Question

The sum of the squares two consecutive multiples of 7 is 1225. Find the multiples. 

Answer 1

Let the required consecutive multiples of 7 be 7x and 7(x + 1).

According to the given condition, 

(7x)² + [7(x + 1)]² = 1225 

⇒ 49x² + 49(x² + 2x + 1) = 1225 

⇒ 49x² + 49x² + 98x + 49 = 1225 

⇒ 98x² + 98x – 1176 = 0 

⇒ x² + x – 12 = 0 

⇒ x² + 4x – 3x – 12 = 0 

⇒ x(x + 4) – 3(x + 4) = 0 

⇒ (x + 4)(x – 3) = 0 

⇒ x + 4 = 0 or x – 3 = 0 

⇒ x = –4 or x = 3 

∴ x = 3   ...(Neglecting the negative value) 

When x = 3,

7x = 7 × 3

= 21 

7(x + 1) = 7(3 + 1)

= 7 × 4

= 28 

Hence, the required multiples are 21 and 28.