what two numbers have a sum and product of 6?

Answer 1

Let the numbers be x and y
product : xy = 6 or y = 6/x
sum : x + y = 6

x + 6/x = 6
x^2 - 6x + 6 = 0
x= [6 ± √(6^2 - 4*1*6) ] / 2
= [6 ± √(36-24)]/2 = (6±√12)/2
= (6 ± 2√3)/2
or
x = 3 +√3 or 3 - √3
if x = 3 + √3 then y = 3 - √3, since x+y=6
(and if x = 3 - √3 then y = 3 + √3, the same two numbers)

so 3 +√3 and 3 - √3 is the answer

checking,
sum: (3 +√3) + (3 - √3) = 6
product: (3 + √3)(3 - √3) = 9 +3√3 - 3√3 - 3 = 6

Answer 2

Let x and y by the numbers. Then

x + y = 6, and
xy = 6

Take the first equation, and note that y = 6 - x, so we substitute this into the second equation.

x(6-x) = 6
6x - x^2 = 6
Bring everything over to the right hand side, to get

x^2 - 6x + 6 = 0

To solve this, we have to either use the quadratic formula, or complete the square. I'm gonna do it the complete the square way (either of them work; one just involves excessive typing).

x^2 - 6x + 9 - 3 = 0
(x-3)^2 - 3 = 0
(x-3)^2 = 3
x - 3 = +/- sqrt(3)
x = 3 +/- sqrt(3)

Therefore, we have two solutions for one number, x.
x = 3 + sqrt(3) or x = 3 - sqrt(3).

Now, we solve for y, using the fact that x + y = 6.
If x = 3 + sqrt(3), then y = 6 - (3 + sqrt(3)) = 3 - sqrt(3)
If x = 3 - sqrt(3), then y = 6 - (3 - sqrt(3)) = 3 + sqrt(3)

Notice that the two answers merge into one because they represent "any two numbers" ("6 and 3" has the same meaning as "3 and 6" should those have been the solution).

x = 3+sqrt(3) , y = 3-sqrt(3) are the two numbers.

Answer 3

let the numbers be a and b

Then a + b = 6 so b = 6 - a
ab = 6

So a(6 - a) = 6
ie 6a - a² = 6
So a² - 6a + 6 = 0
Thus a = (6 ± √(36 - 24))/2
= 6/2 ± ½√(12)
= 3 ± √(3)

So one number is 3 + √(3)
and the other number is 3 - √(3)

So a + b = 3 + √(3) + 3 - √(3) = 6
ab = [3 + √(3)][3 - √(3)] = (3)² - (√(3))² = 9 - 3 = 6

Answer 4

3 & 2

Answer 5

x+y=6
xy=6
x(6-x)=6
6x-x^2-6=0
x^2-6x+6=0
x=(6+/-√(36-24))/2
x=3+/-(1/2)√12=3+/-√3
the numbers are 3+√3 & 3-√3

Answer 6

3 + SQRT(3)

3 - SQRT(3)