Question

The sum of the squares of two consecutive multiples of 7 is 637. Taking the bigger number x as a positive number, find the smaller of these two numbers.

Answer 1

Let the multiplies of 7 be 7x and 7(x + 1).

It is given that the sum of the squares of two consecutive multiplies of 7 is 637.

⇒ (7x)² + [7(x + 1)]² = 637

⇒ 49x² + [7x + 7]² = 637

⇒ 49x² + 49x² + 7² + 98x − 637 = 0

⇒ 98x² + 49 + 98x − 637 = 0

⇒ 98x² + 98x − 588 = 0

⇒ x² + x − 6 = 0

⇒ x² + 3x − 2x − 6 = 0

⇒ x(x + 3) − 2(x + 3) = 0

⇒ (x + 3)(x − 2) = 0

⇒ (x + 3) = 0 or (x − 2) = 0

⇒ x = −3 or x = 2

It is given that x is bigger positive number.

So, the multiplies of 7 is 7 × 2 = 14 and 7(2 + 1) = 7 × 3 = 21

Hence, the smaller angle is 14.