive and the discriminant is negative,$P'(x) > 0$ for all $x \in \mathbb{R}$.
This implies that $P(x)$ is a strictly increasing function.
As a strictly increasing cubic polynomial,it must have exactly one real root and two complex conjugate roots.
Evaluating at endpoints: $P(0) = -2$ and $P(1) = 2(1)^3 + (1)^2 + 3(1) - 2 = 4$.
Since $P(0) < 0$ and $P(1) > 0$,by the Intermediate Value Theorem,there exists a root in the interval $(0, 1)$.
Since there is only one real root and it is positive,statements $(i)$ and $(iii)$ are true.
Statement $(ii)$ is false because there are no negative roots.
Statement $(iv)$ is false because a cubic polynomial with real coefficients must have either one or three real roots.
Statement $(v)$ is false because the only real root is positive.
Statement $(vi)$ is false because a cubic polynomial with one real root must have two complex roots.
Therefore,only $(i)$ and $(iii)$ are true.

(A) Let $x = \frac{p}{q}$ be a rational root in simplest form,where $p, q \in \mathbb{Z}$,$q > 0$,and $\gcd(p, q) = 1$.
According to the Rational Root Theorem,$p$ must be a divisor of the constant term $1$,and $q$ must be a divisor of the leading coefficient $1$.
Thus,$p, q \in \{-1, 1\}$,which implies $x \in \{-1, 1\}$.
Now,test these values in the equation $f(x) = x^{2016} - x^{2015} + x^{1008} + x^{1003} + 1 = 0$:
For $x = 1$: $f(1) = 1^{2016} - 1^{2015} + 1^{1008} + 1^{1003} + 1 = 1 - 1 + 1 + 1 + 1 = 3 \neq 0$.
For $x = -1$: $f(-1) = (-1)^{2016} - (-1)^{2015} + (-1)^{1008} + (-1)^{1003} + 1 = 1 - (-1) + 1 - 1 + 1 = 1 + 1 + 1 - 1 + 1 = 3 \neq 0$.
Since neither $1$ nor $-1$ is a root,the equation has no rational roots.
Therefore,the number of rational roots is $0$.

(B) Given the equation: $\frac{P^2}{x} + \frac{Q^2}{x - 1} = 1$
Multiply by $x(x - 1)$ to clear the denominators:
$P^2(x - 1) + Q^2(x) = x(x - 1)$
$P^2x - P^2 + Q^2x = x^2 - x$
Rearranging into a standard quadratic form $ax^2 + bx + c = 0$:
$x^2 - (P^2 + Q^2 + 1)x + P^2 = 0$
The discriminant $D$ of this quadratic equation is given by $D = b^2 - 4ac$:
$D = (P^2 + Q^2 + 1)^2 - 4(1)(P^2)$
$D = (P^2 + Q^2 + 1)^2 - 4P^2$
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
$D = (P^2 + Q^2 + 1 - 2P)(P^2 + Q^2 + 1 + 2P)$
$D = ((P - 1)^2 + Q^2)((P + 1)^2 + Q^2)$
Since $P$ and $Q$ are non-zero real numbers,$Q^2 > 0$. Thus,$D > 0$ for all non-zero real $P$ and $Q$.
Since the discriminant is strictly positive,the quadratic equation has $2$ distinct real roots.

