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

Given:

The sum of the squares of two consecutive even integers is 340. 

 To do:

We have to find the numbers.

Solution:

Let the two consecutive even integers be $2x$ and $2x+2$.

According to the question,

$(2x)^2+(2x+2)^2=340$

$4x^2+4x^2+8x+4=340$

$8x^2+8x+4-340=0$

$8x^2+8x-336=0$

$8(x^2+x-42)=0$

$x^2+x-42=0$

Solving for $x$ by factorization method, we get,

$x^2+7x-6x-42=0$

$x(x+7)-6(x+7)=0$

$(x+7)(x-6)=0$

$x+7=0$ or $x-6=0$

$x=-7$ or $x=6$

Considering positive values of $x$, we get,

$x=6$, then $2x=2(6)=12$ and $2x+2=2(6)+2=12+2=14$

The required consecutive even integers are $12$ and $14$.