The sum of the squares of two consecutive odd integers is 74. Find the two integers.

x = 1st odd integer

x + 2 = 2nd odd integer

x² + (x + 2)² = 74 {sum of squares is 74}

x² + (x + 2)(x+ 2) = 74 {to square a binomial, multiply it by itself}

x² + x² + 4x + 4 = 74 {used foil method}

2x² + 4x + 4 = 74 {combined like terms}

2x² + 4x - 70 = 0 {subtracted 74 from both sides}

2(x² + 2x - 35) = 0 {factored 2 out}

x² + 2x - 35 = 0 {set in parentheses equal to 0, 2 cannot be 0}

(x + 7)(x - 5) = 0 {factored into two binomials}

x + 7 = 0 or x - 5 = 0 {set each factor equal to 0}

x = -7 or x = 5 {solved each equation for x}

x + 2 = -5 or x + 2 = 7 {substituted -7 and 5, in for x, into x + 2}

x + 2 = 2nd odd integer

x² + (x + 2)² = 74 {sum of squares is 74}

x² + (x + 2)(x+ 2) = 74 {to square a binomial, multiply it by itself}

x² + x² + 4x + 4 = 74 {used foil method}

2x² + 4x + 4 = 74 {combined like terms}

2x² + 4x - 70 = 0 {subtracted 74 from both sides}

2(x² + 2x - 35) = 0 {factored 2 out}

x² + 2x - 35 = 0 {set in parentheses equal to 0, 2 cannot be 0}

(x + 7)(x - 5) = 0 {factored into two binomials}

x + 7 = 0 or x - 5 = 0 {set each factor equal to 0}

x = -7 or x = 5 {solved each equation for x}

x + 2 = -5 or x + 2 = 7 {substituted -7 and 5, in for x, into x + 2}

**-7 and - 5**also**5 and 7**are the two consecutive odd integers**© Algebra House**