Solution of quadratic equations and Nature of roots Questions in English

(C) Given $x = 2 + \sqrt{3}$,so $x - 2 = \sqrt{3}$.
Squaring both sides,we get $(x - 2)^2 = 3$,which simplifies to $x^2 - 4x + 4 = 3$,or $x^2 - 4x + 1 = 0$.
Now,we divide the polynomial $x^3 - 7x^2 + 13x - 12$ by $(x^2 - 4x + 1)$ or rewrite it:
$x^3 - 7x^2 + 13x - 12 = x(x^2 - 4x + 1) - 3x^2 + 12x - 12$
$= x(x^2 - 4x + 1) - 3(x^2 - 4x + 1) - 9$
Since $x^2 - 4x + 1 = 0$,the expression becomes $x(0) - 3(0) - 9 = -9$.

(A) For a quadratic equation $ax^2 + bx + c = 0$,the nature of the roots is determined by the discriminant $D = b^2 - 4ac$.
If $D < 0$,the roots are imaginary (complex conjugates).
Therefore,the condition for the roots to be imaginary is $b^2 - 4ac < 0$.

(C) Let $u = e^{\sin x}$. Since $-1 \le \sin x \le 1$,the range of $u$ is $[e^{-1}, e^1]$,which is approximately $[0.368, 2.718]$.
Substituting $u$ into the equation,we get $u - \frac{1}{u} - 4 = 0$.
Multiplying by $u$,we obtain the quadratic equation $u^2 - 4u - 1 = 0$.
Using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $u = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$.
Since $u = 2 + \sqrt{5} \approx 4.236$ and $u = 2 - \sqrt{5} \approx -0.236$,neither of these values falls within the range $[0.368, 2.718]$.
Therefore,there are no real values of $x$ that satisfy the equation.
Thus,the equation has no real roots.

(B) Given the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.
Multiplying by $r(x+p)(x+q)$,we get $r(x+q) + r(x+p) = (x+p)(x+q)$.
$rx + rq + rx + rp = x^2 + px + qx + pq$.
$2rx + r(p+q) = x^2 + (p+q)x + pq$.
Rearranging into standard quadratic form $ax^2 + bx + c = 0$:
$x^2 + (p+q-2r)x + (pq - rp - rq) = 0$.
Let the roots be $\alpha$ and $-\alpha$ since they are equal in magnitude but opposite in sign.
The sum of the roots is $\alpha + (-\alpha) = 0$.
From the quadratic equation,the sum of roots is $-(p+q-2r) = 0$,which implies $p+q-2r = 0$,so $2r = p+q$.
The product of the roots is $\alpha \times (-\alpha) = -\alpha^2$.
From the equation,the product of roots is $pq - r(p+q)$.
Substituting $r = \frac{p+q}{2}$ into the product:
Product $= pq - (\frac{p+q}{2})(p+q) = pq - \frac{(p+q)^2}{2}$.
Product $= \frac{2pq - (p^2 + 2pq + q^2)}{2} = \frac{2pq - p^2 - 2pq - q^2}{2} = -\frac{p^2 + q^2}{2}$.

(A) For the quadratic equation $ax^2 + bx + c = 0$,the roots are real if the discriminant $D = b^2 - 4ac \geq 0$.
Here,$a = 1, b = 2, c = p$.
Substituting these values into the discriminant formula:
$D = (2)^2 - 4(1)(p) \geq 0$
$4 - 4p \geq 0$
$4 \geq 4p$
$p \leq 1$
Thus,the correct condition is $p \leq 1$.

(A) Let $y = \frac{x}{x^2 - 5x + 9}$.
Then $y(x^2 - 5x + 9) = x$,which implies $yx^2 - (5y + 1)x + 9y = 0$.
Since $x$ is a real number,the discriminant $D \ge 0$.
$D = (-(5y + 1))^2 - 4(y)(9y) \ge 0$.
$(5y + 1)^2 - 36y^2 \ge 0$.
$25y^2 + 10y + 1 - 36y^2 \ge 0$.
$-11y^2 + 10y + 1 \ge 0$.
$11y^2 - 10y - 1 \le 0$.
$(11y + 1)(y - 1) \le 0$.
This inequality holds for $y \in [-\frac{1}{11}, 1]$.
Thus,the maximum value is $1$.

