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The sum of the squares of two consecutive even integers is 136. Find the integers.
The sum of the squares of two consecutive even integers is 136. Find the integers.
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Let the two consecutive even integers be $x$ and $x+2$. According to the problem, $(x)^2+(x+2)^2=136$. Expanding and simplifying the terms give $x^2+x^2+4x+4=136$. Rearranging the terms gives $2x^2+4x-132=0$. Dividing throughout by 2 simplifies the equation to $x^2+2x-66=0$. This equation can be factored as $(x+11)(x-6)=0$. Setting the two factors equal to zero gives $x+11=0$ or $x-6=0$. Solving these two equations respectively gives $x=-11$ or $x=6$. Checking these two values shows that when $x=6$, the two consecutive even integers become $x$ and $x+2$, or 6 and 8 respectively. Thus, the two integers are $\boxed{6\text{ and }8}$.
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