In the following questions two equations numbered I and II are given. Solve both the equations and give answer
I. `x^2-4=0` II. `y^2-9y+20=0`

Options

A. `xgey`
B. `xlty`
C. `xley`
D. relationship can't be established

Answer

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

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

**Solution:**
1. Rearranging the equation gives:
\[ x^2 = 4 \]
2. Taking the square root of both sides:
\[ x = \pm \sqrt{4} \]
3. Therefore, the solutions for \( x \) are:
\[ x = 2 \quad \text{and} \quad x = -2 \]

### Step 2: Solve Equation II
The second equation is:
\[ y^2 - 9y + 20 = 0 \]

**Solution:**
1. We can factor the quadratic equation:
\[ y^2 - 5y - 4y + 20 = 0 \]
This can be grouped as:
\[ (y^2 - 5y) - (4y - 20) = 0 \]
2. Factoring by grouping gives:
\[ y(y - 5) - 4(y - 5) = 0 \]
3. Factoring out the common term:
\[ (y - 5)(y - 4) = 0 \]
4. Setting each factor to zero gives:
\[ y - 5 = 0 \quad \text{or} \quad y - 4 = 0 \]
5. Thus, the solutions for \( y \) are:
\[ y = 5 \quad \text{and} \quad y = 4 \]

### Summary of Solutions
- From Equation I, we have:
\[ x = 2 \quad \text{and} \quad x = -2 \]
- From Equation II, we have:
\[ y = 5 \quad \text{and} \quad y = 4 \]

### Final Comparison
Now we compare the values of \( x \) and \( y \):
- The possible values of \( x \) are \( 2 \) and \( -2 \).
- The possible values of \( y \) are \( 5 \) and \( 4 \).

Since both values of \( y \) (5 and 4) are greater than the positive value of \( x \) (2), we conclude that:
\[ x < y \]

### Answer
The correct answer is \( x < y \).

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