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

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

(B) Let $f(x) = (x - a)(x - c) + 2(x - b)(x - d)$.
We evaluate the function at the given points $a, b, c, d$:
$f(a) = (a - a)(a - c) + 2(a - b)(a - d) = 0 + 2(a - b)(a - d)$. Since $a < b$ and $a < d$,$(a - b) < 0$ and $(a - d) < 0$,so $f(a) > 0$.
$f(b) = (b - a)(b - c) + 2(b - b)(b - d) = (b - a)(b - c)$. Since $b > a$ and $b < c$,$(b - a) > 0$ and $(b - c) < 0$,so $f(b) < 0$.
$f(c) = (c - a)(c - c) + 2(c - b)(c - d) = 0 + 2(c - b)(c - d)$. Since $c > b$ and $c < d$,$(c - b) > 0$ and $(c - d) < 0$,so $f(c) < 0$.
$f(d) = (d - a)(d - c) + 2(d - b)(d - d) = (d - a)(d - c)$. Since $d > a$ and $d > c$,$(d - a) > 0$ and $(d - c) > 0$,so $f(d) > 0$.
Since $f(b) < 0$ and $f(d) > 0$,there exists at least one root in the interval $(b, d)$.
Since $f(a) > 0$ and $f(b) < 0$,there exists at least one root in the interval $(a, b)$.
Since the equation is a quadratic,it has exactly two real and distinct roots.

(A) Let the roots of the equation $ax^2 - bx - c = 0$ be $\alpha$ and $\beta$.
If the roots are shifted by a constant $\lambda$,the new roots become $\alpha + \lambda$ and $\beta + \lambda$.
The difference between the roots remains invariant:
$|(\alpha + \lambda) - (\beta + \lambda)| = |\alpha - \beta|$.
We know that the difference of roots is given by $|\alpha - \beta| = \frac{\sqrt{b^2 - 4ac}}{|a|}$.
Squaring both sides,we get $(\alpha - \beta)^2 = \frac{b^2 - 4ac}{a^2}$.
Therefore,the expression $\frac{b^2 - 4ac}{a^2}$ remains unchanged.

(C) For the quadratic equation $ax^2 + bx + c = 0$ to have no real roots,the discriminant $D = b^2 - 4ac$ must be less than $0$.
Here,$a = (p - 2)$,$b = 2(p - 2)$,and $c = 2$.
If $p = 2$,the equation becomes $0x^2 + 0x + 2 = 0$,which is $2 = 0$,a contradiction. Thus,there are no real roots when $p = 2$.
For $p \neq 2$,$D = [2(p - 2)]^2 - 4(p - 2)(2) < 0$.
$4(p - 2)^2 - 8(p - 2) < 0$.
Divide by $4$: $(p - 2)^2 - 2(p - 2) < 0$.
$(p - 2)(p - 2 - 2) < 0$.
$(p - 2)(p - 4) < 0$.
This inequality holds for $p \in (2, 4)$.
Combining the case $p = 2$ (where the equation is not quadratic and has no roots) and the interval $(2, 4)$,we get $p \in [2, 4)$.
However,checking the options provided,the most appropriate range covering the condition is $p \in (2, 4)$.

(C) Given that $3$ is a root of the equation $x^2 + kx - 24 = 0$.
Substituting $x = 3$ into the equation:
$3^2 + k(3) - 24 = 0$
$9 + 3k - 24 = 0$
$3k - 15 = 0$
$3k = 15 \implies k = 5$.
Now,check the options by substituting $x = 3$ and $k = 5$:
For option $(C)$: $x^2 - kx + 6 = 0$
Substituting $x = 3$ and $k = 5$:
$3^2 - 5(3) + 6 = 9 - 15 + 6 = 0$.
Since the equation is satisfied,$3$ is a root of $x^2 - kx + 6 = 0$.

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

(B) Let $y = x^{1/3}$. Then the equation becomes $y^2 + y - 2 = 0$.
Factoring the quadratic equation,we get $(y + 2)(y - 1) = 0$.
This gives two possible values for $y$: $y = 1$ or $y = -2$.
If $y = 1$,then $x^{1/3} = 1$,which implies $x = 1^3 = 1$.
If $y = -2$,then $x^{1/3} = -2$,which implies $x = (-2)^3 = -8$.
Thus,the roots of the equation are $x = 1$ and $x = -8$.

