Question

Determine two consecutive even positive integers, the sum of whose squares is $100.$

Answer 1

Let the two consecutive even positive integers are $x$ and $(x+2)$

$\therefore (x{)}^2+(x+2{)}^2=100$ [$\because$ Sum of the squares of two number is $100$]

$\Rightarrow x^2+x^2+4x+4=100$

$\Rightarrow 2x^2+4x+4=100$

$\Rightarrow x^2+2x+2=50$

$\Rightarrow x^2+2x=48$

$\Rightarrow x^2+2x-48=0$

$\Rightarrow x^2+8x-6x-48=0$

$\Rightarrow x(x+8)-6(x+8)=0$

$\Rightarrow (x-6)(x+8)=0$

$\Rightarrow (x-6)=0or,(x+8)=0$

$\therefore x=6or,x=-8$ (rejected, as we are considering positive integers)

So, the consecutive even positive integers are $6$ and $8$