Question

Two consecutive positive numbers are such that the sum of their squares is 113. Find the two numbers.

Answer 1

let two consecutive positive numbers be A and (A+1),
then according to the question, we have

A²+(A+1)²=113,

A²+A²+2A+1=113,

2A²+2A=113-1,

2(A²+A)=112,

then

A²+A=56,

A²+A-56=0,

A²+(8-7)A-56=0,

A²+8A-7A-56=0,

A(A+8)-7(A+8)=0,

(A+8)(A-7)=0,

then

A-7=0,

A=7,

therefore

first number=7,

second number=7+1=8

Answer 2

let the 2 consecutive no.s be x,x+1
$x + (x + 1) = 113..........given \\ 2x + 1 = 113 \\ 2x = 113 - 1 \\ 2x = 112 \\ x = 112 \div 2 \\ x = 56$
first no. x=56
sec no. x+1=57