SOLUTION: I need help with this word problem I have tried and so has a tutor at school. Malikka takes a paddleboat 1km upriver against a 4km/hr current then returns downriver with the curre

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Question 170286: I need help with this word problem I have tried and so has a tutor at school.
Malikka takes a paddleboat 1km upriver against a 4km/hr current then returns downriver with the current, back to starting point. Her whole trip takes 2 hours, write a quadratic equation describing this situation. The unknown is the speed, v, that Mallika would paddle with no current present. Solve. What speed (in km/hr) would Mallika paddle by herself with no current present?
Found 2 solutions by stanbon, nerdybill:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Malikka takes a paddleboat 1km upriver against a 4km/hr current then returns downriver with the current, back to starting point. Her whole trip takes 2 hours, write a quadratic equation describing this situation. The unknown is the speed, v, that Mallika would paddle with no current present. Solve. What speed (in km/hr) would Mallika paddle by herself with no current present?
---------------------------
Upriver DATA:
Distance = 1 km ; rate = v-4 km/h ; time = d/r = 1/(v-4) hr.
-----------------------------
Downriver DATA:
Distance = 1 km ; rate = v+4 km/h ; time = d/r = 1/(v+4) hr.
------------------------------
EQUATION:
time up + time down = 2 hrs
1/(v-4) + 1/(v+4) = 2
v+4 + v-4 = 2(v^2-16)
v = v^2-16
v^2 - v - 16 = 0
v = [1 +- sqrt(1 - 4*-16)]/2
v = [1 +- sqrt(65)]/2
Positive solution:
v = 4.53 km/h
======================================
Cheers,
Stan H.

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Let v = speed of paddleboat w/no current
.
Applying "distance formula":
d = rt
where
d is distance
r is rate or speed
t is time
.
Let t = time traveling upriver
then
2-t = time traveling downstream
.
We have two unknowns (v and t) so we'll need two equations:
t(v-4) = 1 (equation 1)
(2-t)(v+5) = 1 (equation 2)
.
Solving equation 1 for 't':
t(v-4) = 1
t = 1/(v-4)
.
Substitute the above into equation 2 and solve for v:
(2-(1/(v-4)))(v+5) = 1
multiply both sides by (v-4):
(2(v-4) - 1)(v+5) = 1
(2v-8 - 1)(v+5) = 1
(2v-9)(v+5) = 1
Applying FOIL:
2v^2 + 10v - 9v -45 = 1
2v^2 + v -45 = 1
2v^2 + v -46 = 0
.
Can't factor so use the quadratic formula. Doing so will yield:
v = {4.552, -5.052}
.
We can toss out the negative solution leaving us with:
v = 4.552 km/hr
.
Details of quadratic below:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation av%5E2%2Bbv%2Bc=0 (in our case 2v%5E2%2B1v%2B-46+=+0) has the following solutons:

v%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A2%2A-46=369.

Discriminant d=369 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+369+%29%29%2F2%5Ca.

v%5B1%5D+=+%28-%281%29%2Bsqrt%28+369+%29%29%2F2%5C2+=+4.55234317807464
v%5B2%5D+=+%28-%281%29-sqrt%28+369+%29%29%2F2%5C2+=+-5.05234317807464

Quadratic expression 2v%5E2%2B1v%2B-46 can be factored:
2v%5E2%2B1v%2B-46+=+2%28v-4.55234317807464%29%2A%28v--5.05234317807464%29
Again, the answer is: 4.55234317807464, -5.05234317807464. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B1%2Ax%2B-46+%29