Question

The difference of squares of two positive integers is 400. Find the integers if twice of the smaller integer is 5 more than the greater integer.

Answer 1

Given: Let the greater integer be x and the smaller integer be y. We are given x² – y² = 400 and 2y = x + 5.

Step-wise calculation:

1. From 2y = x + 5, express x = 2y – 5.

2. Substitute into x² – y² = 400:

(2y – 5)² – y² = 400

3. Expand and simplify:

(4y² – 20y + 25) – y² = 400 

⇒ 3y² – 20y + 25 = 400

⇒ 3y² – 20y – 375 = 0

4. Solve the quadratic:

Discriminant D = (–20)² – 4 × 3 × (–375) 

= 400 + 4500 

= `4900, sqrt(D) = 70` 

`y = (20 ± 70)/6` 

The positive solution is `y = (20 + 70)/6` 

= `90/6`

= 15   ...(The other root is negative, so discard.)

5. Then x = 2y – 5

= 2 × 15 – 5 

= 30 – 5

= 25

The two positive integers are 25 (greater) and 15 (smaller).