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