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