Thread: Rearranging this equation?

I don't see how

100-0.25x^2 =ax^2 - 80ax + 1600a

becomes

x^2 + (80a)/(-0.25-a)x + (1600a+100)/(-0.25-a) = 0

Originally Posted by LandLBoy

I don't see how

100-0.25x^2 =ax^2 - 80ax + 1600a

becomes

x^2 + (80a)/(-0.25-a)x + (1600a+100)/(-0.25-a) = 0

it doesn't ...

can you still get this from what you got

x = - (40a)/(-0.25-a) +- √(1600a^2/(-0.25-a)^2 - (1600a+100)/(-0.25-a))

why is this problem being done in a piece-meal fashion?

mind stating the original problem in its entirety?

http://www.mathhelpforum.com/math-he...em-175040.html

sorry, we're not supposed to use calculus to solve the problem - i found an answer online and i'm trying to interpret it but i'm having some difficulty as u can see

ANSWER:

find the point, where curve 1 and curve 2 touch.

100-0.25x^2 = a(x-40^2)
100-0,25x^2 =ax^2 - 80ax + 40^2*a
...
x^2 + (80a)/(-0.25-a)*x + (-40^2a+100)/(-0.25-a) = 0
x = - (40a)/(-0.25-a) +- √(40^2a^2/(-0.25-a)^2 - (-40^2a+100)/(-0,25-a))

now to have only 1 tangent point, the term under the root must be = 0
40^2a^2/(-0,25-a)^2 = (-40^2a+100)/(-0,25-a)

solution:
a = 0,083333

now find x:
x = - 40*0,08333/(-0,25-0,08333)
x = 10 m

The curves touch at x = 10 m,

and the slope of curve 1 at 10 m is
y' = -1/2x = -5
and of curve 2 at 10 m is
y' = 0,08333(2x-80) = 0,08333*- 60 = - 5

i.e., the slopes are identical --> nice smooth ride

I understand what the answerer did but it's the intermediate steps that are confusing me - that is why I posted a part of the answer trying to figure out what the answerer did

Originally Posted by LandLBoy

I don't see how
100-0.25x^2 =ax^2 - 80ax + 1600a
becomes
x^2 + (80a)/(-0.25-a)x + (1600a+100)/(-0.25-a) = 0

Is there a reason why you're "hanging on" to this weird "-0.25-a"?

You instead could change the term .25x^2 to x^2 / 4, then multiply through by 4.

Quite simpler, and you'll need only half the Tylenols

for a single solution, the discriminant = 0 ...











now go back and determine that