Question 1179118
let x = one of the numbers and y = the other.
if x exceeds y by 5, then x = y + 5
the sum of their squares is 157.
this means x^2 + y^2 = 157
since x = y + 5, then replace x with y + 5 in that equation to get:
(y + 5)^2 + y^2 = 157
simplify to get:
y^2 + 10y + 25 + y^2 = 157
combine like terms to get:
2y^2 + 10y + 25 = 157
subtract 157 from both sides of the equation to get:
2y^2 + 10y - 132 = 0
divide both sides of this equation by 2 to get:
y^2 + 5y - 66 = 0
factor this quadratic equation to get:
(y + 11) * (y - 6) = 0
solve for y to get:
y = -11 or y = 6


when y = 6, x = y + 5 = 11
when y = -11, x = y + 5 = -6


when y = 6 and x = 11, then 6^2 + 11^2 = 36 + 121 = 157
when y = -11 and x = -6, then (-11)^2 + (-6)^2 = 121 + 36 = 157.


your numbers can be:
(6 and 11) or (-11 and -6).