In questions two equations numbered I and II are given you have to solve both equations and give answer
`I.x^2-4x-12=0 II. Y^2-5y-14=0`

Options

A. if `xgty`
B. if`xgey`
C. if`xlty`
D. if X=Y or relation between X and Y can not be established

Answer

To solve the given equations step by step, we will start with each equation separately.

### Step 1: Solve Equation I
The first equation is:
\[ I: x^2 - 4x - 12 = 0 \]

To factor this quadratic equation, we need to find two numbers that multiply to \(-12\) (the constant term) and add up to \(-4\) (the coefficient of \(x\)).

The numbers that satisfy these conditions are \(-6\) and \(2\).

Now we can rewrite the equation as:
\[ (x - 6)(x + 2) = 0 \]

Setting each factor to zero gives us:
1. \( x - 6 = 0 \) → \( x = 6 \)
2. \( x + 2 = 0 \) → \( x = -2 \)

So, the solutions for \(x\) are:
\[ x = 6 \quad \text{and} \quad x = -2 \]

### Step 2: Solve Equation II
The second equation is:
\[ II: y^2 - 5y - 14 = 0 \]

Similarly, we need to find two numbers that multiply to \(-14\) and add up to \(-5\).

The numbers that satisfy these conditions are \(-7\) and \(2\).

Now we can rewrite the equation as:
\[ (y - 7)(y + 2) = 0 \]

Setting each factor to zero gives us:
1. \( y - 7 = 0 \) → \( y = 7 \)
2. \( y + 2 = 0 \) → \( y = -2 \)

So, the solutions for \(y\) are:
\[ y = 7 \quad \text{and} \quad y = -2 \]

### Step 3: Compare the Solutions
Now we have the solutions:
- For \(x\): \(6\) and \(-2\)
- For \(y\): \(7\) and \(-2\)

We can compare the values:
1. When \(x = -2\), \(y = -2\) → \(x = y\)
2. When \(x = 6\), \(y = -2\) → \(x > y\)
3. When \(x = -2\), \(y = 7\) → \(x < y\)

### Conclusion
From the comparisons, we see that:
- In one case, \(x\) is equal to \(y\).
- In another case, \(x\) is greater than \(y\).
- In the last case, \(x\) is less than \(y\).

Thus, we cannot establish a consistent relationship between \(x\) and \(y\).

### Final Answer
The relationship between \(x\) and \(y\) cannot be established.

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