FUN solving Quadratic equations

AI Thread Summary
Marnie walks 1 km/h faster than Jon and completes a 20 km hike one hour earlier. The equations derived from their speeds and times are S + 1 = 20/T for Marnie and S = 20/(T - 1) for Jon. By substituting and manipulating these equations, a quadratic equation in T is formed, which can be solved to find T and subsequently Jon's speed S. An alternative approach involves rearranging the first equation to express T in terms of S, leading to a more complex equation that directly yields S. The discussion highlights different methods for solving the problem, emphasizing the importance of clear variable representation.
sillybilly
Messages
10
Reaction score
0
FUN solving Quadratic equations :)

Homework Statement


Marnie can walk 1km/h faster than Jon. She completes a 20 km hike 1 hour before him. Write an equation ans solve it and find their walking speeds.

Homework Equations


speed= distance/time
the quadratic equation

The Attempt at a Solution


i don't no how to start :(

but i no that for Marnie: S= D/T
and Jon S= D/T-1 because he's an hour slower.
 
Physics news on Phys.org


Be careful- "S" has two different meanings in those equations while "D" and "T" mean the same thing! You should never use one symbol to represent two different things. Instead, since we are told that "Marnie can walk 1km/h faster than Jon", let's use S to represent Jon's speed so that Marnie's speed is S+ 1. Now those two equations are
S+ 1= D/T and S= D/(T-1). Further, we are told that D= 20 km so we actually have

S+ 1= 20/T and S= 20/(T- 1).

One thing you could do is use the second equation to substitute for S in the first:
20/(T-1)+ 1= 20/T. Multiplying both sides of the equation by T(T- 1) gets rid of the fractions and gives 20T+ T(T-1)= 20(T- 1)= 20T- 20, a quadratic equation in T. Solve that equation for T and use S+ 1= 20/T to find S.

That is slightly "inefficient" in that you have to solve for T first when the problem does not ask for T. Another way to do this would be to solve the first equation for T, by inverting to get T/20= 1/(S+ 1) so that T= 20/(S+ 1). Put that into the second equation to get S= 20/(20/(S+1)+ 1). That's a harder equation to solve but gives S directly.
 


ohh wow... thankyou soo much and yes the first one was easier :)
(i did them both)
THANK YOU SOOO MUCH :D
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top