(C) Given the quadratic equation $(p - q)x^2 + (q - r)x + (r - p) = 0$.
Observe 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 zero,then $x = 1$ is always one of the roots.
Let the roots be $x_1$ and $x_2$. We know that the product of the roots is given by $\frac{c}{a}$.
Therefore,$1 \times x_2 = \frac{r - p}{p - q}$.
Thus,the roots are $1$ and $\frac{r - p}{p - q}$.

(B) For the quadratic equation $x^2 - 4x - \log_3 a = 0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
The discriminant $D$ is given by $b^2 - 4ac$.
Here,$a = 1$,$b = -4$,and $c = -\log_3 a$.
$D = (-4)^2 - 4(1)(-\log_3 a) \geq 0$
$16 + 4\log_3 a \geq 0$
$4\log_3 a \geq -16$
$\log_3 a \geq -4$
Since the base $3 > 1$,the inequality direction remains the same:
$a \geq 3^{-4}$
$a \geq \frac{1}{81}$
Thus,the minimum value of $a$ is $\frac{1}{81}$.

(C) Given the quadratic equation $x^2 - m(2x - 8) - 15 = 0$.
Expanding the equation: $x^2 - 2mx + 8m - 15 = 0$.
Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = 1$,$b = -2m$,and $c = 8m - 15$.
For the equation to have equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Substituting the values: $(-2m)^2 - 4(1)(8m - 15) = 0$.
$4m^2 - 32m + 60 = 0$.
Dividing by $4$: $m^2 - 8m + 15 = 0$.
Factoring the quadratic: $(m - 3)(m - 5) = 0$.
Therefore,$m = 3$ or $m = 5$.

(C) Given the quadratic equation $(p - q)x^2 + (q - r)x + (r - p) = 0$.
Let $f(x) = (p - q)x^2 + (q - r)x + (r - p)$.
Observe the sum of the coefficients: $(p - q) + (q - r) + (r - p) = p - q + q - r + r - p = 0$.
Since the sum of the coefficients is $0$,$x = 1$ is always a root of the equation.
Let the roots be $\alpha$ and $\beta$. We know that the product of the roots $\alpha \beta = \frac{c}{a} = \frac{r - p}{p - q}$.
Since $\alpha = 1$,we have $1 \times \beta = \frac{r - p}{p - q}$,so $\beta = \frac{r - p}{p - q}$.
Thus,the roots are $1$ and $\frac{r - p}{p - q}$.

(D) Let $\alpha, \beta$ be the roots of $x^2 - bx + c = 0$ and $\alpha', \beta'$ be the roots of $x^2 - cx + b = 0$.
The difference between the roots of $x^2 - bx + c = 0$ is $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \sqrt{b^2 - 4c}$.
The difference between the roots of $x^2 - cx + b = 0$ is $|\alpha' - \beta'| = \sqrt{(\alpha' + \beta')^2 - 4\alpha'\beta'} = \sqrt{c^2 - 4b}$.
Given that the differences are equal:
$\sqrt{b^2 - 4c} = \sqrt{c^2 - 4b}$
Squaring both sides:
$b^2 - 4c = c^2 - 4b$
$b^2 - c^2 = 4c - 4b$
$(b - c)(b + c) = -4(b - c)$
Assuming $b \neq c$ (if $b = c$,the equations are identical,which is a trivial case),we divide by $(b - c)$:
$b + c = -4$.

(C) The discriminant $D$ of the equation $ax^2 - rx - s = 0$ is given by $D = (-r)^2 - 4(a)(-s) = r^2 + 4as$.
Since $r, s > 0$ and assuming $a > 0$,we have $D = r^2 + 4as > 0$,which implies the roots are real.
Furthermore,the product of the roots is given by $\frac{c}{a} = \frac{-s}{a}$.
Since $s > 0$ and $a > 0$,the product of the roots is negative,which implies that the roots must have opposite signs.

(A) Let $y = x - 2$. The equation becomes $y^2 - 3y + 2 = 0$.
Factoring the quadratic equation,we get $(y - 1)(y - 2) = 0$.
So,$y = 1$ or $y = 2$.
Substituting back $x - 2 = y$:
Case $1$: $x - 2 = 1 \implies x = 3$.
Case $2$: $x - 2 = 2 \implies x = 4$.
The roots of the equation are $3$ and $4$.
The product of the roots is $3 \times 4 = 12$.

