Find two consecutive numbers whose squares have the sum of 85.

1.	Find two consecutive numbers whose squares have the sum of 85.
Answer» Let the two consecutive be considered as (x) and (x +1) respectively. Given that, The sum of their squares is 85. Expressing the same by equation we have, x²+ (x + 1)²= 85 ⇒ x²+ x²+ 2x + 1 = 85 ⇒ 2x²+ 2x + 1 – 85 = 0 ⇒ 2x²+ 2x – 84 = 0 ⇒ 2(x²+ x – 42) = 0 Solving for x by factorization method, we get x²+ 7x – 6x – 42 = 0 ⇒ x(x + 7) – 6(x + 7) = 0 ⇒ (x – 6)(x + 7) = 0 Now, either, x – 6 = 0 ⇒ x = 6 Or, x + 7 = 0 ⇒ x = -7 Thus, the consecutive numbers whose sum of squares can be (6, 7) or (-7, -6).

Answer»

Let the two consecutive be considered as (x) and (x +1) respectively.

Given that,

The sum of their squares is 85.

Expressing the same by equation we have,

x²+ (x + 1)²= 85

⇒ x²+ x²+ 2x + 1 = 85

⇒ 2x²+ 2x + 1 – 85 = 0

⇒ 2x²+ 2x – 84 = 0

⇒ 2(x²+ x – 42) = 0

Solving for x by factorization method, we get

x²+ 7x – 6x – 42 = 0

⇒ x(x + 7) – 6(x + 7) = 0

⇒ (x – 6)(x + 7) = 0

Now, either, x – 6 = 0 ⇒ x = 6

Or, x + 7 = 0 ⇒ x = -7

Thus, the consecutive numbers whose sum of squares can be (6, 7) or (-7, -6).

Discussion