(i) ` x^(2) + 6x + 5 = 0`
(ii) ` y^(2) -10y+21 = 0`

Options

A. x = y or no relation can be established between x and y.
B. ` x le y`
C. ` x ge y`
D. ` x lt y`

Explanation

Let's solve the equations step by step.

### Step 1: Solve the first equation \( x^2 + 6x + 5 = 0 \)

To factor the quadratic equation, we look for two numbers that multiply to \( 5 \) (the constant term) and add up to \( 6 \) (the coefficient of \( x \)).

The numbers \( 1 \) and \( 5 \) satisfy these conditions.

So we can factor the equation as:
\[ (x + 1)(x + 5) = 0 \]

### Step 2: Find the values of \( x \)

Setting each factor to zero gives us:
1. \( x + 1 = 0 \) → \( x = -1 \)
2. \( x + 5 = 0 \) → \( x = -5 \)

Thus, the solutions for \( x \) are:
\[ x = -1 \quad \text{and} \quad x = -5 \]

### Step 3: Solve the second equation \( y^2 - 10y + 21 = 0 \)

Similarly, we look for two numbers that multiply to \( 21 \) and add up to \( -10 \).

The numbers \( -3 \) and \( -7 \) satisfy these conditions.

So we can factor the equation as:
\[ (y - 3)(y - 7) = 0 \]

### Step 4: Find the values of \( y \)

Setting each factor to zero gives us:
1. \( y - 3 = 0 \) → \( y = 3 \)
2. \( y - 7 = 0 \) → \( y = 7 \)

Thus, the solutions for \( y \) are:
\[ y = 3 \quad \text{and} \quad y = 7 \]

### Step 5: Determine the relationship between \( x \) and \( y \)

Now we have the values:
- \( x = -1 \) or \( x = -5 \)
- \( y = 3 \) or \( y = 7 \)

Since both values of \( y \) (3 and 7) are greater than both values of \( x \) (-1 and -5), we can conclude that:
\[ y > x \quad \text{or} \quad x < y \]

### Final Conclusion

The relationship is \( x < y \).

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