QUADRATIC EQUATION Questions in English

(A) Given the quadratic equation $375x^2 - 25x - 2 = 0$.
By the properties of roots,the sum of roots $\alpha + \beta = -(-25)/375 = 25/375 = 1/15$ and the product of roots $\alpha \beta = -2/375$.
We need to evaluate the sum of two infinite geometric series: $S = \sum_{r=1}^{\infty} \alpha^r + \sum_{r=1}^{\infty} \beta^r$.
Since $|\alpha| < 1$ and $|\beta| < 1$,the sum of an infinite geometric series is given by $a/(1-r)$.
Thus,$S = \frac{\alpha}{1-\alpha} + \frac{\beta}{1-\beta}$.
$S = \frac{\alpha(1-\beta) + \beta(1-\alpha)}{(1-\alpha)(1-\beta)} = \frac{\alpha - \alpha\beta + \beta - \alpha\beta}{1 - (\alpha+\beta) + \alpha\beta}$.
Substituting the values: $S = \frac{(\alpha+\beta) - 2\alpha\beta}{1 - (\alpha+\beta) + \alpha\beta}$.
$S = \frac{1/15 - 2(-2/375)}{1 - 1/15 - 2/375} = \frac{25/375 + 4/375}{(375 - 25 - 2)/375} = \frac{29/375}{348/375} = \frac{29}{348} = \frac{1}{12}$.

(C) Given equation: $\cos 2x + \alpha \sin x = 2\alpha - 7$.
Using the identity $\cos 2x = 1 - 2\sin^2 x$,we get:
$1 - 2\sin^2 x + \alpha \sin x = 2\alpha - 7$.
Rearranging the terms to form a quadratic equation in $\sin x$:
$2\sin^2 x - \alpha \sin x + (2\alpha - 8) = 0$.
Using the quadratic formula $\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\sin x = \frac{\alpha \pm \sqrt{\alpha^2 - 4(2)(2\alpha - 8)}}{4} = \frac{\alpha \pm \sqrt{\alpha^2 - 16\alpha + 64}}{4}$.
Since $\alpha^2 - 16\alpha + 64 = (\alpha - 8)^2$,we have:
$\sin x = \frac{\alpha \pm (\alpha - 8)}{4}$.
This gives two possible values for $\sin x$:
$1) \sin x = \frac{\alpha + \alpha - 8}{4} = \frac{2\alpha - 8}{4} = \frac{\alpha - 4}{2}$.
$2) \sin x = \frac{\alpha - \alpha + 8}{4} = \frac{8}{4} = 2$.
Since $\sin x = 2$ is impossible,we must have $\sin x = \frac{\alpha - 4}{2}$.
For a solution to exist,$-1 \leq \sin x \leq 1$ must hold:
$-1 \leq \frac{\alpha - 4}{2} \leq 1$.
$-2 \leq \alpha - 4 \leq 2$.
Adding $4$ to all parts: $2 \leq \alpha \leq 6$.
Thus,$S = [2, 6]$.

(D) Given that $\alpha, \beta, \gamma$ are in $G.P.$,we can write $\beta = \alpha r$ and $\gamma = \alpha r^2$ for some common ratio $r \neq 1$.
Substituting these into the first equation: $\alpha x^2 + 2(\alpha r)x + \alpha r^2 = 0$.
Since $\alpha \neq 0$,we divide by $\alpha$: $x^2 + 2rx + r^2 = 0$,which is $(x + r)^2 = 0$. Thus,the root is $x = -r$.
Since this is a common root with $x^2 + x - 1 = 0$,we substitute $x = -r$ into the second equation:
$(-r)^2 + (-r) - 1 = 0 \implies r^2 - r - 1 = 0$.
From the $G.P.$ properties,$\beta^2 = \alpha\gamma$. Also,$\beta = \alpha r$ and $\gamma = \alpha r^2$.
We need to evaluate $\alpha(\beta + \gamma) = \alpha(\alpha r + \alpha r^2) = \alpha^2 r(1 + r)$.
Since $r^2 = r + 1$,we have $r^2 - r = 1$.
Also,$\beta\gamma = (\alpha r)(\alpha r^2) = \alpha^2 r^3 = \alpha^2 r(r^2) = \alpha^2 r(r + 1) = \alpha^2(r^2 + r)$.
Comparing the coefficients of the two equations $\alpha x^2 + 2\beta x + \gamma = 0$ and $x^2 + x - 1 = 0$,for them to have common roots,the ratio of coefficients must be equal: $\frac{\alpha}{1} = \frac{2\beta}{1} = \frac{\gamma}{-1} = k$.
Thus,$\alpha = k, \beta = k/2, \gamma = -k$.
Since they are in $G.P.$,$\beta^2 = \alpha\gamma \implies (k/2)^2 = k(-k) \implies k^2/4 = -k^2$. This implies $k=0$,which contradicts the non-constant $G.P.$ condition.
Re-evaluating: The condition for a common root in $ax^2+bx+c=0$ and $dx^2+ex+f=0$ is $(af-cd)^2 = (ae-bd)(bf-ce)$.
Substituting $\alpha, 2\beta, \gamma$ and $1, 1, -1$: $(\alpha(-1) - \gamma(1))^2 = (\alpha(1) - 2\beta(1))(2\beta(-1) - \gamma(1))$.
$(-\alpha - \gamma)^2 = (\alpha - 2\beta)(-2\beta - \gamma) \implies \alpha^2 + 2\alpha\gamma + \gamma^2 = -2\alpha\beta - \alpha\gamma + 4\beta^2 + 2\beta\gamma$.
Since $\beta^2 = \alpha\gamma$,$\alpha^2 + 2\alpha\gamma + \gamma^2 = -2\alpha\beta - \beta^2 + 4\beta^2 + 2\beta\gamma = -2\alpha\beta + 3\beta^2 + 2\beta\gamma$.
Substituting $\beta^2 = \alpha\gamma$,we get $\alpha^2 + 2\beta^2 + \gamma^2 = -2\alpha\beta + 3\beta^2 + 2\beta\gamma \implies \alpha^2 - \beta^2 + \gamma^2 = -2\alpha\beta + 2\beta\gamma$.
Using $\alpha = \beta/r, \gamma = \beta r$,we find $\alpha(\beta + \gamma) = \beta\gamma$.

