SOLUTION: Help...spent two nights on these and cannot solve. A) During the first part of a trip, a canoeist travels 63 miles at a certain speed. The canoeist travels 10 miles on the second

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Help...spent two nights on these and cannot solve. A) During the first part of a trip, a canoeist travels 63 miles at a certain speed. The canoeist travels 10 miles on the second      Log On


   

Question 466674: Help...spent two nights on these and cannot solve.
A) During the first part of a trip, a canoeist travels 63 miles at a certain speed. The canoeist travels 10 miles on the second part of a trip at a speed of 5 mph slower. The total time for the trip is 4 hours. What was the speed on each part of the trip?
B) Write a quadratic equation having the given numbers as a solution; -6,-2.
C) Write a quadratic equation in the variable x having the given numbers as solutions. Type the equation in standard form ax^2+bx+c=0; -sqrt 5, 3 sqrt 5.
Thank you so much for your help!!!
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A) During the first part of a trip, a canoeist travels 63 miles at a certain speed. The canoeist travels 10 miles on the second part of a trip at a speed of 5 mph slower. The total time for the trip is 4 hours. What was the speed on each part of the trip?
r = the "certain speed"
t = d/r = 63/r + 10/(r-5) = 4
Multiply thru by r*(r-5)
63(r-5) + 10r = 4r(r-5) = 4r^2 - 20r
63r-315 + 10r = 4r^2 - 20r
4r%5E2+-+93r+%2B+315+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 4x%5E2%2B-93x%2B315+=+0) has the following solutons:

x%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=%28-93%29%5E2-4%2A4%2A315=3609.

Discriminant d=3609 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--93%2B-sqrt%28+3609+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-93%29%2Bsqrt%28+3609+%29%29%2F2%5C4+=+19.1343691479378
x%5B2%5D+=+%28-%28-93%29-sqrt%28+3609+%29%29%2F2%5C4+=+4.11563085206221

Quadratic expression 4x%5E2%2B-93x%2B315 can be factored:
4x%5E2%2B-93x%2B315+=+%28x-19.1343691479378%29%2A%28x-4.11563085206221%29
Again, the answer is: 19.1343691479378, 4.11563085206221. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B-93%2Ax%2B315+%29

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r = x
Ignore the answer that's less than 5.
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B) Write a quadratic equation having the given numbers as a solution; -6,-2.
(x+6)*(x+2) = 0
x%5E2+%2B+8x+%2B+16+=+0
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C) Write a quadratic equation in the variable x having the given numbers as solutions. Type the equation in standard form ax^2+bx+c=0; -sqrt 5, 3 sqrt 5.
%28x+%2B+sqrt%285%29%29%2A%28x+-+3sqrt%285%29%29+=+0
x%5E2+-+2x%2Asqrt%285%29+-+15+=+0