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.