(D) Given the equation $x^{2}-x-1=0$,the roots are $\alpha$ and $\beta$.
By Vieta's formulas,$\alpha+\beta=1$ and $\alpha\beta=-1$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation: $\alpha^{2}-\alpha-1=0 \Rightarrow \alpha^{k}-\alpha^{k-1}-\alpha^{k-2}=0$ and $\beta^{k}-\beta^{k-1}-\beta^{k-2}=0$.
Adding these gives the recurrence relation: $p_{k}=p_{k-1}+p_{k-2}$.
Calculating the first few terms:
$p_{1}=\alpha+\beta=1$
$p_{2}=(\alpha+\beta)^{2}-2\alpha\beta=1^{2}-2(-1)=3$
$p_{3}=p_{2}+p_{1}=3+1=4$
$p_{4}=p_{3}+p_{2}=4+3=7$
$p_{5}=p_{4}+p_{3}=7+4=11$
Checking the options:
$A: p_{1}+p_{2}+p_{3}+p_{4}+p_{5} = 1+3+4+7+11 = 26$ (True).
$B: p_{5}=11$ (True).
$C: p_{5}-p_{4} = 11-7 = 4 = p_{3}$ (True).
$D: p_{2} \cdot p_{3} = 3 \cdot 4 = 12 \neq 11$ (False).
Thus,the statement in option $D$ is not true.

(D) Let $3^{x} = t$,where $t > 0$.
The equation becomes $t(t-1) + 2 = |t-1| + |t-2|$,which simplifies to $t^{2} - t + 2 = |t-1| + |t-2|$.
Case-$I$: $0 < t < 1$
$t^{2} - t + 2 = -(t-1) - (t-2) = -t + 1 - t + 2 = 3 - 2t$.
$t^{2} + t - 1 = 0$.
$t = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Since $t > 0$,we have $t = \frac{\sqrt{5}-1}{2}$. This is in the range $(0, 1)$.
$3^{x} = \frac{\sqrt{5}-1}{2} \Rightarrow x = \log_{3}\left(\frac{\sqrt{5}-1}{2}\right)$,which is one real solution.
Case-$II$: $1 \leq t < 2$
$t^{2} - t + 2 = (t-1) - (t-2) = t - 1 - t + 2 = 1$.
$t^{2} - t + 1 = 0$.
The discriminant $D = (-1)^{2} - 4(1)(1) = 1 - 4 = -3 < 0$. No real solution.
Case-$III$: $t \geq 2$
$t^{2} - t + 2 = (t-1) + (t-2) = 2t - 3$.
$t^{2} - 3t + 5 = 0$.
The discriminant $D = (-3)^{2} - 4(1)(5) = 9 - 20 = -11 < 0$. No real solution.
Thus,there is only one real value for $t$,which gives exactly one real value for $x$. Therefore,$S$ is a singleton.

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

(B) For the equation $a x^{2}-2 b x+5=0$,since it has a repeated root $\alpha$,the discriminant $D = 0$.
$D = (-2b)^{2} - 4(a)(5) = 4b^{2} - 20a = 0 \Rightarrow b^{2} = 5a$.
The root $\alpha$ is given by $\alpha = -(-2b) / (2a) = b/a$.
Since $\alpha$ is a root of $a x^{2}-2 b x+5=0$,we have $a(b/a)^{2} - 2b(b/a) + 5 = 0 \Rightarrow b^{2}/a - 2b^{2}/a + 5 = 0 \Rightarrow b^{2}/a = 5$,which is consistent with $b^{2} = 5a$.
Given $\alpha$ is also a root of $x^{2}-2 b x-10=0$,we have $\alpha^{2} - 2b\alpha - 10 = 0$.
Substituting $\alpha = b/a$,we get $(b/a)^{2} - 2b(b/a) - 10 = 0 \Rightarrow b^{2}/a^{2} - 2b^{2}/a - 10 = 0$.
Since $b^{2} = 5a$,we substitute $b^{2}/a = 5$ and $b^{2} = 5a$ into the equation: $5/a - 2(5) - 10 = 0 \Rightarrow 5/a = 20 \Rightarrow a = 1/4$.
Then $b^{2} = 5(1/4) = 5/4 \Rightarrow b = \pm \sqrt{5}/2$.
Now,$\alpha = b/a = (\pm \sqrt{5}/2) / (1/4) = \pm 2\sqrt{5}$. Thus $\alpha^{2} = (\pm 2\sqrt{5})^{2} = 20$.
For the equation $x^{2}-2bx-10=0$,the product of roots $\alpha\beta = -10$.
So,$\beta = -10/\alpha = -10/(\pm 2\sqrt{5}) = \mp \sqrt{5}$.
Thus $\beta^{2} = (\mp \sqrt{5})^{2} = 5$.
Therefore,$\alpha^{2} + \beta^{2} = 20 + 5 = 25$.

(D) Given equation: $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$.
Divide the entire equation by $e^{2x}$ (since $e^{2x} \neq 0$ for all real $x$):
$e^{2x} + e^x - 4 + e^{-x} + e^{-2x} = 0$.
Rearrange the terms:
$(e^{2x} + e^{-2x}) + (e^x + e^{-x}) - 4 = 0$.
Let $u = e^x + e^{-x}$. Note that $u^2 = (e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x}$,so $e^{2x} + e^{-2x} = u^2 - 2$.
Substitute this into the equation:
$(u^2 - 2) + u - 4 = 0 \Rightarrow u^2 + u - 6 = 0$.
Factor the quadratic equation:
$(u + 3)(u - 2) = 0$.
This gives $u = -3$ or $u = 2$.
Case $1$: $e^x + e^{-x} = -3$. Since $e^x > 0$ and $e^{-x} > 0$,by $AM$-$GM$ inequality,$e^x + e^{-x} \geq 2$. Thus,$u = -3$ has no real solutions.
Case $2$: $e^x + e^{-x} = 2$. By $AM$-$GM$ inequality,$e^x + e^{-x} = 2$ only when $e^x = e^{-x}$,which implies $e^{2x} = 1$,so $x = 0$.
Thus,there is only $1$ real root.

