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:

$\sqrt{x^2}=\sqrt{4}\\\\\\ |x|=2\\\\\\ x=\pm~2$

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.