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 © Algebra House
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