Find two consecutive odd integers such that the sum of their squares is 290.
Let the first odd integer be n.
The next odd integer is n + 2
Square them both:
n^2
(n + 2)^2 = n^2 + 4n + 4 from our expansion calculator
The sum of the squares equals 290
n^2 + n^2 + 4n + 4 = 290
Group like terms:
2n^2 + 4n + 4 = 290
Enter this quadratic into our search engine, and we get:
n = 11, n = -13
Which means the two consecutive odd integer are:
11 and 11 + 2 = 13. (11, 13)
-13 and -13 + 2 = -11 (-13, -11)
Let the first odd integer be n.
The next odd integer is n + 2
Square them both:
n^2
(n + 2)^2 = n^2 + 4n + 4 from our expansion calculator
The sum of the squares equals 290
n^2 + n^2 + 4n + 4 = 290
Group like terms:
2n^2 + 4n + 4 = 290
Enter this quadratic into our search engine, and we get:
n = 11, n = -13
Which means the two consecutive odd integer are:
11 and 11 + 2 = 13. (11, 13)
-13 and -13 + 2 = -11 (-13, -11)