Question 181288
<font face="Times New Roman" size="+2">


Consider the basic formula that relates distance, rate (speed), and time:


*[tex \LARGE \text {          }\math d = rt]


and its variations:


*[tex \LARGE \text {          }\math r = \frac {d}{t}]


and 


*[tex \LARGE \text {          }\math t = \frac {d}{r}]


It will be convenient to use the last one because that will ultimately allow us to solve for the desired quantity, <i>r</i>, directly.


The actual trip can be described thus:


*[tex \LARGE \text {          }\math t = \frac {150}{r}]


And the 'what if' (20 mph faster, 2 hours less) trip can be described:


*[tex \LARGE \text {          }\math t - 2 = \frac {150}{r + 20} \ \ \Rightarrow \ \ t = \left(\frac{150}{r + 20}\right) + 2 \ \ \Rightarrow \ \ t = \left(\frac{150}{r + 20}\right) + \left(\frac {2(r + 20)}{r + 20}\right) \ \ \Rightarrow \ \ t = \frac {2r + 190}{r + 20} ]


Now we have expressed the variable <i>t</i> in two different ways in terms of <i>r</i>, the quantity we seek.  So set these two expressions in <i>r</i> equal to one another.


*[tex \LARGE \text {          }\math \frac{150}{r} = \frac {2r + 190}{r + 20} ]


This is a simple proportion that can be solved by first cross-multiplying and simplifying:


*[tex \LARGE \text {          }\math 150(r + 20) = 2r^2 + 190r \ \ \Rightarrow \ \ -2r^2 -40r + 3000 = 0 \ \ \Rightarrow \ \ r^2 + 20r - 1500 = 0]


Leaving us with a factorable quadratic:


Since *[tex \Large -30 + 50 = 20] and *[tex \Large -30 \times 50 = -1500], we can say:


*[tex \LARGE \text {          }\math r^2 + 20r - 1500 = 0 \ \ \Rightarrow \ \ (r - 30)(r + 50) = 0]


Hence,


*[tex \LARGE \text {          }\math  r = 30]


or


*[tex \LARGE \text {          }\math  r = -50]


But -50 for a speed is absurd and is therefore an extraneous root introduced by the action of squaring the variable during the solution process.  Exclude -50.


That leaves 


*[tex \LARGE \text {          }\math  r = 30] as the answer.


Check:


*[tex \LARGE \text {          }\math  r = 30 \ \ \Rightarrow \ \ t = \frac {150}{30} = 5]


If <i>r</i> were 20 mph faster, then *[tex \LARGE r = 30 + 20 = 50 ] and if <i>t</i> were 2 hours less, then *[tex \LARGE t = 5 - 2 = 3].  If the solution is correct, then a trip lasting 3 hours at 50 miles per hour should cover the same 150 mile distance.


*[tex \LARGE \text {          }\math d = rt = 50 \times 3 = 150]


Answer checks.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>