(A) Since $\alpha$ and $\beta$ are the roots of the equation $5x^{2} + 6x - 2 = 0$,they satisfy the equation.
Therefore,$5\alpha^{2} + 6\alpha - 2 = 0$ and $5\beta^{2} + 6\beta - 2 = 0$.
Multiplying the first equation by $\alpha^{n}$ and the second by $\beta^{n}$,we get:
$5\alpha^{n+2} + 6\alpha^{n+1} - 2\alpha^{n} = 0$ ... $(1)$
$5\beta^{n+2} + 6\beta^{n+1} - 2\beta^{n} = 0$ ... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$5(\alpha^{n+2} + \beta^{n+2}) + 6(\alpha^{n+1} + \beta^{n+1}) - 2(\alpha^{n} + \beta^{n}) = 0$
Given $S_{n} = \alpha^{n} + \beta^{n}$,this simplifies to:
$5S_{n+2} + 6S_{n+1} - 2S_{n} = 0$
For $n = 4$,we substitute into the relation:
$5S_{6} + 6S_{5} - 2S_{4} = 0$
Rearranging the terms gives:
$5S_{6} + 6S_{5} = 2S_{4}$

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

(B) Let $f(x) = (\lambda^{2}+1) x^{2}-4 \lambda x+2$.
For exactly one root to lie in the interval $(0,1)$, the product of the values of the function at the endpoints must be negative, i.e., $f(0) \cdot f(1) < 0$.
$f(0) = 2$
$f(1) = \lambda^{2}+1-4\lambda+2 = \lambda^{2}-4\lambda+3 = (\lambda-1)(\lambda-3)$
So, $2 \cdot (\lambda-1)(\lambda-3) < 0$
This implies $1 < \lambda < 3$.
Now, we check the boundary cases:
Case $1$: If $\lambda = 1$, the equation becomes $2x^{2}-4x+2 = 0$, which simplifies to $2(x-1)^{2} = 0$. The roots are $x=1, 1$. Since neither root lies strictly inside $(0,1)$, $\lambda = 1$ is not included.
Case $2$: If $\lambda = 3$, the equation becomes $10x^{2}-12x+2 = 0$, which simplifies to $2(5x^{2}-6x+1) = 0$, or $2(5x-1)(x-1) = 0$. The roots are $x = 1/5$ and $x = 1$. Since $1/5$ lies in $(0,1)$, $\lambda = 3$ is included.
Thus, the set of values is $\lambda \in (1, 3]$.

(C) Given $\alpha, \beta$ are roots of $x^{2}+px+2=0$. Thus,$\alpha+\beta = -p$ and $\alpha\beta = 2$.
Since $\frac{1}{\alpha}, \frac{1}{\beta}$ are roots of $2x^{2}+2qx+1=0$,they are also roots of $x^{2}+qx+\frac{1}{2}=0$.
Since $\frac{1}{\alpha}, \frac{1}{\beta}$ are roots of $x^{2}+px+2=0$ transformed by $x \to 1/x$,we have $2(1/x)^{2} + p(1/x) + 1 = 0$,which simplifies to $2+px+x^{2}=0$,or $x^{2}+px+2=0$. Comparing $2x^{2}+2qx+1=0$ with $x^{2}+\frac{p}{2}x+\frac{1}{2}=0$,we get $q = p/2$,so $p = 2q$.
Now,consider the expression $E = \left(\frac{\alpha^{2}-1}{\alpha}\right)\left(\frac{\beta^{2}-1}{\beta}\right)\left(\frac{\alpha\beta+1}{\beta}\right)\left(\frac{\alpha\beta+1}{\alpha}\right)$.
Since $\alpha^{2}+p\alpha+2=0$,$\alpha^{2}-1 = -p\alpha-3$. Similarly,$\beta^{2}-1 = -p\beta-3$.
$E = \frac{(-p\alpha-3)(-p\beta-3)(\alpha\beta+1)^{2}}{(\alpha\beta)^{2}} = \frac{(p\alpha+3)(p\beta+3)(2+1)^{2}}{2^{2}} = \frac{9}{4}(p^{2}\alpha\beta + 3p(\alpha+\beta) + 9)$.
Substituting $\alpha\beta=2$ and $\alpha+\beta=-p$:
$E = \frac{9}{4}(2p^{2} - 3p^{2} + 9) = \frac{9}{4}(9-p^{2})$.

(D) For the equation $x^{2}-x+2 \lambda=0$,the sum of roots is $\alpha+\beta=1$ and the product of roots is $\alpha \beta=2 \lambda$.
For the equation $3x^{2}-10x+27 \lambda=0$,the sum of roots is $\alpha+\gamma=\frac{10}{3}$ and the product of roots is $\alpha \gamma=\frac{27 \lambda}{3}=9 \lambda$.
Subtracting the sum of roots: $(\alpha+\gamma)-(\alpha+\beta)=\frac{10}{3}-1 \Rightarrow \gamma-\beta=\frac{7}{3}$.
Dividing the product of roots: $\frac{\alpha \gamma}{\alpha \beta}=\frac{9 \lambda}{2 \lambda} \Rightarrow \frac{\gamma}{\beta}=\frac{9}{2} \Rightarrow \gamma=\frac{9}{2} \beta$.
Substitute $\gamma$ into the difference equation: $\frac{9}{2} \beta-\beta=\frac{7}{3} \Rightarrow \frac{7}{2} \beta=\frac{7}{3} \Rightarrow \beta=\frac{2}{3}$.
Then $\gamma=\frac{9}{2} \times \frac{2}{3}=3$.
Since $\alpha+\beta=1$,we have $\alpha=1-\frac{2}{3}=\frac{1}{3}$.
Using $\alpha \beta=2 \lambda$: $\frac{1}{3} \times \frac{2}{3}=2 \lambda \Rightarrow \frac{2}{9}=2 \lambda \Rightarrow \lambda=\frac{1}{9}$.
Finally,$\frac{\beta \gamma}{\lambda}=\frac{\frac{2}{3} \times 3}{\frac{1}{9}}=\frac{2}{\frac{1}{9}}=18$.

