Directions: In the following question, two quadratic equations labeled I and II are provided. You are required to solve both equations to determine the relationship between x and y.

I. 2x² + 16x + 32 = 0
II. 2y² + 45y - 98 = 0

The correct option is x = y or no relation can be established between x and y.

To determine the relationship between $x$ and $y$, we solve both quadratic equations step-by-step.

Step 1: Solve Equation I for $x$

Given Equation I:
$2x^2+16x+32=0$
Divide the entire equation by 2:
$x^2+8x+16=0$
Factorize the perfect square quadratic expression:
${(x+4)}^2=0$
$x+4=0$
$x=-4$

Thus, the value of $x$ is $-4$.

Step 2: Solve Equation II for $y$

Given Equation II:
$2y^2+45y-98=0$
Split the middle term $45y$ into $49y-4y$ since $49\times (-4)=-196$ (which equals $2\times (-98)$):
$2y^2+49y-4y-98=0$
Factor by grouping:
$y(2y+49)-2(2y+49)=0$
$(2y+49)(y-2)=0$
Solving for $y$:
$2y+49=0\Rightarrow y=-24.5$
$y-2=0\Rightarrow y=2$

Thus, the values of $y$ are $2$ and $-24.5$.

Step 3: Compare values of $x$ and $y$

1. When $x=-4$ and $y=2$: $-4<2$, so $x<y$.
2. When $x=-4$ and $y=-24.5$: $-4>-24.5$, so $x>y$.

Since $x<y$ in one case and $x>y$ in another case, no relation can be established between $x$ and $y$.