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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 $14$.
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