Question

find the two consecutive numbers whose sum of square is613​

Answer 1

Step-by-step explanation:

Let two consecutive positive integers be X, x + 1

according to the question

x² + (x + 1)² = 613

x² + x² + 1 + 2x = 613

2x² + 2x = 613 - 1

2x² + 2x - 612 = 0

2 ( x² + x - 306 ) = 0

x² + x - 306 = 0

x² - 18x + 17x - 306 = 0

x (x - 18) + 17 (x - 18) = 0

(x - 18) (x + 17) = 0

x = 18

x = -17

Hope my answer helps you!!

Answer 2

Answer:

let the two consecutive number be x and x+1

now

x^2+(x+1) ^2=613

x^2+x^2+2x+1=613

2x^2+2x-612=0

2(x^2+x-306) =0

x^2+18x-17x-306=0

x(x+18) -17(x+18) =0

(x+18) =0, (x-17) =0

-18and 17

now x= 17 and 17+1= 18 ans....