(A) For the quadratic expression $f(x) = x^2 - ax + (1 - 2a^2)$ to be always positive for all real $x$,the coefficient of $x^2$ must be positive (which is $1 > 0$,true) and the discriminant $D$ must be less than $0$.
The discriminant $D = b^2 - 4ac = (-a)^2 - 4(1)(1 - 2a^2)$.
Setting $D < 0$:
$a^2 - 4(1 - 2a^2) < 0$
$a^2 - 4 + 8a^2 < 0$
$9a^2 - 4 < 0$
Factoring the expression:
$(3a - 2)(3a + 2) < 0$
Solving the inequality,we get:
$ - \frac{2}{3} < a < \frac{2}{3} $

(B) For a quadratic equation $ax^2 + bx + c = 0$ to have equal roots,the discriminant $D$ must be equal to $0$.
The discriminant $D = b^2 - 4ac$.
Given the equation $x^2 + px + 64 = 0$,we have $a = 1$,$b = p$,and $c = 64$.
Setting $D = 0$:
$p^2 - 4(1)(64) = 0$
$p^2 - 256 = 0$
$p^2 = 256$
$p = \pm 16$.
Since it is given that $p > 0$,we take $p = 16$.

(D) The given equation is $a - bx - x^2 = 0$,which can be rewritten as $x^2 + bx - a = 0$.
Comparing this with the standard quadratic equation $Ax^2 + Bx + C = 0$,we have $A = 1$,$B = b$,and $C = -a$.
The discriminant $D$ is given by $D = B^2 - 4AC = b^2 - 4(1)(-a) = b^2 + 4a$.
Since $a > 0$ and $b > 0$,$D = b^2 + 4a > 0$,so the roots are real and distinct.
Let the roots be $\alpha$ and $\beta$. The product of the roots is $\alpha \beta = \frac{C}{A} = -a$.
Since $a > 0$,the product of the roots is negative,which implies the roots have opposite signs.
The sum of the roots is $\alpha + \beta = -\frac{B}{A} = -b$.
Since $b > 0$,the sum of the roots is negative. For the sum of two numbers with opposite signs to be negative,the numerically larger root must be negative.

(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 have $x^2 = 2 + x$
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 function must yield a non-negative value,$x \neq -1$
Therefore,$x = 2$.

(B) For the roots of a quadratic equation $ax^2 + bx + c = 0$ to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square.
Here,$a = 6$,$b = -7$,and $c = k$.
$D = (-7)^2 - 4(6)(k) = 49 - 24k$.
For $D$ to be a perfect square,we test values of $k$:
If $k = 1$,$D = 49 - 24(1) = 25 = 5^2$ (a perfect square).
If $k = 2$,$D = 49 - 24(2) = 49 - 48 = 1 = 1^2$ (a perfect square).
Thus,the values of $k$ for which the roots are rational are $1$ and $2$.

(B) Let the roots of the equation $x^2 + px + 3 = 0$ be $\alpha$ and $\beta$.
Given that $|\alpha - \beta| = \sqrt{p}$.
Squaring both sides,we get $(\alpha - \beta)^2 = p$.
We know that $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.
Substituting the coefficients from the quadratic equation,$\alpha + \beta = -p$ and $\alpha\beta = 3$.
So,$(-p)^2 - 4(3) = p$.
$p^2 - 12 = p \implies p^2 - p - 12 = 0$.
Factoring the quadratic equation: $(p - 4)(p + 3) = 0$.
Thus,$p = 4$ or $p = -3$.
Since the difference between the roots is $\sqrt{p}$,$p$ must be non-negative for the roots to be real and the expression to be defined in the real number system. Therefore,$p = 4$.

(A) Since the roots of the equation $x^2 + \lambda x + \mu = 0$ are equal,the discriminant $D = 0$.
Thus,$\lambda^2 - 4\mu = 0 \implies \lambda^2 = 4\mu$.
Given that $x = 2$ is a root of the equation $x^2 + \lambda x - 12 = 0$,we substitute $x = 2$ into the equation:
$2^2 + \lambda(2) - 12 = 0
4 + 2\lambda - 12 = 0
2\lambda = 8
\lambda = 4$.
Now,substitute $\lambda = 4$ into $\lambda^2 = 4\mu$:
$4^2 = 4\mu
16 = 4\mu
\mu = 4$.
Therefore,$(\lambda, \mu) = (4, 4)$.

