Question 944234: Find two positive numbers that differ by eight and whose reciprocals sum is 1/6 Found 2 solutions by macston, MathLover1:Answer by macston(5194) (Show Source):
You can put this solution on YOUR website! X=one positive number, Y is the other positive number ;
X=Y+8 Add Y to each side of Equation 1 and substitute in Equation 2 Multiply each side by 6 Subtract 6/Y from each side convert right side to fraction Multiply each side by Y+8 Multiply each side by Y Multiply right side through Subtract 6Y from each side
Quadratic equation (in our case ) has the following solutons:
For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant : .
Discriminant d=208 is greater than zero. That means that there are two solutions: .
Quadratic expression can be factored:
Again, the answer is: 9.21110255092798, -5.21110255092798.
Here's your graph:
The positive answer is 9.2111 so ANSWER 1 Y=9.2111
X-Y=8
X-9.2111=8 Add 9.2111 to each side
X=17.2111 ANSWER 2 X=17.2111
FINAL ANSWER THE TWO POSITIVE NUMBERS ARE 17.2111 and 9.2111
CHECK
1/X+1/Y=1/6
1/17.2111+1/9.2111=1/6
0.16667=1/6
0.16667=0.16667
You can put this solution on YOUR website!
let two positive numbers be and
if they differ by eight, we have
=> ......eq.1
and if their reciprocals sum is , we have
.
.....cross multiply
.....eq.2 ....substitute from eq.1
.....use quadratic formula
y = 2-2 sqrt(13)=-5.2 we need positive number only
find :
=>