Question 447847: Find two consecutive positive integers such that the sum of their squares is 41.
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! "Find two consecutive positive integers such that the sum of their squares is 41."
x = 1st positive integer
x + 1 = 2nd positive integer
x² + (x + 1)² = 41 {sum of squares is 41}
x² + (x + 1)(x + 1) = 41 {when you square a binomial, multiply it by itself}
x² + x² + 2x + 1 = 41 {used foil method}
2x² + 2x + 1 = 41 {combined like terms}
2x² + 2x - 40 = 0 {subtracted 41 from both sides}
2(x² + x - 20) = 0 {factored 2 out}
2(x + 5)(x - 4) = 0 {factored into two binomials}
x + 5 = 0 or x - 4 = 0 {set each factor equal to 0, excluding the 2}
x = -5 or x = 4 {solved each equation for x}
x = 4 {only positive integer}
x + 1 = 5 {substituted 4, in for x, into x + 1}
4 and 5 are the consecutive positive integers
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