(D) Given the equation $[x^2] = 2x - 1$.
Since $[x^2]$ is an integer,$2x - 1$ must be an integer,which implies $2x$ is an integer.
By the definition of the greatest integer function,$x^2 - 1 < [x^2] \le x^2$.
Substituting $[x^2] = 2x - 1$,we get $x^2 - 1 < 2x - 1 \le x^2$.
From $x^2 - 1 < 2x - 1$,we have $x^2 - 2x < 0$,so $x(x - 2) < 0$,which gives $0 < x < 2$.
From $2x - 1 \le x^2$,we have $x^2 - 2x + 1 \ge 0$,which is $(x - 1)^2 \ge 0$,which is true for all $x$.
Case $1$: $0 \le x^2 < 1 \Rightarrow 0 \le x < 1$. Then $[x^2] = 0$. The equation becomes $0 - 2x + 1 = 0$,so $x = 1/2$. Since $0 \le 1/2 < 1$,this is a solution.
Case $2$: $1 \le x^2 < 2 \Rightarrow 1 \le x < \sqrt{2}$. Then $[x^2] = 1$. The equation becomes $1 - 2x + 1 = 0$,so $2x = 2$,$x = 1$. Since $1 \le 1 < \sqrt{2}$,this is a solution.
Case $3$: $2 \le x^2 < 3 \Rightarrow \sqrt{2} \le x < \sqrt{3}$. Then $[x^2] = 2$. The equation becomes $2 - 2x + 1 = 0$,so $2x = 3$,$x = 3/2$. Since $1.414 \le 1.5 < 1.732$,this is a solution.
Case $4$: $3 \le x^2 < 4 \Rightarrow \sqrt{3} \le x < 2$. Then $[x^2] = 3$. The equation becomes $3 - 2x + 1 = 0$,so $2x = 4$,$x = 2$. But $x < 2$,so this is not a solution.
The solutions are $x = 1/2, 1, 3/2$. The sum is $1/2 + 1 + 3/2 = 3$.

(C) Given that the numbers $a$ and $b$ differ from their reciprocals by $1$,we have the equation $|x - \frac{1}{x}| = 1$.
This implies $x - \frac{1}{x} = 1$ or $x - \frac{1}{x} = -1$.
For $x - \frac{1}{x} = 1$,we get $x^2 - x - 1 = 0$. The roots are $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Since $a, b > 0$,we take $a = \frac{1 + \sqrt{5}}{2}$.
For $x - \frac{1}{x} = -1$,we get $x^2 + x - 1 = 0$. The roots are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Since $a, b > 0$,we take $b = \frac{\sqrt{5} - 1}{2}$.
Thus,$a + b = \frac{1 + \sqrt{5}}{2} + \frac{\sqrt{5} - 1}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}$.

(B) Given that $\alpha, \beta, \gamma, \delta$ are the roots of $x^4 + x^2 + 1 = 0$.
Let $y = x^2$.
Since $x$ is a root of the given equation,we have $x^4 + x^2 + 1 = 0$.
Substituting $x^2 = y$,we get $y^2 + y + 1 = 0$.
Since the roots of the new equation are $y_1 = \alpha^2, y_2 = \beta^2, y_3 = \gamma^2, y_4 = \delta^2$,and each root of the original equation satisfies $x^4 + x^2 + 1 = 0$,we observe that the equation $y^2 + y + 1 = 0$ is satisfied by these squared values.
Since the original equation is of degree $4$,it has $4$ roots. The equation $y^2 + y + 1 = 0$ is of degree $2$,but it represents the squared values.
Specifically,$x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1) = 0$.
The roots of $x^2 + x + 1 = 0$ are $\omega, \omega^2$ and the roots of $x^2 - x + 1 = 0$ are $-\omega, -\omega^2$.
Squaring these,we get $\omega^2, \omega^4 = \omega$ and $(-\omega)^2 = \omega^2, (-\omega^2)^2 = \omega^4 = \omega$.
Thus,the roots $\alpha^2, \beta^2, \gamma^2, \delta^2$ are $\omega, \omega^2, \omega, \omega^2$.
These are the roots of $(x^2 + x + 1)^2 = 0$.

