Question 458488
If the area of the square is *numerically* five more than its perimeter, then we would say that


*[tex \LARGE A = P + 5]


*[tex \LARGE s^2 = 4s + 5]


*[tex \LARGE s^2 - 4s - 5 = 0 \Rightarrow (s - 5)(s + 1) = 0 \Rightarrow s = 5] (discard the negative solution -1).