A89: Re: Re: Re: Re: C question... actually two C questions...
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A89: Re: Re: Re: Re: C question... actually two C questions...
I found this to be very entertaining, in an odd, geeky kinda way. It
probably belongs on the TI-Math list instead, but I sure would like to see
more of them.
> gah! i can't figure it out! surely you must have a divide by 0 somewhere,
> but it's escaping me...
Sure you wanna know? You'll be kicking yourself afterwords. . .
Everything done is valid EXCEPT raising both sides of an equation to a
power. Doing so risks creating extraneous roots, which is what we're doing
here. Take a look at this step:
> (3 - 7/2)^2 = (4 - 7/2)^2 / sqrt
By taking the square root of both sides, we're raising then to the 1/2
power.
We're creating an extraneous root, which has been cleverly dropped because
you've incorrectly assumed that, if a^2=b^2, a=b.
The "correct" answer to that is (+/-)a=(+/-)b [that's plus or minus].
That can be simplified to the equivalent a=(+/-)b. We'll use that
simplification below - we're taking two square roots, meaning we should be
creating a total of four "possible" answers, but you'll have the same two
possibilities repeated twice; hence we can ditch the (+/-) on one side.
The correct way to evaluate that - without skipping a step - would be as
follows:
> (3 - 7/2)^2 = (4 - 7/2)^2
(3 - 7/2) = (+/-) sqrt((4 - 7/2)^2)
(3 - 7/2) = (+/-) (4 - 7/2)
One of those evaluates into the correct -1/2 = -1/2; the other gives the
incorrect 3=4. That's our extraneous root being slipped in.
The trick is that our a and b in a^2=b^2 are 1/2 and -1/2 and NOT equal,
although their squares are.
I _THINK_ that you'll only create a possible extraneous root if either the
numerator or denominator is even, because that's the only time you have to
worry about positive and negative bases giving the same result when raised
to the power. In other words, if we had a^3=b^3, I think we could make the
assumption that a=b - but I'll leave that to others here to confirm of
refute.
-Scott