Question

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

Answer 1

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 :
\[a=2n=2\times2=4\]