Question 255744
Let x = first integer and x+1 = consecutive integer.
(i) {{{x^2 + (x+1)^2 = 41}}}
step 1 - expand (i) to get
(ii) {{{x^2 + x^2 + 2x + 1 = 41}}}
step 2 - set (i) =0 and factor as
(iii) {{{2x^2 + 2x -40 = 0}}}
divide by 2 to get
{{{(x - 4 )(x + 5) = 0}}}
x-4 = 0 - - -> x = 4
x+5=0 - - - > x = -5
SInce we want positive, we have
x = 4 and x+1 = 5.