(D) Given the equation: $[x]^{2} + 2[x + 2] - 7 = 0$.
Using the property $[x + n] = [x] + n$ for any integer $n$,we have $[x + 2] = [x] + 2$.
Substituting this into the equation:
$[x]^{2} + 2([x] + 2) - 7 = 0$
$[x]^{2} + 2[x] + 4 - 7 = 0$
$[x]^{2} + 2[x] - 3 = 0$
Let $y = [x]$. Then $y^{2} + 2y - 3 = 0$.
$(y + 3)(y - 1) = 0$.
So,$[x] = 1$ or $[x] = -3$.
If $[x] = 1$,then $x \in [1, 2)$.
If $[x] = -3$,then $x \in [-3, -2)$.
Since the solution set is $[1, 2) \cup [-3, -2)$,there are infinitely many real values of $x$ that satisfy the equation.

(A) Given the quadratic equation $7x^{2}-3x-2=0$.
From the properties of roots,the sum of roots $\alpha+\beta = -(-3)/7 = 3/7$ and the product of roots $\alpha\beta = -2/7$.
We need to evaluate the expression $E = \frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$.
$E = \frac{\alpha(1-\beta^{2}) + \beta(1-\alpha^{2})}{(1-\alpha^{2})(1-\beta^{2})} = \frac{\alpha - \alpha\beta^{2} + \beta - \alpha^{2}\beta}{1 - \beta^{2} - \alpha^{2} + \alpha^{2}\beta^{2}} = \frac{(\alpha+\beta) - \alpha\beta(\alpha+\beta)}{1 - (\alpha^{2}+\beta^{2}) + (\alpha\beta)^{2}}$.
Using $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta = (3/7)^{2} - 2(-2/7) = 9/49 + 4/7 = 9/49 + 28/49 = 37/49$.
Substituting the values:
Numerator $= (3/7) - (-2/7)(3/7) = 3/7 + 6/49 = 21/49 + 6/49 = 27/49$.
Denominator $= 1 - 37/49 + (-2/7)^{2} = 1 - 37/49 + 4/49 = 49/49 - 33/49 = 16/49$.
Therefore,$E = \frac{27/49}{16/49} = \frac{27}{16}$.

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

(C) The given equation is $2x(2x + 1) = 1$,which simplifies to $4x^{2} + 2x - 1 = 0$.
Since $\alpha$ and $\beta$ are the roots of this quadratic equation,the sum of the roots is $\alpha + \beta = -b/a = -2/4 = -1/2$.
From this,we can express $\beta$ as $\beta = -1/2 - \alpha$.
Also,since $\alpha$ is a root,it satisfies the equation $4\alpha^{2} + 2\alpha - 1 = 0$,which implies $4\alpha^{2} + 2\alpha = 1$,or $2\alpha^{2} + \alpha = 1/2$.
Substituting $1/2 = 2\alpha^{2} + \alpha$ into the expression for $\beta$:
$\beta = -(2\alpha^{2} + \alpha) - \alpha$.
$\beta = -2\alpha^{2} - \alpha - \alpha$.
$\beta = -2\alpha^{2} - 2\alpha$.
$\beta = -2\alpha(\alpha + 1)$.

(D) Given the quadratic equation $x^{2}-64x+256=0$.
From the properties of roots,the sum of roots $\alpha+\beta = 64$ and the product of roots $\alpha\beta = 256$.
We need to evaluate the expression $\left(\frac{\alpha^{3}}{\beta^{5}}\right)^{\frac{1}{8}}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{\frac{1}{8}}$.
This simplifies to $\frac{\alpha^{3/8}}{\beta^{5/8}} + \frac{\beta^{3/8}}{\alpha^{5/8}}$.
Taking the common denominator,we get $\frac{\alpha^{3/8} \cdot \alpha^{5/8} + \beta^{3/8} \cdot \beta^{5/8}}{(\alpha\beta)^{5/8}}$.
Since $\alpha^{3/8} \cdot \alpha^{5/8} = \alpha^{(3+5)/8} = \alpha^1 = \alpha$,the expression becomes $\frac{\alpha+\beta}{(\alpha\beta)^{5/8}}$.
Substituting the values $\alpha+\beta = 64$ and $\alpha\beta = 256 = 2^8$,we get $\frac{64}{(2^8)^{5/8}}$.
This simplifies to $\frac{64}{2^5} = \frac{64}{32} = 2$.

(D) For the first equation $x^{2}+5x+6=0$:
Factoring the quadratic,we get $(x+2)(x+3)=0$.
Thus,the roots are $x_1 = -2$ and $x_2 = -3$.
For the second equation $y^{2}+3y+2=0$:
Factoring the quadratic,we get $(y+1)(y+2)=0$.
Thus,the roots are $y_1 = -1$ and $y_2 = -2$.
Comparing the roots:
For $x = -2$,we have $y = -1$ (where $x < y$) and $y = -2$ (where $x = y$).
For $x = -3$,we have $y = -1$ (where $x < y$) and $y = -2$ (where $x < y$).
In all cases,$x \le y$.