(A) Let $P(x) = ax^3 + bx^2 + cx + d$.
Since $x^2 + x + 1$ is a factor of $P(x)$,we can write $P(x) = (x^2 + x + 1)(ax + k)$ for some constant $k$.
Expanding this,we get $P(x) = ax^3 + (a+k)x^2 + (a+k)x + k$.
Comparing the constant term with the original polynomial,we have $k = d$.
Thus,$P(x) = (x^2 + x + 1)(ax + d)$.
Setting $P(x) = 0$,we get $(x^2 + x + 1)(ax + d) = 0$.
The roots of $x^2 + x + 1 = 0$ are complex (discriminant $D = 1^2 - 4(1)(1) = -3 < 0$).
The real root comes from $ax + d = 0$,which gives $x = -\frac{d}{a}$.

(D) Given the equation $x^4 - 4x^3 + ax^2 + bx + 1 = 0$ has four real roots $\alpha, \beta, \gamma, \delta$.
By Vieta's formulas,we have:
$\sum \alpha = 4$
$\sum \alpha \beta = a$
$\sum \alpha \beta \gamma = -b$
$\alpha \beta \gamma \delta = 1$
For real positive roots,the Arithmetic Mean $(AM)$ is greater than or equal to the Geometric Mean $(GM)$:
$\frac{\alpha + \beta + \gamma + \delta}{4} \geq (\alpha \beta \gamma \delta)^{1/4}$
Substituting the values: $\frac{4}{4} \geq (1)^{1/4} \implies 1 \geq 1$.
Since the $AM$ equals the $GM$,all roots must be equal: $\alpha = \beta = \gamma = \delta = 1$.
Now,calculate $a$ and $b$:
$a = \sum \alpha \beta = \binom{4}{2} \times (1 \times 1) = 6 \times 1 = 6$.
$-b = \sum \alpha \beta \gamma = \binom{4}{3} \times (1 \times 1 \times 1) = 4 \times 1 = 4 \implies b = -4$.
Thus,$a = 6$ and $b = -4$.

(A) For the given equation $(a^2 - 3a + 2)x^2 + (a^2 - 4)x + a^2 - a - 2 = 0$ to be a linear equation,the coefficient of $x^2$ must be zero,while the coefficient of $x$ must not be zero.
Step $1$: Set the coefficient of $x^2$ to zero.
$a^2 - 3a + 2 = 0$
$(a - 1)(a - 2) = 0$
So,$a = 1$ or $a = 2$.
Step $2$: Check the condition for $x$ coefficient $(a^2 - 4) \neq 0$ for these values.
If $a = 1$: Coefficient of $x$ is $(1)^2 - 4 = -3 \neq 0$. This is valid.
If $a = 2$: Coefficient of $x$ is $(2)^2 - 4 = 0$. This makes the equation $0x^2 + 0x + (4 - 2 - 2) = 0$,which is $0 = 0$. This is not a linear equation.
Thus,only $a = 1$ satisfies the condition.

(C) For the quadratic equation $x^2 - 8x + (a^2 - 6a) = 0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = b^2 - 4ac \geq 0$
$(-8)^2 - 4(1)(a^2 - 6a) \geq 0$
$64 - 4(a^2 - 6a) \geq 0$
Divide by $4$:
$16 - (a^2 - 6a) \geq 0$
$16 - a^2 + 6a \geq 0$
$a^2 - 6a - 16 \leq 0$
Factor the quadratic expression:
$(a - 8)(a + 2) \leq 0$
This inequality holds when $a$ lies between the roots of the equation $(a - 8)(a + 2) = 0$,which are $a = 8$ and $a = -2$.
Thus,$-2 \leq a \leq 8$.

(A) Since $p(x) = f(x) - g(x)$,we have $p(x) = (a - a_1)x^2 + (b - b_1)x + (c - c_1)$.
Let $A = a - a_1$,$B = b - b_1$,and $C = c - c_1$. Then $p(x) = Ax^2 + Bx + C$.
Given that $p(x) = 0$ only for $x = -1$,it implies that $x = -1$ is a repeated root of the quadratic equation $p(x) = 0$.
Therefore,$p(x) = A(x + 1)^2$.
We are given $p(-2) = 2$.
Substituting $x = -2$ into the expression: $p(-2) = A(-2 + 1)^2 = A(-1)^2 = A$.
Thus,$A = 2$.
Now,we need to find $p(2)$:
$p(2) = A(2 + 1)^2 = 2(3)^2 = 2(9) = 18$.

