# Math Help - Problem Solving with Quadratics

1. ## Problem Solving with Quadratics

Hi guys,
I need some help with a question that i just cant figure out how to solve. I have tried a lot but it does not seem to help. I am looking around and there is some complex math so i am guessing this might be a no brain-er for you guys.
Anyways the question is,

Two Trains travel a 105 km track each day. The express travels 10kmph faster and takes 30 minutes less than the normal train does. Find the speed of the express.
The answer is : 51.1 km/h ( the answer book tells me that)

I need to know how to solve to get to that answer. I cant find a way.
Thanks in advance, please get to me as soon as you guys can.
Thanks.
- Cluelessgeorge

2. Use $d = vt$.

So: Let $v$ and $t$ be the speed and time of the original train.

Then $(v+10)(t-0.5) = 105$ and $vt = 105$. So $v = \frac{105}{t}$. Substitute this into the equation to solve for $v,t$.

3. Hey Turkeywilliams,
Thanks for the suggestion, I tried it out and it did not seem to generate the answer i needed.
Look below at my solution and see if anything else comes to your mind. Or anybody else, please consider looking and see if you can give me a solution.

SOLUTION :

d= vt
Let v and t represent speed and time.

(1) (v + 10)(t - 0.5) = 105
(2) vt = 105 ...v = 105/t

Sub Equation 2 into 1

(105/t + 10)(t - 0.5) = 105
105 - 55/t + 10t - 5 = 105
105t - 5 +10t^2 - 5t = 105t
10t^2 - 5t- 55 = 0
5 ( 2t^2 - t - 11) = 0

2t^2 - t - 11= 0

Using a graphing calculator...
t = -2.108 OR t = 2.608
As this is a real life situation, negative roots are eliminated as there is no negative time. Hence the only solution for time is 2.608.

Substitute t = 2.608 into Equation 1

(v + 10)(2.608 - 0.5) = 105
(v + 10)(2.108) = 105
v + 10 = 49.81
v = 39.81

This is my solution and according to my book it is wrong. Because my book says that the answer to this question should be 51.1 km/h.

So please reconsider your original thought process and see what you can figure out.
Thanks guys,
I really appreciate it.

- cluelessgeorge

4. You are trying to find $v+10$ or the speed of the express train. You solved for the speed of the original train. Also your $t$ is a bit off. Adding $10$ to your $v$ we get $49.81$ which is close to $51.1$.

I get $t = 2.554$. Plug this into the equation and solve for $v+10$. You should get $51.1 \ \frac{\text{km}}{\text{h}}$.

5. ## Thanks

Thank you very much
I did not consider that i was looking for V +10.
and I am a little surprised that you got different values for x than me.
I will trust you but i was wondering, why would that happen?
Thanks again for the help.

- cluelessgeorge

6. I used Maple to get the values. Did you just graph the equation and 'guess' what the value was?

7. I dont know what maple is but I didn't guess the values.
I have this computer graphing program which graphs and tells me the intercepts.
It could have been wrong, I am guessing i should use a graphing calculator, a proper one for accuracy.

8. Originally Posted by tukeywilliams
Use $d = vt$.

So: Let $v$ and $t$ be the speed and time of the original train.

Then $(v+10)(t-0.5) = 105$ and $vt = 105$. So $v = \frac{105}{t}$. Substitute this into the equation to solve for $v,t$.
Why use a graphing calculator for an estimate?
$\left ( \frac{105}{t} + 10 \right ) \left ( t - \frac{1}{2} \right ) = 105$<-- Multiply both sides by 2t

$(105 + 10 t) ( 2t - 1) = 210t$

$210t - 105 + 20t^2 - 10t = 210t$

$20t^2 - 10t - 105 = 0$

$4t^2 - 2t - 21 = 0$ <-- Use the quadratic formula

$t = \frac{1 \pm \sqrt{85}}{4}$

Use the "=" solution, as the "-" solution gives a negative time.
$t = \frac{1 + \sqrt{85}}{4}$

So what's v + 10?
$v = \frac{105}{t}$

$v = \frac{105}{ \frac{1 + \sqrt{85}}{4} }$

$v = \frac{420}{1 + \sqrt{85}} \approx 41.0977$

So $v + 10 \approx 51.0977$

-Dan