Trivial equation. Note: it's 6:20am

peteris.krumins_at_gmail.com
Date: 12/31/04 

Date: 30 Dec 2004 20:22:06 -0800

Hello.

Here is a trivial equation: f(t) == (v*t - gt^2) / 2
I need to find the roots when f(t) == 0;

After factoring to make things easier:
t(v - gt/2) == 0;
t1_1 = 0 or v - gt/2 == 0
v == gt/2
2v == gt
t1_2 = 2v /g

So far so good. This is trivial.

f(t) is in form ax + bx^2, quadratic equation so it could also
be solved without factoring but with quadratic formula:
x_1, x_2 = (-b +- sqrt(b^2 - 4ac)) / 2a

Ok, let's do it:
a == v
b == -g/2
c == 0

t2_1, t2_2 = (g/2 +- sqrt((-(g/2))^2 - 4*v*0)) / 2v
= (g/2 +- sqrt(g^2/4)) / 2v
= (g/2 +- g/2) / 2v

and this leads to roots:
t2_1 = (g/2 - g/2) / 2v = 0
t2_2 = (g/2 + g/2) / 2v = g/2v

Clearly t1_1 matches t2_2, t1_1 == t2_2; 0 = 0;
But t2_1 does not match t2_2: 2v/g != g/2v
I cant see my mistake. :(

 
P.Krumins

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