SOLUTION: Missing numbers. Find two real numbers that have a sum of 6 and a product of 4.

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Question 181815This question is from textbook Algebra for college students
: Missing numbers. Find two real numbers that have a sum of 6 and a product of 4.This question is from textbook Algebra for college students

Found 2 solutions by user_dude2008, edjones:
Answer by user_dude2008(1825) (Show Source):
You can put this solution on YOUR website!
product of 4---> x*y=4 ---> y=4/x
sum of 6 ----> x+y=6 ---> x+4/x=6 ---> (x^2+4)/x=6

----> x^2-6x+4=0

Solve for x

Use quadratic formula to get

x=3+sqrt(5) or x=3-sqrt(5)

---> y=4/(3+sqrt(5)) ----> y=3-sqrt(5)

---> y=4/(3-sqrt(5)) ----> y=3+sqrt(5)

Answer: Two numbers are 3+sqrt(5) and 3-sqrt(5)

Answer by edjones(7311) (Show Source):
You can put this solution on YOUR website!
x+y=6
y=6-x
.
x*y=4
x(6-x)=4 substitute 6-x for y
6x-x^2=4
x^2-6x=-4 multiply each side by -1.
x^2-6x+4=0
x=sqrt(5)+3, x=-sqrt(5)+3
.
Ed