In each question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
I `x^2-2x-8=0`
II `y^2+15y+54=0`

Options

A. relation can’t be established
B. `xley`
C. `xgty`
D. `xgey`

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:
\[ x^2 - 2x - 8 = 0 \]

This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 \)
- \( b = -2 \)
- \( c = -8 \)

To factor this equation, we need to find two numbers that multiply to \( ac = 1 \cdot (-8) = -8 \) and add to \( b = -2 \).

The numbers that satisfy this are \( -4 \) and \( 2 \).

Now, we can rewrite the equation:
\[ x^2 - 4x + 2x - 8 = 0 \]
Grouping the terms:
\[ (x^2 - 4x) + (2x - 8) = 0 \]
Factoring by grouping:
\[ x(x - 4) + 2(x - 4) = 0 \]
Factoring out the common term \( (x - 4) \):
\[ (x - 4)(x + 2) = 0 \]

Setting each factor to zero gives us:
1. \( x - 4 = 0 \) → \( x = 4 \)
2. \( x + 2 = 0 \) → \( x = -2 \)

So, the solutions for Equation I are:
\[ x = 4 \quad \text{and} \quad x = -2 \]

### Step 2: Solve Equation II
The second equation is:
\[ y^2 + 15y + 54 = 0 \]

This is also a quadratic equation where:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 54 \)

We need to find two numbers that multiply to \( ac = 1 \cdot 54 = 54 \) and add to \( b = 15 \).

The numbers that satisfy this are \( 6 \) and \( 9 \).

Now, we can rewrite the equation:
\[ y^2 + 6y + 9y + 54 = 0 \]
Grouping the terms:
\[ (y^2 + 6y) + (9y + 54) = 0 \]
Factoring by grouping:
\[ y(y + 6) + 9(y + 6) = 0 \]
Factoring out the common term \( (y + 6) \):
\[ (y + 6)(y + 9) = 0 \]

Setting each factor to zero gives us:
1. \( y + 6 = 0 \) → \( y = -6 \)
2. \( y + 9 = 0 \) → \( y = -9 \)

So, the solutions for Equation II are:
\[ y = -6 \quad \text{and} \quad y = -9 \]

### Summary of Solutions
- For Equation I: \( x = 4 \) and \( x = -2 \)
- For Equation II: \( y = -6 \) and \( y = -9 \)