solving applications using quadric formula

Question

 Using quadric formula, If Stan's Subaru travels 280 miles averaging a certain speed. If the car had gone 5 mph faster, the trip would have taken 1 hour less. Find Stan's average speed. I need to know how to do this using the quadric formula.

Answer 1

Richard's approach is valid and efficient, but doesn't employ the quadratic equation.

 

We can do so, however, building upon the good foundation he has laid. Take the formulas he showed:

 

1) v = 280/t

2) v+5 = 280/(t-1)

 

We are solving for v, so let's work t out of the picture:

 

1) t = 280/v [sub this into (2)]

2) v + 5 = 280 ((280/v) -1) [multiply both sides by ((280/v) -1)]

 

(v+5) x ((280/v) - 1) = 280

- v - 5 + 280 + (1400/v) = 280

- v + 1400/v = 5

-v² +1400 = 5v

v² + 5v - 1400 = 0

 

(We have kind of a hint at the answer here, because we can factor this down to (v + 40) x (v - 35) = 0. From this we know already that the solution is when v = -40 or v = 35, but we haven't used the dreaded QUADRATIC EQUATION!)

 

NOW we can bring the quadratic equation into play:

 

v = (-b +/- sqrt(b² - 4ac)) / 2a

v = (-5 +/- sqrt(25 - 4 x 1 x -1400)) / 2

v = (-5 +/- sqrt(5625))/2

v = (-5 +/- 75) / 2

v = -40, 35

 

I guess you'll need to figure out whether -40 or 35 is the right speed. :)

Answer 2

There are 2 unknowns that you need to solve for here; speed and time, let's name them v and t.

 

The question gives you 2 scenarios.

 

In the first scenario:

 

v=280/t, you get this from knowing speed=distance/time.

 

The second scenario is:

 

(v+5) = 280/(t-1),

 

Rearranged:

v=(280/(t-1)) - 5

 

Now you need to solve for either v or t first. I'd pick t. 

 

Here is the first step: 280/t=280/(t-1) - 5

 

You should be able to do it from here. If not, let me know.