SOLUTION: At 10:00 AM, a snowstorm is 250 miles west of St. Louis. The storm travels 150 miles eastward at a constant rate, and then its speed decreases by 5 miles per hour. If the storm r

Question 170780: At 10:00 AM, a snowstorm is 250 miles west of St. Louis. The storm travels 150 miles eastward at a constant rate, and then its speed decreases by 5 miles per hour. If the storm reaches St. Louis at 8:00 PM, what was the storm's original speed?
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
distance equals rate times time
lets call the 1st 150 miles rate=r and time=t
and we will call the last 100 miles rate r-5 and time 10-t
:
150=rt--->t=150/r.....eq 1
100=(r-5)(10-t).......eq 2
:
take value of t from eq 1 and plug it into eq 2
:
100=(r-5)(10-(150/r))--->100=(r-5)((10r-150)/r)...multiply both sides by r
:
100r=(r-5)(10r-150)
:

divide by 10
:

r=27.25 and 2.75......throw out the 2.75 rate because r-5 is negative.
so mph-the original speed of the storm

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=600 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 27.2474487139159, 2.75255128608411. Here's your graph:

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