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*