(D) For equation $I$: $5x^2 + 3x - 14 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-3 \pm \sqrt{3^2 - 4(5)(-14)}}{2(5)} = \frac{-3 \pm \sqrt{9 + 280}}{10} = \frac{-3 \pm \sqrt{289}}{10} = \frac{-3 \pm 17}{10}$
$x_1 = \frac{14}{10} = 1.4$ and $x_2 = \frac{-20}{10} = -2$.
For equation $II$: $10y^2 - 3y - 27 = 0$
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{3 \pm \sqrt{(-3)^2 - 4(10)(-27)}}{2(10)} = \frac{3 \pm \sqrt{9 + 1080}}{20} = \frac{3 \pm \sqrt{1089}}{20} = \frac{3 \pm 33}{20}$
$y_1 = \frac{36}{20} = 1.8$ and $y_2 = \frac{-30}{20} = -1.5$.
Comparing the values:
$x = \{1.4, -2\}$ and $y = \{1.8, -1.5\}$.
Since $1.4 < 1.8$ and $1.4 > -1.5$,the relationship between $x$ and $y$ cannot be established.

(D) For equation $I$: $x^{2}+5x+6=0$
Factorizing the quadratic equation: $x^{2}+3x+2x+6=0$
$x(x+3)+2(x+3)=0$
$(x+3)(x+2)=0$
Therefore,the roots are $x_{1}=-3$ and $x_{2}=-2$.
For equation $II$: $y^{2}+3y+2=0$
Factorizing the quadratic equation: $y^{2}+2y+y+2=0$
$y(y+2)+1(y+2)=0$
$(y+2)(y+1)=0$
Therefore,the roots are $y_{1}=-2$ and $y_{2}=-1$.
Comparing the values:
$x$ values are $\{-3, -2\}$
$y$ values are $\{-2, -1\}$
Comparing each pair:
$-3 < -2$,$-3 < -1$,$-2 = -2$,$-2 < -1$.
In all cases,$x \le y$ is satisfied.

(B) For equation $I: 2x^2 + 3x + 1 = 0$
Factoring the quadratic equation: $2x^2 + 2x + x + 1 = 0$
$2x(x + 1) + 1(x + 1) = 0$
$(2x + 1)(x + 1) = 0$
Thus,$x_1 = -1$ and $x_2 = -0.5$.
For equation $II: 12y^2 + 7y + 1 = 0$
Factoring the quadratic equation: $12y^2 + 4y + 3y + 1 = 0$
$4y(3y + 1) + 1(3y + 1) = 0$
$(4y + 1)(3y + 1) = 0$
Thus,$y_1 = -1/4 = -0.25$ and $y_2 = -1/3 \approx -0.33$.
Comparing the values:
$x$ values are $\{-1, -0.5\}$
$y$ values are $\{-0.25, -0.33\}$
Since both values of $x$ are less than both values of $y$ (e.g.,$-0.5 < -0.33$),we conclude that $x < y$.

(B) For equation $I$: $2x^2 + 23x + 63 = 0$
We factorize the quadratic equation: $2x^2 + 14x + 9x + 63 = 0$
$2x(x + 7) + 9(x + 7) = 0$
$(2x + 9)(x + 7) = 0$
So,$x = -4.5$ or $x = -7$.
For equation $II$: $4y^2 + 19y + 21 = 0$
We factorize the quadratic equation: $4y^2 + 12y + 7y + 21 = 0$
$4y(y + 3) + 7(y + 3) = 0$
$(4y + 7)(y + 3) = 0$
So,$y = -1.75$ or $y = -3$.
Comparing the values:
$x_1 = -4.5, x_2 = -7$
$y_1 = -1.75, y_2 = -3$
Since both values of $x$ are less than both values of $y$ ($-7 < -3$ and $-4.5 < -1.75$),we conclude that $x < y$.

(D) For equation $I$: $4x^2 - 29x + 45 = 0$
Using the splitting the middle term method: $4x^2 - 20x - 9x + 45 = 0$
$4x(x - 5) - 9(x - 5) = 0$
$(4x - 9)(x - 5) = 0$
So,$x = 5$ or $x = 9/4 = 2.25$.
For equation $II$: $3y^2 - 19y + 28 = 0$
Using the splitting the middle term method: $3y^2 - 12y - 7y + 28 = 0$
$3y(y - 4) - 7(y - 4) = 0$
$(3y - 7)(y - 4) = 0$
So,$y = 4$ or $y = 7/3 \approx 2.33$.
Comparing the values:
If $x = 5$,then $x > y$ (since $y$ is $4$ or $2.33$).
If $x = 2.25$,then $x < y$ (since $y = 4$) and $x < y$ (since $y = 2.33$).
Since we get different results ($x > y$ and $x < y$),the relationship between $x$ and $y$ cannot be established.

(C) For equation $I$: $2x^{2} - 13x + 21 = 0$
$2x^{2} - 6x - 7x + 21 = 0$
$2x(x - 3) - 7(x - 3) = 0$
$(2x - 7)(x - 3) = 0$
So,$x = 3$ or $x = 3.5$.
For equation $II$: $5y^{2} - 22y + 21 = 0$
$5y^{2} - 15y - 7y + 21 = 0$
$5y(y - 3) - 7(y - 3) = 0$
$(5y - 7)(y - 3) = 0$
So,$y = 3$ or $y = 1.4$.
Comparing the values:
If $x = 3$,then $y = 3$ $(x = y)$ or $y = 1.4$ $(x > y)$.
If $x = 3.5$,then $y = 3$ $(x > y)$ or $y = 1.4$ $(x > y)$.
In all cases,$x \ge y$.

(D) For equation $I$: $12 x^{2} + 11 x - 56 = 0$.
Splitting the middle term: $12 x^{2} + 32 x - 21 x - 56 = 0$.
$4 x(3 x + 8) - 7(3 x + 8) = 0$.
$(4 x - 7)(3 x + 8) = 0$.
So,$x = \frac{7}{4} = 1.75$ or $x = -\frac{8}{3} \approx -2.67$.
For equation $II$: $4 y^{2} - 15 y + 14 = 0$.
Splitting the middle term: $4 y^{2} - 8 y - 7 y + 14 = 0$.
$4 y(y - 2) - 7(y - 2) = 0$.
$(4 y - 7)(y - 2) = 0$.
So,$y = \frac{7}{4} = 1.75$ or $y = 2$.
Comparing the values:
$x = \{1.75, -2.67\}$ and $y = \{1.75, 2\}$.
Since $1.75 \le 1.75$,$1.75 < 2$,$-2.67 < 1.75$,and $-2.67 < 2$,we conclude that $x \le y$.

