College Algebra, Chapter 1, 1.3, Section 1.3, Problem 80

The sum of the squares of two consecutive even integers is $1252$. Find the integers.

$
\begin{array}{|c|c|}
\hline\\
\text{Words} & \text{Algebra} \\
\text{First even number} & x \\
\text{Second even number} & x + 2 \\
\text{Square of the first number} & x^2 \\
\text{Square of the second number} & (x + 2)^2 \\
\text{Sum of the square of the numbers} & x^2 + (x + 2)^2\\
\hline
\end{array}
$

From the table,


$
\begin{equation}
\begin{aligned}

x^2 (x + 2)^2 =& 1252
&& \text{Model}
\\
\\
x^2 + x^2 + 4x + 4 =& 1252
&& \text{Expand}
\\
\\
2x^2 + 4x =& 1248
&& \text{Combine like terms}
\\
\\
x^2 + 2x =& 624
&& \text{Divide both sides by 2}
\\
\\
x^2 + 2x - 624 =& 0
&& \text{Subtract 624}
\\
\\
(x + 26)(x - 24) =& 0
&& \text{Factor}
\\
\\
x + 26 =& 0 \text{ and } x - 24 = 0
&& \text{Zero Product Property}
\\
\\
x =& -26 \text{ and } x = 24
&& \text{Solve for } x
\\
\\
x =& 24
&& \text{Choose } x > 0

\end{aligned}
\end{equation}
$


Thus, the even numbers are 24 and 26.