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.
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