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Question 151132: The lenght of Jim's leader is always 2 feet shorter than lenght of this fishing rod.If the product of the lenght of the leader and lenght of the rod equals the strenght of his line in pounds and he uses a 16 lb line,what is the lenght of his leader?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The lenght of Jim's leader is always 2 feet shorter than lenght of this fishing rod.If the product of the lenght of the leader and lenght of the rod equals the strenght of his line in pounds and he uses a 16 lb line,what is the lenght of his leader?
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Let the length of the leader be "x".
Let the length of his rod be "r".
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EQUATIONS:
x = r-2
x*r = 16
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Substitute to solve for "r":
(r-2)r = 16
r^2-2r-16 = 0
r = [2 +- sqrt(4-4*1*-16)]/2
r = [2 +- sqrt(4+64)]/2
r = [2 +- sqrt(4*17)]/2
r = [2 +- 2sqrt(17)]/2
Positive answer:
r = [2+2sqrt(17)]/2 = 5.123 ft is the rod length
x = r-2 so leader length = 3.123 ft
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Cheers,
Stan H.
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