Question

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

Answer 1

Let the 2 consecutive multiples of 7 be 7x and 7(x + 1)

According to the Question;

(7x)² + (7(x + 1))² = 1225

⇒ 49x² + (7x + 7)² = 1225

⇒ 49x² + 49x² + 49 + 2(7x)(7) = 1225

⇒ 49x² + 49x² + 49 + 2(49)x = 1225

⇒ 49(x² + x² + 1 + 2x) = 1225

⇒ 49(2x² + 2x + 1) = 1225

⇒ 2x² + 2x + 1 = 1225 ÷ 49

⇒ 2x² + 2x + 1 = 25

⇒ 2x² + 2x + 1 - 25 = 0

⇒ 2x² + 2x - 24 = 0

⇒ 2(x² + x - 12) = 0

⇒ x² + x - 12 = 0 ÷ 2

⇒ x² + x - 12 = 0

Factorizing the above Equation

x² + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x - 3)(x + 4) = 0

∴ x = 3 or -4

Multiples cannot be negative

So x = 3

Multiples are 7x and 7(x + 1)

7x = 7(3) = 21

7(x + 1) = 7(3 + 1) = 7(4) = 28

Hence the 2 consecutive Multiples are 21 and 28