Question

find the two consecutive positive integers sum of whose square is 613

Answer 1

Let x be one integer. Therefore x +1 is the next integer.
x^2+(x+1)^2= 613 ................
x^2 +(x+1)^2 =613
.
x^2 +x^2 +2x+ 1 =613
.
2x^2 +2x -612 =0
.
x^2 +x -306=0
.
(x+18)(x-17) =0
.
x = 17, -18
.
Integers,,17,18,,,,,,,check,,17^2 +18^2 =613,,,ok
.
also -18,,-17,,,,,,,check, (-17)^2 +(-18)^2 = 613,,,ok

Answer 2

Heya, find the answer in the attached image.

Hope this helps. . . . . if it really helped, then please mark it as the BRAINLIEST and please say a thanks.

Thanks.