(B) To solve the system of linear equations:
$I.$ $7x - 3y = 13$
$II.$ $5x + 4y = 40$
Multiply equation $(I)$ by $4$ and equation $(II)$ by $3$ to eliminate $y$:
$(7x - 3y) \times 4 = 13 \times 4 \Rightarrow 28x - 12y = 52$
$(5x + 4y) \times 3 = 40 \times 3 \Rightarrow 15x + 12y = 120$
Adding the two resulting equations:
$(28x - 12y) + (15x + 12y) = 52 + 120$
$43x = 172$
$x = 172 / 43 = 4$
Substitute $x = 4$ into equation $(I)$:
$7(4) - 3y = 13$
$28 - 3y = 13$
$3y = 28 - 13$
$3y = 15$
$y = 5$
Comparing the values,$x = 4$ and $y = 5$,we find that $x < y$.

(A) Step $1$: Solve equation $I$.
$\sqrt{1225} x + \sqrt{4900} = 0$
$35x + 70 = 0$
$35x = -70$
$x = -70 / 35 = -2$
Step $2$: Solve equation $II$.
$(81)^{1/4} y + (343)^{1/3} = 0$
$(3^4)^{1/4} y + (7^3)^{1/3} = 0$
$3y + 7 = 0$
$3y = -7$
$y = -7 / 3 \approx -2.33$
Step $3$: Compare $x$ and $y$.
Since $-2 > -2.33$,therefore $x > y$.

(B) Step $1$: Solve equation $I$.
$\frac{18}{x^2} + \frac{6}{x} - \frac{12}{x^2} = \frac{8}{x^2}$
Multiply the entire equation by $x^2$ (assuming $x \neq 0$):
$18 + 6x - 12 = 8$
$6 + 6x = 8$
$6x = 2$
$x = \frac{2}{6} = \frac{1}{3} \approx 0.333$
Step $2$: Solve equation $II$.
$y^3 + 9.68 + 5.64 = 16.95$
$y^3 + 15.32 = 16.95$
$y^3 = 16.95 - 15.32$
$y^3 = 1.63$
Since $1^3 = 1$ and $2^3 = 8$,$y$ must be slightly greater than $1$ (specifically $y \approx 1.177$).
Step $3$: Compare $x$ and $y$.
$x \approx 0.333$ and $y \approx 1.177$.
Therefore,$x < y$.

(C) Step $1$: Solve equation $I$ for $x$.
$x = \sqrt[3]{2197} = 13$.
Step $2$: Solve equation $II$ for $y$.
$y^2 = 169 \Rightarrow y = \pm 13$.
So,$y$ can be $13$ or $-13$.
Step $3$: Compare $x$ and $y$.
If $y = 13$,then $x = y$.
If $y = -13$,then $x > y$.
Combining these cases,we get $x \ge y$.

(C) Step $1$: Solve equation $I$.
$x = \sqrt{2304} = 48$.
Since the square root symbol denotes the principal (positive) square root,$x = 48$.
Step $2$: Solve equation $II$.
$y^2 = 2304$.
Taking the square root on both sides,$y = \pm \sqrt{2304} = \pm 48$.
So,$y = 48$ or $y = -48$.
Step $3$: Compare $x$ and $y$.
If $x = 48$ and $y = 48$,then $x = y$.
If $x = 48$ and $y = -48$,then $x > y$.
Combining these two cases,we get $x \ge y$.

(D) Step $1$: Solve equation $I$.
$\frac{15}{\sqrt{x}} - \frac{9}{\sqrt{x}} = \sqrt{x}$
$\frac{6}{\sqrt{x}} = \sqrt{x}$
$6 = \sqrt{x} \cdot \sqrt{x}$
$x = 6$
Step $2$: Solve equation $II$.
$y^{10} - (36)^{5} = 0$
$y^{10} = (36)^{5}$
Since $36 = 6^2$,we have $y^{10} = (6^2)^5 = 6^{10}$.
Taking the $10^{th}$ root on both sides,$y = \pm 6$.
Step $3$: Compare $x$ and $y$.
$x = 6$ and $y = 6$ or $y = -6$.
If $y = 6$,then $x = y$.
If $y = -6$,then $x > y$.
Since both conditions are possible,the relationship between $x$ and $y$ cannot be established.

(D) For equation $I$: $7x^2 + 16x - 15 = 0$
$7x^2 + 21x - 5x - 15 = 0$
$7x(x + 3) - 5(x + 3) = 0$
$(7x - 5)(x + 3) = 0$
So,$x = 5/7 \approx 0.71$ or $x = -3$.
For equation $II$: $y^2 - 6y - 7 = 0$
$y^2 - 7y + y - 7 = 0$
$y(y - 7) + 1(y - 7) = 0$
$(y + 1)(y - 7) = 0$
So,$y = -1$ or $y = 7$.
Comparing the values:
If $x = 0.71$,then $x > y$ (for $y = -1$) and $x < y$ (for $y = 7$).
If $x = -3$,then $x < y$ (for both $y = -1$ and $y = 7$).
Since the relationship changes depending on the values chosen,the relationship between $x$ and $y$ cannot be established.

