If the sum of two numbers is `4` and their product is `3`, then what is the sum of their squares?

The two numbers are `1` and `3`, being the two zeros of:

`x^2-4x+3 = (x-1)(x-3)`

In general we find:

`(x-a)(x-b) = x^2-(a+b)x+ab`

Notice that the coefficient of the middle term is `-(a+b)` and the constant term is `ab`.

Given that the two numbers are `1` and `3`, the sum of their squares is:

`1^2+3^2 = 1+9 = 10`

First, let's call the two numbers `m` and `n`.

We can then write:

`m + n = 4`
`mn = 3`

We can solve the first equation for `m`:

`m + n - color(red)(n) = 4 - color(red)(n)`

`m + 0 = 4 - n`

`m = 4 - n`

Next, we can substitute `4 - n` for `m` in the second equation and solve for `n`:

`(4 - n)n = 3`

`4n - n^2 - 3 = 0`

`n^2 - 4n + 3 = 0`

`(n - 1)(n - 3) = 0`

Solution 1)
`n - 1 = 0`

`n = 1`

Solution 2)
`n - 3 = 0`

`n = 3`

Substituting these back into the solution to the first equation gives:

Solution 1)

`m = 4 - 1`

`m = 3`

Solution 2)

`m = 4 - 3`

`m = 1`

The two numbers therefore are `3` and `1`.

The sum of their squares is therefore:

`3^2 + 1^2 = 9 + 1 = 10`

Calling the two numbers `x` and `y`, we are given:

`{ (x+y=4), (xy=3) :}`

and we find:

`x^2+y^2 = x^2+2xy+y^2-2xy`

`color(white)(x^2+y^2) = (x+y)^2-2xy`

`color(white)(x^2+y^2) = 4^2-2(3)`

`color(white)(x^2+y^2) = 16-6`

`color(white)(x^2+y^2) = 10`