The sum of two numbers is 6 and their product is 4. How do you find the larger of the two numbers?

Let the two numbers be `x` and `y`

We are told that
[1]`color(white)("XXXX")``x+y=6`
and
[2]`color(white)("XXXX")``xy = 4`

Rearranging [1] we have
[3]`color(white)("XXXX")``y = 6-x`

Substituting [3] into [2]
[4]`color(white)("XXXX")``x(6-x) = 4`

Which simplifies as
[5]`color(white)("XXXX")``x^2-6x+4 = 0`

Using the quadratic formula `x= (-b+-sqrt(b^2-4ac))/(2a)`

[6]`color(white)("XXXX")``x = (6+-sqrt(36-16))/2`

[7]`color(white)("XXXX")``x= 3+-sqrt(5)`

Since in [1] and [2] `x` and `y` are symmetric, they share the same solution possibilities.

The larger of these possibilities is `3+sqrt(5)`

It is possible to do this using one variable.
If two numbers add up to 6, they can be written as `x and (6 - x)`

Their product is 4 `rArr x(6-x) = 4`

`6x - x^2 = 4 " "rArr x^2 - 6x + 4 = 0" a quadratic"`

This does not factorise, but it is a good example for using completing the square because `a = 1 and "b is even"`

`x^2 - 6x + " " = -4 " + move the constant"`

`x^2 - 6x + "??? " = -4 " + ???"`
`x^2 - 6x + 9" " = -4 + 9" "`add `(b/2)^2 "to both sides"`
`(x - 3)^2 = 5`
`x - 3 = +-sqrt5`

`x = 3 + sqrt5 =5.236" " or x = 3 - sqrt5 = 0.764`

5.236 is the larger.