Question

find the two consecutive positive integers whose sum of their squares is 365​

Answer 1

Answer:

x²+(x+1)²=365

x²+x²+2x+1=365

2x²+2x-364=0

u can solve it by substitution or quadratic equation method

Answer 2

: Required Answer

$\longrightarrow$ Let us say, the two consecutive positive integers be x and x + 1.

Therefore, as per the given questions,

x2 + (x + 1)2 = 365

⇒ x2 + x2 + 1 + 2x = 365

⇒ 2x2 + 2x – 364 = 0

⇒ x2 + x – 182 = 0

⇒ x2 + 14x – 13x – 182 = 0

⇒ x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

⇒ x = – 14 or x = 13

since, the integers are positive, so x can be 13, only.

x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.