Question

The sum of the squares of two numbers is 8 . The product of the two numbers is 4 . Find the numbers.

Answer 1

Hello there.

First, assume the numbers
`x,~y` such that they satisties both affirmations:

• The sum of the squares of two numbers is
  `8`.
• The product of the two numbers is
  `4`.

With these informations, we can set the following equations:

`\begin{center}\align x^2+y^2=8\\ x\cdot y=4\\\end{center}`

Multiply both sides of the second equation by a factor of
`2`:

`2\cdot x\cdot y = 2\cdot 4\\\\\\ 2xy=8~~~~~(2)^(\ast)`

Make
`(1)-(2)^(\ast)`

`x^2+y^2-2xy=8-8\\\\\\ x^2-2xy+y^2=0`

We can rewrite the expression on the left hand side using the binomial expansion in reverse:
`(a-b)^2=a^2-2ab+b^2`, such that:

`(x-y)^2=0`

The square of a number is equal to
`0` if and only if such number is equal to
`0`, thus:

`x-y=0\\\\\\ x=y~~~~~~(3)`

Substituting that information from
`(3)` in
`(2)`, we get:

`x\cdot x = 4\\\\\\ x^2=4`

Calculate the square root on both sides of the equation:

`√(x^2)=√(4)\\\\\\ |x|=2\\\\\\ x=\pm~2`

Once again with the information in
`(3)`, we have that:

`y=\pm~2`

The set of solutions of that satisfies both affirmations is:

`S=\{(x,~y)\in\mathbb{R}^2~|~(x,~y)=(-2,\,-2),~(2,~2)\}`

This is the set we were looking for.