Does the order of algebraic steps (not operations) matter?


by Darren73
Tags: algebra, order, steps
Darren73
Darren73 is offline
#1
Aug31-13, 02:05 PM
P: 3
I know that the example I use has been done online many times, but my question isn't about how to get the answer, that is obvious, my question goes a little deeper.

For example when deriving the equation for an ellipse we have the constraint that the ellipse equation describe the set of (x,y) such that the sum of the distances from (x,y) to the foci [(-c,0), (c,0)] is constant, lets say 2a for convenience.

After setting up the equality that c^2+b^2=a^2 (Where b is the y value when (x,y) is in between the foci) we quickly establish the following equation which is used in various derivations online:

sqrt((x+c)^2+y^2)+sqrt((x-c)^2+y^2)=2a

If we square both sides first and then isolate and simplify we arrive at an impasse:

2x^2+2y^2+2c^2+2*sqrt(c^4-2c^2(x^2-y^2)+(x^2-y^2)^2)=4a^2

But if we isolate one of the radicals first and then square we can simplify down to an expression

cx-a^2=a*sqrt((x-c)^2+y^2)

And then if we square both sides again and simplify we can get an expression that allows us to use the c^2+b^2=a^2 relation and obtain the formula.

I'm confused as to why one sequence of steps (squaring, isolating, simplifying,...) does not allow you to find the solution whereas another set of steps (isolating, squaring, simplifying, squaring,...) does allow you to find the solution. It seems to me that since any of the expressions along the way to finding the solution are transformable into each other by simple techniques (adding 0, multiplying by 1, isolating, squaring, etc.) that any path should allow us to arrive at the solution. Now maybe my algebra is rusty and maybe you're supposed to always isolate first, but I'm not sure. Can someone help explain this to me?
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eigenperson
eigenperson is offline
#2
Aug31-13, 06:24 PM
P: 133
First of all, you squared it wrong. You should get ##\sqrt{c^4-2c^2(x^2-y^2)+(x^2+y^2)^2}## as the radical.

And this is not an impasse. All you have to do is move some terms to the other side:

##\sqrt{c^4-2c^2(x^2-y^2)+(x^2+y^2)^2} = 2a^2 - c^2 - x^2 - y^2 = (a^2 + b^2) - (x^2 + y^2)##

Now you can square again to get

##c^4 - 2c^2(x^2 - y^2) + (x^2+y^2)^2 = (a^2 + b^2)^2 - 2(a^2+b^2)(x^2 + y^2) + (x^2 + y^2)^2##

and this simplifies properly.

Note that in either case, you end up squaring, rearranging terms, then squaring again. So your intuition that it should be possible both ways was correct; you just didn't try hard enough to make it work the first way :P


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