(B) For the quadratic equation $ax^2 + bx + 1 = 0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = b^2 - 4ac \geq 0$
Given $c = 1$,we have $b^2 - 4a \geq 0$,or $b^2 \geq 4a$.
We test values for $b \in \{1, 2, 3, 4\}$:
$1$. If $b = 1$,$1^2 \geq 4a \Rightarrow 1 \geq 4a$,which has no solution for $a \in \{1, 2, 3, 4\}$.
$2$. If $b = 2$,$2^2 \geq 4a$ $\Rightarrow 4 \geq 4a$ $\Rightarrow a \leq 1$. Thus,$a = 1$ ($1$ solution).
$3$. If $b = 3$,$3^2 \geq 4a$ $\Rightarrow 9 \geq 4a$ $\Rightarrow a \leq 2.25$. Thus,$a \in \{1, 2\}$ ($2$ solutions).
$4$. If $b = 4$,$4^2 \geq 4a$ $\Rightarrow 16 \geq 4a$ $\Rightarrow a \leq 4$. Thus,$a \in \{1, 2, 3, 4\}$ ($4$ solutions).
Total number of equations = $0 + 1 + 2 + 4 = 7$.

(D) Given the equation $ax^2 + bx + c = 0$ with $a < 0, b > 0, c > 0$.
Multiply by $-1$ to get $-ax^2 - bx - c = 0$,where $-a > 0, -b < 0, -c < 0$.
Let $f(x) = ax^2 + bx + c$. Since $a < 0$,the parabola opens downwards.
Since $f(0) = c > 0$,the $y$-intercept is positive.
Since the product of roots $\alpha \beta = \frac{c}{a} < 0$,the roots have opposite signs.
Since $\alpha < \beta$,we have $\alpha < 0 < \beta$.
The sum of roots $\alpha + \beta = -\frac{b}{a}$. Since $a < 0$ and $b > 0$,$-\frac{b}{a} > 0$,so $\alpha + \beta > 0$.
This implies $|\beta| > |\alpha|$.
Thus,$\alpha < 0 < \beta < |\beta|$ is incorrect,but checking the magnitude condition: $\alpha + \beta > 0$ means the positive root $\beta$ has a larger absolute value than the negative root $\alpha$,i.e.,$|\beta| > |\alpha|$.
Given $\alpha < 0 < \beta$,and $|\beta| > |\alpha|$,the correct relation is $\alpha < 0 < |\alpha| < \beta$.

(B) Given $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$.
This simplifies to $\frac{x_1 x_2 + x_2 x_3 + x_3 x_1}{x_1 x_2 x_3} = \frac{1}{x_1 + x_2 + x_3}$.
$(x_1 + x_2 + x_3)(x_1 x_2 + x_2 x_3 + x_3 x_1) = x_1 x_2 x_3$.
Let $x_1, x_2, x_3$ be roots of the cubic equation $t^3 + \alpha t^2 + \beta t + \gamma = 0$,where $\alpha = -(x_1 + x_2 + x_3)$,$\beta = x_1 x_2 + x_2 x_3 + x_3 x_1$,and $\gamma = -x_1 x_2 x_3$.
The equation becomes $(-\alpha)(\beta) = -(-\gamma) \implies \gamma = \alpha \beta$.
Substituting this into the cubic equation: $t^3 + \alpha t^2 + \beta t + \alpha \beta = 0$.
$t^2(t + \alpha) + \beta(t + \alpha) = 0 \implies (t + \alpha)(t^2 + \beta) = 0$.
Thus,the roots are $t = -\alpha, \sqrt{-\beta}, -\sqrt{-\beta}$.
Let $x_1 = -\alpha$,$x_2 = k$,and $x_3 = -k$ where $k = \sqrt{-\beta}$.
Then $x_1^n + x_2^n + x_3^n = x_1^n + k^n + (-k)^n$.
If $n$ is an odd integer,$k^n + (-k)^n = 0$,so $x_1^n + x_2^n + x_3^n = x_1^n$.
Also,$\frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n} = \frac{1}{x_1^n} + \frac{1}{k^n} + \frac{1}{(-k)^n} = \frac{1}{x_1^n} + 0 = \frac{1}{x_1^n}$.
Thus,the equality holds for all odd integers $n$.

