Given:
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.
Value of x:
From the given, the expression can be written as,
[tex]x^2+(x+2)^2=2+(x+4)[/tex]
Simplifying, we get;
[tex]x^2+x^2+4x+4=2+x+4[/tex]
[tex]2x^2+4x+4=x+6[/tex]
[tex]2x^2+3x+4=6[/tex]
[tex]2x^2+3x-2=0[/tex]
Factoring the middle term, we get;
[tex]2x^2+4x-x-2=0[/tex]
[tex]2x(x+2)-1(x+2)=0[/tex]
[tex](2x-1)(x+2)=0[/tex]
[tex]2x-1=0 \ or \ x+2=0[/tex]
[tex]x=\frac{1}{2} \ or \ x=-2[/tex]
Since, the fraction [tex]x=\frac{1}{2}[/tex] is neither a even nor odd integer.
Thus, the value of x is -2.
Consecutive integers:
Substituting the value of x in the consecutive integers x, x + 2 and x + 4, we get;
[tex]x=-2[/tex]
[tex]x+2=-2+2=0[/tex]
[tex]x+4=-2+4=2[/tex]
Thus, the 3 consecutive integers are -2, 0 and 2.