(C) For equation $I$: $x^{2}+5x+6=0$
Factoring the quadratic: $x^{2}+3x+2x+6=0$
$x(x+3)+2(x+3)=0$
$(x+2)(x+3)=0$
So,$x = -2$ or $x = -3$.
For equation $II$: $y^{2}+7y+12=0$
Factoring the quadratic: $y^{2}+4y+3y+12=0$
$y(y+4)+3(y+4)=0$
$(y+3)(y+4)=0$
So,$y = -3$ or $y = -4$.
Comparing the values:
If $x = -2$,then $x > y$ (since $-2 > -3$ and $-2 > -4$).
If $x = -3$,then $x \ge y$ (since $-3 = -3$ and $-3 > -4$).
Combining these,we get $x \ge y$.

(B) For equation $I$: $x^{2}-9x+20=0$.
We need two numbers whose product is $20$ and sum is $9$. These are $5$ and $4$.
So,$(x-5)(x-4)=0$,which gives $x = 5$ or $x = 4$.
For equation $II$: $y^{2}-13y+42=0$.
We need two numbers whose product is $42$ and sum is $13$. These are $7$ and $6$.
So,$(y-7)(y-6)=0$,which gives $y = 7$ or $y = 6$.
Comparing the values:
If $x=4$,then $y=7$ $(x < y)$ or $y=6$ $(x < y)$.
If $x=5$,then $y=7$ $(x < y)$ or $y=6$ $(x < y)$.
In all cases,$x < y$.

(B) Given equations are:
$I.$ $12x + 3y = 14$
$II.$ $4x + 2y = 16$
To eliminate $x$,multiply equation $II$ by $3$:
$3 \times (4x + 2y) = 3 \times 16$
$12x + 6y = 48$ (Equation $III$)
Subtract equation $I$ from equation $III$:
$(12x + 6y) - (12x + 3y) = 48 - 14$
$3y = 34$
$y = \frac{34}{3} \approx 11.33$
Substitute $y = \frac{34}{3}$ into equation $II$:
$4x + 2(\frac{34}{3}) = 16$
$4x + \frac{68}{3} = 16$
$4x = 16 - \frac{68}{3}$
$4x = \frac{48 - 68}{3} = -\frac{20}{3}$
$x = -\frac{5}{3} \approx -1.67$
Comparing the values,$x = -1.67$ and $y = 11.33$.
Since $-1.67 < 11.33$,therefore $x < y$.

(B) Given equations are:
$I.$ $x = \sqrt{625} = 25$
$II.$ $y = \sqrt{676} = 26$
Comparing the values of $x$ and $y$:
Since $25 < 26$,we have $x < y$.

(B) For equation $I$: $x^{2}+4x+4=0$
This is a perfect square: $(x+2)^{2}=0$
Therefore,$x = -2$.
For equation $II$: $y^{2}-8y+16=0$
This is a perfect square: $(y-4)^{2}=0$
Therefore,$y = 4$.
Comparing the values: $x = -2$ and $y = 4$.
Since $-2 < 4$,we have $x < y$.

(D) For equation $I: x^{2}-19x+84=0$
We need two numbers whose product is $84$ and sum is $19$. These numbers are $12$ and $7$.
$x^{2}-12x-7x+84=0$
$x(x-12)-7(x-12)=0$
$(x-12)(x-7)=0$
So,$x = 12$ or $x = 7$.
For equation $II: y^{2}-25y+156=0$
We need two numbers whose product is $156$ and sum is $25$. These numbers are $13$ and $12$.
$y^{2}-13y-12y+156=0$
$y(y-13)-12(y-13)=0$
$(y-13)(y-12)=0$
So,$y = 13$ or $y = 12$.
Comparing the values:
If $x=12$,then $y=13$ $(x < y)$ or $y=12$ $(x = y)$.
If $x=7$,then $y=13$ $(x < y)$ or $y=12$ $(x < y)$.
In all cases,$x \le y$.

(C) Step $1$: Solve for $x$ from equation $I$.
$x^{3} - 468 = 1729$
$x^{3} = 1729 + 468$
$x^{3} = 2197$
$x = \sqrt[3]{2197} = 13$
Step $2$: Solve for $y$ from equation $II$.
$y^{2} - 1733 + 1564 = 0$
$y^{2} - 169 = 0$
$y^{2} = 169$
$y = \pm 13$
Step $3$: Compare $x$ and $y$.
We have $x = 13$ and $y = 13$ or $y = -13$.
In the first case,$x = y$ $(13 = 13)$.
In the second case,$x > y$ $(13 > -13)$.
Combining both cases,we get $x \ge y$.

(D) For equation $I$:
$\frac{9}{\sqrt{x}} + \frac{19}{\sqrt{x}} = \sqrt{x}$
$\frac{9+19}{\sqrt{x}} = \sqrt{x}$
$28 = \sqrt{x} \times \sqrt{x}$
$x = 28$
For equation $II$:
$y^{5} = \frac{(28)^{11/2}}{\sqrt{y}}$
$y^{5} \times y^{1/2} = (28)^{11/2}$
$y^{5 + 1/2} = (28)^{11/2}$
$y^{11/2} = (28)^{11/2}$
$y = 28$
Comparing the values,$x = 28$ and $y = 28$,therefore $x = y$.

(A) For equation $I$:
$\sqrt{784} x + 1234 = 1486$
$28 x = 1486 - 1234$
$28 x = 252$
$x = 252 / 28 = 9$
For equation $II$:
$\sqrt{1089} y + 2081 = 2345$
$33 y = 2345 - 2081$
$33 y = 264$
$y = 264 / 33 = 8$
Comparing the values,$x = 9$ and $y = 8$,therefore $x > y$.

(B) For equation $I$:
$\frac{12 - 23}{\sqrt{x}} = 5\sqrt{x}$
$-\frac{11}{\sqrt{x}} = 5\sqrt{x}$
$-11 = 5x \implies x = -2.2$
For equation $II$:
$\frac{\sqrt{y} - 5\sqrt{y}}{12} = -\frac{1}{\sqrt{y}}$
$\frac{-4\sqrt{y}}{12} = -\frac{1}{\sqrt{y}}$
$-\frac{\sqrt{y}}{3} = -\frac{1}{\sqrt{y}}$
$\sqrt{y} \cdot \sqrt{y} = 3$
$y = 3$
Comparing the values: $x = -2.2$ and $y = 3$.
Since $-2.2 < 3$,therefore $x < y$.

