Using Identity find (i) `(4p-3q)^2` (ii) `(49)^2`

Options

A. (i)`19p^2+9q^2-24pq`
(ii)`2481`
B. (i)`16p^2+9q^2-24pq`
(ii)`2401`
C. (i)`6p^2+8q^2-24pq`
(ii)`2501`
D. (i)`10p^2+9q^8-24pq`
(ii)`2401`

Explanation

To solve the given problems using algebraic identities, we will apply the identity for the square of a binomial, which states:

\[ (a - b)^2 = a^2 - 2ab + b^2 \]

### Solution for (i) \( (4p - 3q)^2 \)

1. **Identify \( a \) and \( b \)**:
- Let \( a = 4p \) and \( b = 3q \).

2. **Apply the identity**:
\[ (4p - 3q)^2 = (4p)^2 - 2(4p)(3q) + (3q)^2 \]

3. **Calculate each term**:
- \( (4p)^2 = 16p^2 \)
- \( 2(4p)(3q) = 24pq \)
- \( (3q)^2 = 9q^2 \)

4. **Combine the terms**:
\[ (4p - 3q)^2 = 16p^2 - 24pq + 9q^2 \]

Thus, the final answer for \( (4p - 3q)^2 \) is:
\[ \boxed{16p^2 - 24pq + 9q^2} \]

### Solution for (ii) \( (49)^2 \)

1. **Rewrite 49**:
- We can express \( 49 \) as \( 50 - 1 \).

2. **Apply the identity**:
\[ (50 - 1)^2 = (50)^2 - 2(50)(1) + (1)^2 \]

3. **Calculate each term**:
- \( (50)^2 = 2500 \)
- \( 2(50)(1) = 100 \)
- \( (1)^2 = 1 \)

4. **Combine the terms**:
\[ (50 - 1)^2 = 2500 - 100 + 1 = 2401 \]

Thus, the final answer for \( (49)^2 \) is:
\[ \boxed{2401} \]