(A) Let the roots of the equation be $x_1, x_2, x_3 > 0$.
From Vieta's formulas:
$x_1 + x_2 + x_3 = 2a$
$x_1x_2 + x_2x_3 + x_3x_1 = 3b$
$x_1x_2x_3 = 8$
Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality for the terms $\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}$:
$\frac{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3}}{3} \geq \sqrt[3]{\frac{1}{x_1} \cdot \frac{1}{x_2} \cdot \frac{1}{x_3}}$
$\frac{\frac{x_1x_2 + x_2x_3 + x_3x_1}{x_1x_2x_3}}{3} \geq \left(\frac{1}{8}\right)^{1/3}$
$\frac{\frac{3b}{8}}{3} \geq \frac{1}{2}$
$\frac{b}{8} \geq \frac{1}{2}$
$b \geq 4$
Thus,the minimum value of $b$ is $4$.

(C) For the equation $ax^2 + bx + c = 0$,the roots are $r_1 = \alpha - \beta$ and $r_2 = \gamma - \delta$. The difference of the roots is $|r_1 - r_2| = |(\alpha - \beta) - (\gamma - \delta)| = |\alpha - \beta - \gamma + \delta| = \frac{\sqrt{D_1}}{|a|}$.
For the equation $Ax^2 + Bx + C = 0$,the roots are $R_1 = \alpha + \delta$ and $R_2 = \beta + \gamma$. The difference of the roots is $|R_1 - R_2| = |(\alpha + \delta) - (\beta + \gamma)| = |\alpha - \beta + \delta - \gamma| = \frac{\sqrt{D_2}}{|A|}$.
Since $|\alpha - \beta - \gamma + \delta| = |\alpha - \beta + \delta - \gamma|$,we have $\frac{\sqrt{D_1}}{|a|} = \frac{\sqrt{D_2}}{|A|}$.
Therefore,$\left| \frac{a}{A} \right| = \sqrt{\frac{D_1}{D_2}}$.

(A) Let $f(x) = \frac{3}{x - a^3} + \frac{5}{x - a^5} + \frac{7}{x - a^7}$.
Given $a > 1$,we have $a^3 < a^5 < a^7$.
As $x \to (a^3)^+$,$f(x) \to \infty$. As $x \to (a^5)^-$,$f(x) \to -\infty$. Thus,there is a root in $(a^3, a^5)$.
As $x \to (a^5)^+$,$f(x) \to \infty$. As $x \to (a^7)^-$,$f(x) \to -\infty$. Thus,there is a root in $(a^5, a^7)$.
Since $a > 1$,all intervals $(a^3, a^5)$ and $(a^5, a^7)$ consist of positive values.
Therefore,the equation has two real and positive roots.

(B) Given $P(x) = (x - \alpha)(x - \beta)(x - \gamma)$ where $\alpha, \beta, \gamma$ are integral roots.
Substituting $x = 6$ into the equation,we get:
$P(6) = (6 - \alpha)(6 - \beta)(6 - \gamma) = 3$.
Since $\alpha, \beta, \gamma$ are integers,$(6 - \alpha), (6 - \beta),$ and $(6 - \gamma)$ must be integer factors of $3$. The factors of $3$ are $\{1, -1, 3, -3\}$.
Let $x_1 = 6 - \alpha, x_2 = 6 - \beta, x_3 = 6 - \gamma$. Then $x_1 x_2 x_3 = 3$.
Possible sets for $(x_1, x_2, x_3)$ are:
$1) \{1, 1, 3\} \implies \alpha = 5, \beta = 5, \gamma = 3$. Then $a = \alpha + \beta + \gamma = 5 + 5 + 3 = 13$.
$2) \{1, -1, -3\} \implies \alpha = 5, \beta = 7, \gamma = 9$. Then $a = \alpha + \beta + \gamma = 5 + 7 + 9 = 21$.
$3) \{-1, -1, 3\} \implies \alpha = 7, \beta = 7, \gamma = 3$. Then $a = \alpha + \beta + \gamma = 7 + 7 + 3 = 17$.
Comparing these values with the options,$a$ can be $13, 17,$ or $21$. Thus,$a$ cannot be $15$.

