x² + ( x - 5 )² = 157
x² + x² - 10 x + 25 = 157
2 x² - 10 x - 132 = 0
2 ( x² - 5 x - 66 ) = 0
2 ( x² - 11 x + 6 x - 66 ) = 0
2 ( x ( x - 11 ) + 6 ( x - 11 )) = 0
2 ( x - 11 ) ( x + 6 ) = 0 (The zero-product property)
x - 11 = 0, x = 11
x + 6 = 0, x = - 6
There are two solutions:
1. Two numbers that differ by 5 and have squares that sum to 157 are 11 and 6 ( 6 = 11-5, 11² + 6² = 121 + 36 = 157 )
2. Same is for the numbers -6 and -11
( - 11 = -6 - 5, (-6)² + (-11 )² = 36 + 121 = 157 )
