SOLUTION: I need help with this word problem: A plane flies 720 miles against a steady 30mi/h head wind and then returns to the same point with the wind. If the entire trip takes 10h, wha

Question 88835: I need help with this word problem:
A plane flies 720 miles against a steady 30mi/h head wind and then returns to the same point with the wind. If the entire trip takes 10h, what is the planes speed in still air?
Thank you for the help
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the blank rate equation

Now plug in the given values. This is the rate equation with the wind.

Solve for t

Now plug in the given values. This is the rate equation against the wind.

Solve for t

Now combine the two "t" equations and set them equal to 10

Plug in and

Multiply both sides by

Distribute the left side

Foil the left side

Multiply the right side

Distribute and multiply

Distribute and multiply

Combine like terms on the left side

Subtract 1440x from both sides

Now let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=10, b=-1440, and c=-9000

Negate -1440 to get 1440

Square -1440 to get 2073600

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root

Multiply 2 and 10 to get 20

So now the expression breaks down into two parts

or

Lets look at the first part:

Add the terms in the numerator
Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator
Divide

So another answer is

So our solutions are:
or

Since a negative speed doesn't make any sense, our solution is

So the plane's speed in still air is 150 mph