Asked by Tom
                There are three consecutive integers the square of the largest one equals the sum of the squares of the two other.Find the integers
            
            
        Answers
                    Answered by
            Kathleen
            
    The numbers are represented by x, x+1 and X+2.
X^2 + (X+1)^2=(X+2)^2
X^2 + X^2 +2X +1=X^2 +4X + 4
2X^2+2X+1=X^2+4X+4
X^2-2X-3=0
(X-3)(X+1)=0
X=3 X=-1
The numbers are X=3, X+1=4 X+2=5
Check: 3^2 + 4^2=5^2
9+16=25
A second solution is X=-1 X=0 X=1
-1^2 + 0^2=1^2
1+0=1
    
X^2 + (X+1)^2=(X+2)^2
X^2 + X^2 +2X +1=X^2 +4X + 4
2X^2+2X+1=X^2+4X+4
X^2-2X-3=0
(X-3)(X+1)=0
X=3 X=-1
The numbers are X=3, X+1=4 X+2=5
Check: 3^2 + 4^2=5^2
9+16=25
A second solution is X=-1 X=0 X=1
-1^2 + 0^2=1^2
1+0=1
                    Answered by
            Reiny
            
    Three consecutive integers (x-1), x, (x+1)
(could have said x,x+1,x+2 or x+10,x+11,x+12 etc)
"the square of the largest one equals the sum of the squares of the two other"
----> (x+1)^2 = x^2 + (x-1)^2
x^2 + 2x + 1 = x^2 + x^2 - 2x + 1
0 = x^2 - 4x
x(x-4) = 0
x = 0 or x = 4
Case 1: x = 0
The integers are -1, 0, and 1
Case 2: x = 4
The integers are 3,4,and 5
Both solutions work, check them out
    
(could have said x,x+1,x+2 or x+10,x+11,x+12 etc)
"the square of the largest one equals the sum of the squares of the two other"
----> (x+1)^2 = x^2 + (x-1)^2
x^2 + 2x + 1 = x^2 + x^2 - 2x + 1
0 = x^2 - 4x
x(x-4) = 0
x = 0 or x = 4
Case 1: x = 0
The integers are -1, 0, and 1
Case 2: x = 4
The integers are 3,4,and 5
Both solutions work, check them out
                    Answered by
            Tom
            
    thank so much
    
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