Question 372443: Two positive numbers have a difference of 2. Their reciprocals have a sum of 20/21. What are the numbers? Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! let the numbers be x
the other number will be x+2
1/x + 1/(x+2) = 20/21
((x+2)+x)/x(x+2)= 20/21
(2x+2)/(x(x+2)= 20/21
21(2x+2)= 20x(x+2)
42x+42=20x^2+40x
20x^2-2x-42=0
/2
10x^2-x-21=0
10x^2-15x+14x-21=0
10x(x-1.5)+14(x-1.5)=0
(10x+14)(x-1.5)=0
10(x+1.4)(x-1.5)=0
(x+1.4)(x-1.5)=0
x =-1.4
OR
x=1.5
..
1.5 & 3.5
OR
-1.4 & 0.6
...
m.ananth@hotmail.ca
