Simultaneous equations?

Dec 2014
9
1
United Kingdom
Hi

I've got two equations and I want to work out the X value where they intersect.

\(\displaystyle y=exp(-ax+bx^2)\)
\(\displaystyle y=exp((-cx+log(e))\)

a, b, c and e are constants. Since the lines intersect at y, I've tried to do:
\(\displaystyle exp(-ax+bx^2)=exp((-cx+log(e))\)
\(\displaystyle -ax+bx^2=-cx+log(e)\)
\(\displaystyle -ax+bx^2 +cx=log(e)\)
\(\displaystyle (c-a)x+bx^2 =log(e)\)
\(\displaystyle (c-a)x+bx^2 - log(e) = 0\)
\(\displaystyle bx^2+(c-a)x - log(e) = 0\)

This, I believe, can be solved by the quadratic formula. But when I use sample values of
\(\displaystyle a = 0.041,
b=0.010,
c= 0.205,
e=1.210\)
I get answers of 1.09 or 6.18. I've got a graph of the intersection between the two curves, and it should be at around x=15.2 (see attached graph)

I don't understand where this is going wrong. I realise the graph is logarithmic and doesn't pass through the x axis. Is that why this is going wrong - if so, why can't the quadratic equation be used this way? Or am I just handling the quadratic wrongly?

Thanks for your help
L
 

Attachments

Nov 2019
23
4
Boston
If the graphs are of the exp functions then they are not supposed to cross the x axis. Please post the solution of the quadratic equation.
 
Dec 2014
9
1
United Kingdom
Thanks for the replies. It seems like this isn't the right approach to work out the solution to:

\(\displaystyle −ax+bx^2=−cx+log(e)\)

...but why not? The curves intersect at a point, how else can I find the intersection point X from the equations?
 
Dec 2014
9
1
United Kingdom
I think I've solved this problem, using the 'completing the square' method given at Completing the Square

I'm not sure why this method gives the answer and the quadratic formula doesn't, but at least I've something that works. Thanks to everyone for their help on this!
 
Nov 2019
23
4
Boston
I think I've solved this problem, using the 'completing the square' method given at Completing the Square

I'm not sure why this method gives the answer and the quadratic formula doesn't, but at least I've something that works. Thanks to everyone for their help on this!
According to post #3 1.09 is correct. Not sure where the graph and x=15.2 come from.
 
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