The sum of the squares of two consecutive positive even numbers is 52.
What is the smaller number? @tcarroll010

Question

The sum of the squares of two consecutive positive even numbers is 52.
What is the smaller number?  @tcarroll010

anonymous · Answer

Let be a and b the two consecutive positive even numbers.
So :
$$a=2n~~~ b=a+2=2n+2$$
And then :
$$a^2+b^2=52\$$
Then :
$$4n^2+(2n+2)^2=52$$
So :
$$4n^2+4n^2+8n+4=52$$
So :
$$8n^2+8n-48=0$$
We divide by 8 :
$$n^2+n-6=0$$
SO :
$$\Delta=1-4\times(-6)+25$$
So :
$$n=\frac{-1+\sqrt{25}}{2}=2$$
So the smallest number is :
$$a=2n=2\times2=4$$