In the given questions, two equations (I) & (II) are given . You have to solve both the equations and mark the answer accordingly
I. ` x^(2) - 11 x + 30 = 0`
II. ` y^(2) + 12 y + 36 = 0 `

Options

A. ` x lt y`
B. ` x gt y `
C. ` x le y `
D. ` x ge y `

Answer

To solve the given equations step by step, we will start with Equation I and then move on to Equation II.

### Step 1: Solve Equation I
The first equation is:
\[ x^2 - 11x + 30 = 0 \]

To solve this quadratic equation, we will factor it. We need to find two numbers that multiply to \(30\) (the constant term) and add up to \(-11\) (the coefficient of \(x\)).

The factors of \(30\) that satisfy this condition are \(-5\) and \(-6\), because:
\[ -5 \times -6 = 30 \quad \text{and} \quad -5 + -6 = -11 \]

Now we can factor the equation:
\[ (x - 5)(x - 6) = 0 \]

### Step 2: Find the values of \(x\)
Setting each factor equal to zero gives us:
1. \(x - 5 = 0 \Rightarrow x = 5\)
2. \(x - 6 = 0 \Rightarrow x = 6\)

Thus, the solutions for Equation I are:
\[ x = 5 \quad \text{and} \quad x = 6 \]

### Step 3: Solve Equation II
The second equation is:
\[ y^2 + 12y + 36 = 0 \]

This is also a quadratic equation. We will factor it as well. We need to find two numbers that multiply to \(36\) (the constant term) and add up to \(12\) (the coefficient of \(y\)).

The factors of \(36\) that satisfy this condition are \(6\) and \(6\), because:
\[ 6 \times 6 = 36 \quad \text{and} \quad 6 + 6 = 12 \]

Now we can factor the equation:
\[ (y + 6)(y + 6) = 0 \quad \text{or} \quad (y + 6)^2 = 0 \]

### Step 4: Find the value of \(y\)
Setting the factor equal to zero gives us:
\[ y + 6 = 0 \Rightarrow y = -6 \]

Thus, the solution for Equation II is:
\[ y = -6 \]

### Step 5: Compare the values of \(x\) and \(y\)
We have found:
- \(x = 5\) or \(x = 6\)
- \(y = -6\)

Now we compare the values:
- \(5 > -6\)
- \(6 > -6\)

### Conclusion
Both values of \(x\) are greater than \(y\). Therefore, we can conclude:
\[ x > y \]

### Final Answer
The answer is that \(x\) is greater than \(y\).

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