(A) For equation $I: 6x^{2} - 49x + 99 = 0$
We need two numbers whose product is $6 \times 99 = 594$ and sum is $49$.
The numbers are $22$ and $27$.
$6x^{2} - 22x - 27x + 99 = 0$
$2x(3x - 11) - 9(3x - 11) = 0$
$(2x - 9)(3x - 11) = 0$
$x = \frac{9}{2} = 4.5$ or $x = \frac{11}{3} \approx 3.67$
For equation $II: 5y^{2} + 17y + 14 = 0$
We need two numbers whose product is $5 \times 14 = 70$ and sum is $17$.
The numbers are $10$ and $7$.
$5y^{2} + 10y + 7y + 14 = 0$
$5y(y + 2) + 7(y + 2) = 0$
$(5y + 7)(y + 2) = 0$
$y = -\frac{7}{5} = -1.4$ or $y = -2$
Comparing the values:
$x$ values are $4.5$ and $3.67$.
$y$ values are $-1.4$ and $-2$.
Since all values of $x$ are greater than all values of $y$,we have $x > y$.

(A) Step $1$: Solve for $x$.
$x = (1331)^{1/3} = (11^3)^{1/3} = 11$.
Step $2$: Solve the quadratic equation $2y^2 - 17y + 36 = 0$.
Using the splitting the middle term method:
$2y^2 - 9y - 8y + 36 = 0$
$y(2y - 9) - 4(2y - 9) = 0$
$(y - 4)(2y - 9) = 0$
So,$y = 4$ or $y = 9/2 = 4.5$.
Step $3$: Compare $x$ and $y$.
Since $x = 11$ and the values of $y$ are $4$ and $4.5$,it is clear that $11 > 4$ and $11 > 4.5$.
Therefore,$x > y$.

(B) For equation $I$: $2x^2 + 3x + 1 = 0$
$2x^2 + 2x + x + 1 = 0$
$2x(x + 1) + 1(x + 1) = 0$
$(2x + 1)(x + 1) = 0$
So,$x = -0.5$ or $x = -1$.
For equation $II$: $12y^2 + 7y + 1 = 0$
$12y^2 + 4y + 3y + 1 = 0$
$4y(3y + 1) + 1(3y + 1) = 0$
$(4y + 1)(3y + 1) = 0$
So,$y = -0.25$ or $y = -0.33$.
Comparing the values:
$x$ values are $\{-1, -0.5\}$ and $y$ values are $\{-0.33, -0.25\}$.
Since both values of $x$ are less than both values of $y$,we conclude that $x < y$.

(B) Given equations are:
$I. 7x - 3y = 13$
$II. 5x + 4y = 40$
To eliminate $y$,multiply equation $(I)$ by $4$ and equation $(II)$ by $3$:
$(7x - 3y) \times 4 = 13 \times 4 \Rightarrow 28x - 12y = 52$
$(5x + 4y) \times 3 = 40 \times 3 \Rightarrow 15x + 12y = 120$
Adding the two resulting equations:
$(28x - 12y) + (15x + 12y) = 52 + 120$
$43x = 172$
$x = 172 / 43 = 4$
Substitute $x = 4$ into equation $(I)$:
$7(4) - 3y = 13$
$28 - 3y = 13$
$3y = 28 - 13$
$3y = 15$
$y = 5$
Comparing the values,$x = 4$ and $y = 5$,so $x < y$.

(B) Given equations are:
$2x + 5y = 6$ --- $(1)$
$5x + 11y = 9$ --- $(2)$
To solve for $x$ and $y$,multiply equation $(1)$ by $5$ and equation $(2)$ by $2$:
$(2x + 5y = 6) \times 5 \implies 10x + 25y = 30$ --- $(3)$
$(5x + 11y = 9) \times 2 \implies 10x + 22y = 18$ --- $(4)$
Subtract equation $(4)$ from equation $(3)$:
$(10x + 25y) - (10x + 22y) = 30 - 18$
$3y = 12$
$y = 4$
Substitute $y = 4$ into equation $(1)$:
$2x + 5(4) = 6$
$2x + 20 = 6$
$2x = 6 - 20$
$2x = -14$
$x = -7$
Comparing the values,$x = -7$ and $y = 4$. Since $-7 < 4$,we have $x < y$.

(A) For equation $I: 6x^{2} + 29x + 35 = 0$
We factorize the quadratic equation: $6x^{2} + 14x + 15x + 35 = 0$
$2x(3x + 7) + 5(3x + 7) = 0$
$(2x + 5)(3x + 7) = 0$
Thus,$x = -2.5$ or $x = -2.33$.
For equation $II: 3y^{2} + 19y + 30 = 0$
We factorize the quadratic equation: $3y^{2} + 9y + 10y + 30 = 0$
$3y(y + 3) + 10(y + 3) = 0$
$(3y + 10)(y + 3) = 0$
Thus,$y = -3.33$ or $y = -3$.
Comparing the values:
Since $-2.5 > -3.33$,$-2.5 > -3$,$-2.33 > -3.33$,and $-2.33 > -3$,we conclude that $x > y$.

(A) Step $1$: Solve equation $I$.
$\sqrt{1225} x + \sqrt{4900} = 0$
$35 x + 70 = 0$
$35 x = -70$
$x = \frac{-70}{35} = -2$
Step $2$: Solve equation $II$.
$(81)^{1/4} y + (343)^{1/3} = 0$
Since $81 = 3^4$ and $343 = 7^3$,we have:
$(3^4)^{1/4} y + (7^3)^{1/3} = 0$
$3 y + 7 = 0$
$3 y = -7$
$y = -\frac{7}{3} \approx -2.33$
Step $3$: Compare $x$ and $y$.
$x = -2$
$y \approx -2.33$
Since $-2 > -2.33$,therefore $x > y$.