(B) Let $f(x) = x^3 + x^2 - 5x - 1$.
To find the intervals where the roots lie,we examine the values of $f(x)$ at various points:
$f(1) = 1 + 1 - 5 - 1 = -4 < 0$
$f(2) = 8 + 4 - 10 - 1 = 1 > 0$
Since $f(1) < 0$ and $f(2) > 0$,one root $\alpha$ lies in $(1, 2)$. Thus,$[\alpha] = 1$.
$f(0) = -1 < 0$
$f(-1) = -1 + 1 + 5 - 1 = 4 > 0$
Since $f(-1) > 0$ and $f(0) < 0$,one root $\beta$ lies in $(-1, 0)$. Thus,$[\beta] = -1$.
$f(-2) = -8 + 4 + 10 - 1 = 5 > 0$
$f(-3) = -27 + 9 + 15 - 1 = -4 < 0$
Since $f(-3) < 0$ and $f(-2) > 0$,one root $\gamma$ lies in $(-3, -2)$. Thus,$[\gamma] = -3$.
Therefore,$[\alpha] + [\beta] + [\gamma] = 1 + (-1) + (-3) = -3$.

(B) Given the identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$.
This can be written as $\frac{1}{2}(a + b + c)[(a - b)^2 + (b - c)^2 + (c - a)^2] = 0$.
Since $a, b, c$ are distinct,$(a - b)^2 + (b - c)^2 + (c - a)^2 \neq 0$.
Therefore,we must have $a + b + c = 0$.
For the quadratic equation $ax^2 + bx + c = 0$,if we substitute $x = 1$,we get $a(1)^2 + b(1) + c = a + b + c$.
Since $a + b + c = 0$,$x = 1$ is one root of the equation.
Let the roots be $\alpha$ and $\beta$. We know $\alpha \cdot \beta = \frac{c}{a}$.
Since $\alpha = 1$,the other root $\beta = \frac{c}{a}$.

(D) The given equation is $x^4 + x^3 = 2(x^2 + 3ax + 2a^2)$.
Rearranging the terms,we get $x^4 + x^3 - 2x^2 - 6ax - 4a^2 = 0$.
This can be factored as $(x^2 + 2x + 2a)(x^2 - x - 2a) = 0$.
For the equation to have four real solutions,both quadratic factors must have real roots.
For $x^2 + 2x + 2a = 0$,the discriminant $D_1 = 2^2 - 4(1)(2a) = 4 - 8a \geq 0$,which implies $a \leq \frac{1}{2}$.
For $x^2 - x - 2a = 0$,the discriminant $D_2 = (-1)^2 - 4(1)(-2a) = 1 + 8a \geq 0$,which implies $a \geq -\frac{1}{8}$.
Combining these conditions,the set of values for $a$ is $[-\frac{1}{8}, \frac{1}{2}]$.

(B) Let $2^x = t$. Since $x > 0$,we have $t = 2^x > 2^0 = 1$.
Substituting $t$ into the equation,we get $t^2 - (k + 2)t + 2k = 0$.
Factoring the quadratic equation: $(t - k)(t - 2) = 0$.
So,the roots are $t_1 = k$ and $t_2 = 2$.
We require exactly one positive root for $x$,which means exactly one root for $t$ must be greater than $1$.
We already have one root $t_2 = 2$,which is greater than $1$.
For the equation to have exactly one root $t > 1$,the other root $t_1 = k$ must satisfy $k \le 1$.
Also,if $k = 2$,the roots are $t_1 = 2, t_2 = 2$,which gives only one distinct root $t = 2 > 1$,which is valid.
Thus,the condition is $k \le 1$ or $k = 2$.