Question

Find the two consecutive integers, whose sum of their squares is 365.

Guy 's answer this​

Answer 1

Answer:

let one integer be 'x' ; as the other one is consecutive the second integer be (x+1)

as their sum of square is 365

x²+(x+1)² = 365

x²+x²+2x+1=365      {as (a+b)² = a²+b²+2ab)

2x²+2x+1  = 365

2x²+2x+1-365=0

2x²+2x-364=0

x²+x - 182 = 0    {on dividing the whole term by 2}

x²+14x-13x-182 = 0    {on splitting the middile term}

x(x+14)-13(x+14)=0

(x+14)(x-13)  = 0

x+14=0   or   x-13=0

  x  = -14       or   x = 13

if x=(-14) other integer is -14+1 = -13

if x = 13  other integer is 13+1 = 14

hence the required pair of integers can be (13,14) or (-13,-14)