# Does not compute!

• Jul 16th 2007, 06:03 AM
earachefl
Does not compute!
A word problem is given which I've answered correctly, but a second question doesn't add up. The problem is:

A North-South highway intersects an East-West highway at a point P. An automobile crosses P at 10:00 AM, traveling east at a constant rate of 20 mph. At that same instant another automobile is two miles north of P, traveling south at 50 mph. Find a formula which expresses the distance d between the automobiles at time t (hours) after 10:00 AM. At what time will the automobiles be 104 miles apart?

My answer, which the book confirms, is

$\displaystyle d = \sqrt{2900t^2 - 100t + 4}$

To answer the second question, I started with

$\displaystyle 104 = \sqrt{2900t^2 - 100t + 4}$

$\displaystyle \Rightarrow 10816 = 2900t^2 -100t + 4$

$\displaystyle \Rightarrow 2704 = 725t^2 - 25t + 1$

$\displaystyle \Rightarrow 725t^2 - 25t - 2703 = 0$

Using the Quadratic Formula, I get

$\displaystyle \frac {25 \pm \sqrt {625 + 7838700}} {1450}$

which I can only reduce to

$\displaystyle \frac {5 \pm \sqrt {313573}} {290}$

But the answer given in the book is:

$\displaystyle \frac {1 + \sqrt 30} {29}$

So how do I get there? Did I make a mistake along the way?
• Jul 16th 2007, 10:47 AM
Soroban
Hello,earachefl!

Quote:

A North-South highway intersects an East-West highway at a point $\displaystyle P$.
An automobile crosses $\displaystyle P$ at 10:00 AM, traveling east at a constant rate of 20 mph.
At that same instant another automobile is two miles north of $\displaystyle P$, traveling south at 50 mph.
(a) Find a formula which expresses the distance $\displaystyle d$ between the automobiles
. . .at time $\displaystyle t$ (hours) after 10:00 AM.
(b) At what time will the automobiles be 104 miles apart?

I don't agree with either of their answers . . .
Code:

          Q *             |         50t |             |           B *             |  *  d       2-50t |    *             |        *       - - P * - - - - - * - -             |    20t    A             |

At 10 AM, the first car is at point $\displaystyle P$.
. . $\displaystyle t$ hours later, it is at point $\displaystyle A\!:\;PA = 20t$

At 10 AM, the second car is at point $\displaystyle Q\!:\;QP = 2$.
. . $\displaystyle t$ hours later, it is at point $\displaystyle B\!:\;QB = 50t$.
. . Hence: .$\displaystyle BP = 2-50t$

Pythagorus says: .$\displaystyle d^2\:=\:(20t)^2 + (2-50t)^2$

(a) Then we have: .$\displaystyle \boxed{d \:=\:\sqrt{2900t^2 - {\color{red}200}t + 4}}$

If $\displaystyle d = 104\!:\;\;\sqrt{2900t^2 - 200t + 4} \:=\:104$

Square both sides: .$\displaystyle 2900t^2 - 200t + 4 \:=\:10816\quad\Rightarrow\quad 2900t^2 - 200t - 10812 \:=\:0$

Divide by 4: .$\displaystyle 725t^2 - 50t - 2703 \:=\:0$

Quadratic Formula: .$\displaystyle t \;=\;\frac{-(-50) \pm\sqrt{(-50)^2 - 4(725)(-2703)}}{2(725)}$

. . $\displaystyle t \;=\;\frac{50\pm\sqrt{7841200}}{1450} \;=\;\frac{50 \pm 20\sqrt{19603}}{1450} \;=\;\frac{5 \pm 2\sqrt{19603}}{145}$

And the positive root is: .$\displaystyle \boxed{t \:=\:\frac{5 + 2\sqrt{19603}}{145} \:\approx\:1.966\text{ hours.}}$