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 answer is x = y or no relation can be established between x and y.

To determine the relationship between the two variables, we solve both quadratic equations step-by-step to find the roots for
$x$
and
$y$
.

Step 1: Solve Equation I for
$x$

Equation I is:

$2x^2+16x+32=0$
Divide the entire equation by
$2$
to simplify:

$x^2+8x+16=0$
This expression is a perfect square trinomial:

${(x+4)}^2=0$
Taking the square root on both sides:

$x+4=0$

$x=-4$
Thus, the roots for Equation I are
$x=-4,-4$
.

Step 2: Solve Equation II for
$y$

Equation II is:

$2y^2+45y-98=0$
We factor the equation by splitting the middle term. We need two numbers whose product is
$2\times (-98)=-196$
and whose sum is
$45$
.
The numbers are
$49$
and
$-4$
, since
$49\times (-4)=-196$
and
$49+(-4)=45$
.
Rewrite the middle term:

$2y^2+49y-4\mathrm{y<mo>-</mo><mn>98</mn><mo>=</mo><mn>0</mn>}$
Factor by grouping:

$y(2y+49)-2(2y+49)=0$

$(2y+49)(y-2)=0$
Equating each factor to zero:

$2y+49=0\Rightarrow y=-\frac{49}{2}=-24.5$

$y-2=0\Rightarrow y=2$
Thus, the roots for Equation II are
$y=2$
and
$y=-24.5$
.

Step 3: Compare
$x$
and
$y$

Now we compare the root
$x=-4$
with both values of
$y$
:
1. Comparing
$x=-4$
with
$y=2$
:

$-4<2\Rightarrow x<y$
2. Comparing
$x=-4$
with
$y=-24.5$
:

$-4>-24.5\Rightarrow x>y$

Since we obtain both
$x<y$
and
$x>y$
, no fixed relationship can be established between
$x$
and
$y$
.
Therefore, the correct option is x = y or no relation can be established between x and y.