Question

Find two consecutive even positive integers sum of whose squares is 290

Answer 1

Answer:

Let the integers be x and x + 1

Step-by-step explanation:

${x }^{2} + {(x + 1)}^{2} = 290 \\ {x}^{2} + x {}^{2} + 2x + 1 = 290 \\ 2 {x}^{2} + 2x - 289 = 0$