SOLUTION: Two airplanes leave simultaneously from an airport. One flies due south; the other flies due east at a rate 20 mph faster than the first airplane. After 1.5 hours, radar indicate

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Two airplanes leave simultaneously from an airport. One flies due south; the other flies due east at a rate 20 mph faster than the first airplane. After 1.5 hours, radar indicate      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 598103: Two airplanes leave simultaneously from an airport. One flies due south; the other flies due east at a rate 20 mph faster than the first airplane. After 1.5 hours, radar indicates that the airplanes are 450 miles apart. What is the ground speed of each airplane?
Please help - I am so lousy at word problems and stuggle with how to set up and solve. My instructor requires that we show all the work. So please help me with this homework problem step by step as I need to really understand how to pull out necessary information, then set up the equation to be solved. You are a lifesaver... Thank you so much!!!

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Two airplanes leave simultaneously from an airport. One flies due south; the other flies due east at a rate 20 mph faster than the first airplane. After 1.5 hours, radar indicates that the airplanes are 450 miles apart. What is the ground speed of each airplane?
.
Let x = ground speed of slower plane
then
x+20 = ground speed of faster plane
.
Because of the directions the two planes are flying, they form "right angle" (90 degrees). Therefore, you can use Pythagorean Theorem to solve:
applying the distance formula: d=rt
(1.5x)^2 + (1.5(x+20))^2 = 450^2
2.25x^2 + (1.5x+30)^2 = 202500
2.25x^2 + (1.5x+30)(1.5x+30) = 202500
2.25x^2 + 2.25x+90x+900 = 202500
4.5x^2+90x-201600 = 0
.5x^2+10x-22400 = 0
apply the quadratic formula to solve, which gives us:
x = {201.8962, -221.8962}
throw out the negative solution (extraneous) leaving
x = 202 mph (slower plane)
.
faster plane:
x+20 = 202+20 = 222 mph
.
details of quadratic follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 0.5x%5E2%2B10x%2B-22400+=+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=%2810%29%5E2-4%2A0.5%2A-22400=44900.

Discriminant d=44900 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-10%2B-sqrt%28+44900+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2810%29%2Bsqrt%28+44900+%29%29%2F2%5C0.5+=+201.896201004171
x%5B2%5D+=+%28-%2810%29-sqrt%28+44900+%29%29%2F2%5C0.5+=+-221.896201004171

Quadratic expression 0.5x%5E2%2B10x%2B-22400 can be factored:
0.5x%5E2%2B10x%2B-22400+=+0.5%28x-201.896201004171%29%2A%28x--221.896201004171%29
Again, the answer is: 201.896201004171, -221.896201004171. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+0.5%2Ax%5E2%2B10%2Ax%2B-22400+%29