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

(B) For the quadratic equation $Ax^2 + Bx + C = 0$ to have real roots,the discriminant $D = B^2 - 4AC$ must be greater than or equal to $0$.
Here,$A = 1$,$B = -8$,and $C = a^2 - 6a$.
Substituting these values into the discriminant formula:
$D = (-8)^2 - 4(1)(a^2 - 6a) \geq 0$
$64 - 4(a^2 - 6a) \geq 0$
$64 - 4a^2 + 24a \geq 0$
Dividing by $-4$ (and reversing the inequality sign):
$a^2 - 6a - 16 \leq 0$
Factoring the quadratic expression:
$(a - 8)(a + 2) \leq 0$
The roots of the inequality are $a = 8$ and $a = -2$.
For the expression to be $\leq 0$,$a$ must lie between the roots inclusive:
$-2 \leq a \leq 8$.

(B) Let $f(x, y, z) = x^2 + 4y^2 + 3z^2 - 2x - 12y - 6z + 14$.
Rearranging the terms to complete the squares:
$f(x, y, z) = (x^2 - 2x + 1) + (4y^2 - 12y + 9) + (3z^2 - 6z + 3) + 1$
$= (x - 1)^2 + (2y - 3)^2 + 3(z - 1)^2 + 1$.
For the expression to have a minimum value,each squared term must be zero:
$(x - 1)^2 = 0 \implies x = 1$
$(2y - 3)^2 = 0 \implies y = \frac{3}{2}$
$(z - 1)^2 = 0 \implies z = 1$
Substituting these values,the minimum value is $f(1, \frac{3}{2}, 1) = 0 + 0 + 0 + 1 = 1$.

(A) Let the two consecutive integers be $n$ and $n+1$.
For a quadratic equation $x^2 - bx + c = 0$,the roots are $\alpha$ and $\beta$.
Given $\alpha = n$ and $\beta = n+1$.
The discriminant $D = b^2 - 4c$.
We know that $D = (\alpha - \beta)^2$.
Substituting the values,$D = (n - (n+1))^2 = (-1)^2 = 1$.
Therefore,$b^2 - 4c = 1$.

(C) The curve $y = x^2 + ax + 25$ touches the $x-$axis,which means the quadratic equation $x^2 + ax + 25 = 0$ has equal roots.
For equal roots,the discriminant $D$ must be zero.
$D = b^2 - 4ac = 0$
Substituting the values $a=1, b=a, c=25$:
$a^2 - 4(1)(25) = 0$
$a^2 - 100 = 0$
$a^2 = 100$
$a = \pm 10$

(C) Given the equation $x^2 + x + 1 = 0$.
Since $\alpha$ is a root,$\alpha^2 + \alpha + 1 = 0$,which implies $\alpha^3 = 1$.
Also,the sum of roots $\alpha + \alpha^2 = -1$.
We need to find the equation with roots $\alpha^{31}$ and $\alpha^{62}$.
Sum of new roots: $\alpha^{31} + \alpha^{62} = \alpha^{30} \cdot \alpha + (\alpha^3)^{20} \cdot \alpha^2 = (1)^{10} \cdot \alpha + (1)^{20} \cdot \alpha^2 = \alpha + \alpha^2 = -1$.
Product of new roots: $\alpha^{31} \cdot \alpha^{62} = \alpha^{93} = (\alpha^3)^{31} = 1^{31} = 1$.
The required quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values: $x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.

(C) Given the expression $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}$.
Squaring both sides,we get $x^2 = 2 + \sqrt{2 + \sqrt{2 + \dots}}$.
Since the inner part is the same as $x$,we can write $x^2 = 2 + x$.
Rearranging the terms,we get the quadratic equation $x^2 - x - 2 = 0$.
Factoring the equation: $(x - 2)(x + 1) = 0$.
This gives $x = 2$ or $x = -1$.
Since $x$ is the square root of a positive value,$x$ must be positive.
Therefore,$x = 2$.

(A) Given the equation: $x - \frac{2}{x - 1} = 1 - \frac{2}{x - 1}$.
Adding $\frac{2}{x - 1}$ to both sides,we get $x = 1$.
However,for the original equation to be defined,the denominator $x - 1$ must not be zero,which implies $x \neq 1$.
Since the only potential solution $x = 1$ is excluded from the domain,the equation has no roots.

