<u>Given</u>:
Let the 3 consecutive even integers be x, x + 2 and x + 4.
Given that the sum of the squares of the first and the second integers is 2 more than the third integer.
We need to determine the 3 consecutive integers.
<u>Value of x:</u>
From the given, the expression can be written as,
![x^2+(x+2)^2=2+(x+4)](https://tex.z-dn.net/?f=x%5E2%2B%28x%2B2%29%5E2%3D2%2B%28x%2B4%29)
Simplifying, we get;
![x^2+x^2+4x+4=2+x+4](https://tex.z-dn.net/?f=x%5E2%2Bx%5E2%2B4x%2B4%3D2%2Bx%2B4)
![2x^2+4x+4=x+6](https://tex.z-dn.net/?f=2x%5E2%2B4x%2B4%3Dx%2B6)
![2x^2+3x+4=6](https://tex.z-dn.net/?f=2x%5E2%2B3x%2B4%3D6)
![2x^2+3x-2=0](https://tex.z-dn.net/?f=2x%5E2%2B3x-2%3D0)
Factoring the middle term, we get;
![2x^2+4x-x-2=0](https://tex.z-dn.net/?f=2x%5E2%2B4x-x-2%3D0)
![2x(x+2)-1(x+2)=0](https://tex.z-dn.net/?f=2x%28x%2B2%29-1%28x%2B2%29%3D0)
![(2x-1)(x+2)=0](https://tex.z-dn.net/?f=%282x-1%29%28x%2B2%29%3D0)
![2x-1=0 \ or \ x+2=0](https://tex.z-dn.net/?f=2x-1%3D0%20%5C%20or%20%5C%20x%2B2%3D0)
![x=\frac{1}{2} \ or \ x=-2](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B2%7D%20%5C%20or%20%5C%20x%3D-2)
Since, the fraction
is neither a even nor odd integer.
Thus, the value of x is -2.
<u>Consecutive integers:</u>
Substituting the value of x in the consecutive integers x, x + 2 and x + 4, we get;
![x=-2](https://tex.z-dn.net/?f=x%3D-2)
![x+2=-2+2=0](https://tex.z-dn.net/?f=x%2B2%3D-2%2B2%3D0)
![x+4=-2+4=2](https://tex.z-dn.net/?f=x%2B4%3D-2%2B4%3D2)
Thus, the 3 consecutive integers are -2, 0 and 2.