There are three consecutive integers the square of the largest one equals the sum of the squares of the two other.Find the integers
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
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
thank so much
To solve this problem, let's first represent the three consecutive integers as x, x+1, and x+2.
According to the problem, the square of the largest integer (x+2) equals the sum of the squares of the other two integers (x^2 + (x+1)^2).
So, we can write the equation as:
(x + 2)^2 = x^2 + (x + 1)^2
Now let's solve this equation step by step:
Expanding both sides of the equation:
x^2 + 4x + 4 = x^2 + x^2 + 2x + 1
Combine like terms:
x^2 + 4x + 4 = 2x^2 + 2x + 1
Move all terms to one side of the equation:
0 = 2x^2 - x^2 + 2x - 4x + 1 - 4
Simplify:
0 = x^2 - 2x - 3
Now, let's factor the quadratic equation:
0 = (x - 3)(x + 1)
Setting each factor equal to zero and solving for x:
x - 3 = 0 or x + 1 = 0
If x - 3 = 0, then x = 3.
If x + 1 = 0, then x = -1.
So, the two possible values for x are 3 and -1.
If x = 3, the three consecutive integers would be 3, 4, and 5.
If x = -1, the three consecutive integers would be -1, 0, and 1.
Therefore, the solution to the problem is either the set of integers {3, 4, 5} or the set of integers {-1, 0, 1}.