Question

find two consecutive positive integer, sum of whose square is 365​

Answer 1

Let the consecutive integers be x and x+1

Atq

x²+(x+1)²=365

X²+x²+1+2x=365

2x²+2x+1-365=0

2x²+2x-364=0

Divide by 2

x²+x-182=0

Now you can solve it

Answer 2

Step-by-step explanation:

let them be X and x+1

so,

x^2+x^2+1+2x=365

2x^2+2x=364

2x(x+1)=364

2x²+2x-364=0

now just find 2 numbers whose multiplication is 728 and plus or minus is 2. and do the factorisation