I can't do word problems, please help me :)?

Question

The length of Jim's leader is always 2 ft. shorter than the length of his fishing rod. If the product of the length of the leader and the length of the rod equals strength of his line in lbs. and he uses 16 lb line, what is the length of his leader?

Answer 1

Let L = length of leader and F = length of fishing rod.
Since the leader is 2 ft shorter than the rod, L = F - 2.
Since the product of the length of the leader and the length of the rod is 16 (the strength of the line in lbs) you need to find the solution to the equation
F x L = F x (F - 2) = 16.
This is a quadratic equation so put it in standard form:
F^2 - 2F = 16 so F^2 - 2F - 16 = 0.

If you try to solve this by factoring, this is the same as coming up with two whole numbers which differ by 2 and multiply to 16, the terms of the problem. The factors of 16 are 4x4, 2x6, and 16x1, none of these differ by 2, so you can't solve by factoring and you will have to put this into the quadratic formula....
F = (-1 +/- sqrt ((-2)^2 - (4x 1 x -16)) / (2x1)
...= (-1 +/- sqrt (4 + 64))/2
...= (-1 +/- sqrt (68))/2
BTW, you CAN do word problems if you understand them. Here's a link to help you understand how to do them...
http://www.purplemath.com/modules/translat.htm
See also the other topics under the heading solving word problems
http://www.purplemath.com/modules/index.htm
see also
http://www.purplemath.com/modules/why_math.htm
PS -- Cattbarf, below, has a more elegant solution, involving completing the square. For info on how to do this, see...
http://www.purplemath.com/modules/sqrquad.htm

Answer 2

That's a fine kettle of fish. Let Y be the length of the leader. The length of the rod is Y+2.
"If the product,......" (Y)(Y+2)=16
Pushing stuff around Y^2+2Y-16= 0
Solving... (Y+1)^2 = 17 (this is a complete the square routine,which is easier to do than grinding out the quadratic with these numbers)
Y+1 = +/- sqrt 17
we dont need the negative root, so
Y= sqrt(17)-1
Then Y (which is what we want) is appx 3.12 ft.