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

I. x² - 20x + 96 = 0
II. y² - 3y - 28 = 0

The correct answer is If x > y.

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

Step 1: Solve Equation I for $x$

The first quadratic equation is:

$x^2-20x+96=0$

We factorize the equation by splitting the middle term $-20x$ into two factors whose sum is $-20$ and product is $96$. The numbers are $-12$ and $-8$.

$x^2-12x-8x+96=0$

$x(x-12)-8(x-12)=0$

$(x-12)(x-8)=0$

Setting each factor to zero gives:

$x-12=0\Rightarrow x=12$

$x-8=0\Rightarrow x=8$

So, the roots for $x$ are $x=12$ and $x=8$.

Step 2: Solve Equation II for $y$

The second quadratic equation is:

$y^2-3y-28=0$

We factorize by splitting the middle term $-3y$ into two factors whose sum is $-3$ and product is $-28$. The numbers are $-7$ and $4$.

$y^2-7y+4y-28=0$

$y(y-7)+4(y-7)=0$

$(y-7)(y+4)=0$

Setting each factor to zero gives:

$y-7=0\Rightarrow y=7$

$y+4=0\Rightarrow y=-4$

So, the roots for $y$ are $y=7$ and $y=-4$.

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

Now, we compare every value of $x$ with every value of $y$:

1. When $x=12$ and $y=7$: $12>7\Rightarrow x>y$

2. When $x=12$ and $y=-4$: $12>-4\Rightarrow x>y$

3. When $x=8$ and $y=7$: $8>7\Rightarrow x>y$

4. When $x=8$ and $y=-4$: $8>-4\Rightarrow x>y$

Since in all possible comparisons, $x$ is strictly greater than $y$, the correct relationship is If x > y.