Question

Given the sum of two numbers equal 16 and the sum of their squares equal 160 find the numbers

Answer 1

We have

$\begin{cases}x+y=16\\x^2+y^2=160\end{cases}$

From the first equation, we deduce $y=16-x$

Plug this in the second equation to get

$x^2+(16-x)^2=160 \iff x^2+(256-32x+x^2)=160 \iff 2x^2-32x+96=0$

We can divide both sides by 2 to get

$x^2-16x+48=0$

Use the quadratic formula to solve this equation, and you'll have the solutions 4 and 12.

So, if x=4, y must be 12, so that their sum is 16.

Similarly, if x=12, y must be 4.

The two numbers are 4 and 12.