Asked by juliet
find four consecutive integers such that the sum of the squares of the first two is 11 less than the square of the fourth
Answers
Answered by
Marth
Let x, x+1, x+2, x+3 be the four consecutive integers.
"the sum of the squares of the first two is 11 less than the square of the fourth"
x^2 + (x+1)^2 + 11 = (x+3)^2
x^2 + x^2 + 2x + 1 + 11 = x^2 + 6x + 9
2x^2 + 2x + 12 = x^2 + 6x + 9
x^2 - 4x + 3 = 0
(x-3)(x-1) = 0
x = 1, 3
So the four consecutive integers are 1, 2, 3, 4; or 3, 4, 5, 6.
"the sum of the squares of the first two is 11 less than the square of the fourth"
x^2 + (x+1)^2 + 11 = (x+3)^2
x^2 + x^2 + 2x + 1 + 11 = x^2 + 6x + 9
2x^2 + 2x + 12 = x^2 + 6x + 9
x^2 - 4x + 3 = 0
(x-3)(x-1) = 0
x = 1, 3
So the four consecutive integers are 1, 2, 3, 4; or 3, 4, 5, 6.