Alg3.2.20

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

Let the first even number be $x\,$

The consecutive number is $x+2\,$

The sum of the squares is $x^2+(x+2)^2\,$

Given $x^2+x^2+4x+4=100,2x^2+4x-96=0,x^2+2x-48=0\,$

Solving the equation

$x^2+8x-6x-48=0,x(x+8)-6(x+8)=0,(x+8)(x-6)=0\,$

Hence the values of x are

$x=-8,6\,$

Therefore the number is $6\,$ and the other number is $8\,$