Find the consecutive even integers whose squares have the sum 340.

Hint: In this question, we will be letting the two consecutives even integers as variables. Then form a quadratic equation with the given data. Further simplify it by splitting and grouping the common terms to get the required values.

Complete step-by-step answer:
Let us consider the two consecutives even integers as \[2x\] and \[2x + 2\] where \[x\] is a positive integer.
Given that these integers sum of squares is 34. So, we have
\[ \Rightarrow {\left( {2x} \right)^2} + {\left( {2x + 2} \right)^2} = {\left( {340} \right)^2}\]
We know that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. By using this formula, we have
$ \Rightarrow 4{x^2} + 4{x^2} + 4 + 8x = 340$
And hence on further simplification, we have
$
   \Rightarrow 8{x^2} + 8x + 400 - 4 = 0 \\
   \Rightarrow 8{x^2} + 8x - 336 = 0 \\
$
Now on taking 8 common from the whole equation, we have
$
   \Rightarrow 8\left( {{x^2} + x - 42} \right) = 0 \\
   \Rightarrow {x^2} + x - 42 = 0 \\
$
By splitting and grouping the common terms, we have
$
   \Rightarrow {x^2} + 7x - 6x - 42 = 0 \\
   \Rightarrow x\left( {x + 7} \right) - 6\left( {x + 7} \right) = 0 \\
   \Rightarrow \left( {x + 7} \right)\left( {x - 6} \right) = 0 \\
  \therefore x = 6, - 7 \\
$
Since, $x$ is a positive integer we have $x = 6$
Hence, the two consecutives even integers are $2x = 2 \times 6 = 12$ and $2x + 2 = 2 \times 6 + 2 = 14$.
Thus, the required values are 12, 14.

Note: Here the formed equation ${x^2} + x - 42 = 0$ is a quadratic equation. A quadratic equation is an equation in one variable whose degree is 2 and has two roots or zeros. A number which is exactly divisible by 2 without leaving any remainder is called an even number.