Find two consecutive odd integers such that the sum of their squares is 130. x = 1st odd integer x + 2 = 2nd odd integer {odd integers increase by two each time} x² + (x + 2)² = 130 {sum of their squares is 130} x² + (x + 2)(x + 2) = 130 {when you square a binomial, multiply it by itself} x² + x² + 4x + 4 = 130 {used foil method} 2x² + 4x + 4 = 130 {combined like terms} 2x² + 4x  126 = 0 {subtracted 130 from both sides} 2(x² + 2x  63) = 0 {factored 2 out} 2(x + 9)(x  7) = 0 {factored into two binomials} x + 9 = 0 or x  7 = 0 {set each factor equal to 0, excluding the 2} x = 9 or x = 7 {solved each equation for x} x + 2 = 7 or x + 2 = 9 {substituted 9 and 7, in for x, into x + 2} 9 and 7, also 7 and 9 are the two consecutive odd integers © Algebra House
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