(A) For a quadratic equation $Ax^2 + Bx + C = 0$ to have more than two solutions,it must be an identity,which implies $A = 0$,$B = 0$,and $C = 0$.
$1$. $A = a^2 - a - 2 = (a - 2)(a + 1) = 0 \implies a = 2, -1$
$2$. $B = a^2 - 4 = (a - 2)(a + 2) = 0 \implies a = 2, -2$
$3$. $C = a^2 - 3a + 2 = (a - 1)(a - 2) = 0 \implies a = 1, 2$
The common value of $a$ that satisfies all three conditions is $a = 2$.

(D) Let $y = \frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$.
$y = \frac{(3x^2 + 9x + 7) + 10}{3x^2 + 9x + 7} = 1 + \frac{10}{3x^2 + 9x + 7}$.
For $y$ to be maximum,the denominator $p = 3x^2 + 9x + 7$ must be minimum.
The minimum value of a quadratic expression $ax^2 + bx + c$ (where $a > 0$) is given by $\frac{-D}{4a} = \frac{-(b^2 - 4ac)}{4a}$.
Here,$a = 3$,$b = 9$,$c = 7$.
$D = b^2 - 4ac = 9^2 - 4(3)(7) = 81 - 84 = -3$.
$p_{\text{min}} = \frac{-(-3)}{4(3)} = \frac{3}{12} = \frac{1}{4}$.
Thus,$y_{\text{max}} = 1 + \frac{10}{1/4} = 1 + 10 \times 4 = 1 + 40 = 41$.

(A) Given the equation $x^2 - 3|x| + 2 = 0$.
Since $x^2 = |x|^2$,we can rewrite the equation as $|x|^2 - 3|x| + 2 = 0$.
Let $t = |x|$,where $t \geq 0$. 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 $1$: $|x| = 1$ implies $x = 1$ or $x = -1$.
Case $2$: $|x| = 2$ implies $x = 2$ or $x = -2$.
Thus,the real solutions are $x \in \{-2, -1, 1, 2\}$.
The total number of real solutions is $4$.

(B) For the equation $px^2 + 2qx + r = 0$ to have real roots,the discriminant must be non-negative:
$D_1 = (2q)^2 - 4pr \geq 0 \implies 4q^2 - 4pr \geq 0 \implies q^2 \geq pr \dots (i)$
For the equation $qx^2 - 2\sqrt{pr}x + q = 0$ to have real roots,the discriminant must be non-negative:
$D_2 = (-2\sqrt{pr})^2 - 4(q)(q) \geq 0 \implies 4pr - 4q^2 \geq 0 \implies pr \geq q^2 \dots (ii)$
From $(i)$ and $(ii)$,we have $q^2 \geq pr$ and $pr \geq q^2$,which implies $q^2 = pr$.

(D) The given equation is $2x^2 - (k - 1)x + 8 = 0$.
For equal and real roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Here,$a = 2$,$b = -(k - 1)$,and $c = 8$.
Substituting these values into $D = 0$:
$(-(k - 1))^2 - 4(2)(8) = 0$
$(k - 1)^2 - 64 = 0$
$(k - 1)^2 = 64$
Taking the square root on both sides:
$k - 1 = \pm 8$
Case $1$: $k - 1 = 8 \Rightarrow k = 9$
Case $2$: $k - 1 = -8 \Rightarrow k = -7$
Therefore,the values of $k$ are $9$ and $-7$.

(A) Let $y = \frac{1}{x - 2}$. Then $x - 2 = \frac{1}{y}$,so $x = \frac{1}{y} + 2 = \frac{1 + 2y}{y}$.
Substitute $x = \frac{2y + 1}{y}$ into the original equation $x^2 - 3x + 1 = 0$:
$(\frac{2y + 1}{y})^2 - 3(\frac{2y + 1}{y}) + 1 = 0$
$\frac{4y^2 + 4y + 1}{y^2} - \frac{6y + 3}{y} + 1 = 0$
Multiply by $y^2$:
$(4y^2 + 4y + 1) - y(6y + 3) + y^2 = 0$
$4y^2 + 4y + 1 - 6y^2 - 3y + y^2 = 0$
$-y^2 + y + 1 = 0$
$y^2 - y - 1 = 0$.
Thus,the required equation is $x^2 - x - 1 = 0$.