Find three consecutives integers such that the sum of their squares is 77
2007-08-25 11:35:30 · 4 answers · asked by luicanna 1 in Science & Mathematics ➔ Mathematics
x² + (x + 1)² + (x + 2)² = 77
x² + x² + 2x + 1 + x² + 4x + 4 = 77
3x² + 6x - 72 = 0
x² + 2x - 24 = 0
(x + 6) (x - 4) = 0
x = - 6 , x = 4
Integers are 4 , 5 , 6 OR - 6 , - 5 , - 4
2007-08-25 11:49:38 · answer #1 · answered by Como 7 · 2⤊ 0⤋
(x-1)^2 + x^2 + (x+1)^2 = 77
x^2 - 2x + 1 + x^2 + x^2 + 2x + 1 = 77
3x^2 + 2 = 77
3x^2 = 75
x^2 = 25
x = 5 or -5
So the consecutive integers could be:
4, 5, 6 or -6, -5, -4
2007-08-25 18:39:22 · answer #2 · answered by Kemmy 6 · 0⤊ 0⤋
(a-1)^2 + a^2 + (a+1)^2 = 77
a^2 - 2a + 1 + a^2 + a^2 + 2a + 1 = 77
3a^2 = 75
a = 5
The integers are 4, 5 and 6
2007-08-25 18:41:21 · answer #3 · answered by ? 5 · 0⤊ 0⤋
x^2 + (x+1)^2 + (x+2)^2 = 77.
4,5,6
2007-08-25 18:38:49 · answer #4 · answered by Anonymous · 0⤊ 0⤋