Why can't I solve this dirt-simple equation??

(No, it’s not homework :stuck_out_tongue: )

This ought to be extremely easy, but I’m just not getting it: A² -2A -1 =0. I even have the solution: A =√2 +1. I can confirm that that value of A fits the equation, but I can’t figure out how you show A from the equation. What am I missing?

The quadratic formula? And you have one of two solutions.

Try completing the square or the quadratic formula.

A² -2A -1 =0
Add 2 to each side:
A² -2A +1 =2
So the first part is a square:
(A -1)² =2
Take the square root:
A-1 = √2 or - √2
Add 1 to each side:
A = 1+√2 or 1 - √2

Ah, so in more general terms, the trick is to add or subtract from the polynomial expression on one side of the equation to convert it into a form that can be factored. Got it. Thanks all!

More specifically, factored as a perfect square.

If you have x[sup]2[/sup]+ax, the constant term that will make this factor as a perfect square is (a/2)[sup]2[/sup].

By the way, if it’s not homework, what is it? :slight_smile:

It’s related to some tinkering with Pythagorean triples I’ve been doing. The whole-number solutions to A²+B²=C² can be derived from two other numbers, let’s say P and Q, where A=P²-Q², B=2PQ, and C=P²+Q².
(Example P=2, Q=1 gives 3, 4, and 5).

I was interested in the cases where A and B differ as little as possible, by ±1. If A≈B, then you get P²-Q²=2PQ.
If you want to express the larger in terms of the smaller, then you set Q as 1 and you get P²-1=2P, or P²-2P-1=0.
I’d worked out by sheer trial and error that the answer had to be √2 +1, but I couldn’t remember enough high-school algebra to prove it.

OK, as the OP’s query seems to be solved, why can’t I figure out what’s wrong with this seemingly straightforward problem?

ETA: Dammit. That’s been bugging me for two weeks, and I figured it out five seconds after I clicked “Submit.” Never mind. Duh.

Mind sharing the answer for those of us who stink at math and logic and foolishly clicked the link? It seemed similar to the $30 Hotel Room Problem but as far as I can tell the credits and debits are all in the right columns. What am I missing?

There is a certain similarity to the $30 Hotel Room Problem, indeed. You may want to ask yourself why on earth the sums of the remaining balances over time would be at all meaningful. (For example, if he’d stopped after the first two withdraws, would you be shocked that the amount he withdrew ($35) didn’t match up with the sums of the balances along the way ($30 + $15 = $45; note how this “double counts” money (and eventually “triple counts”, etc.) the longer it’s left in the account)?)

Very similar to the $30 hotel room in that the intent is trick you into adding numbers that shouldn’t be added and are only close to the expected result by chance.

It’s true that the second column adds up to 51 dollars, but that’s simply a coincidence based on the amounts withdrawn. For example, if the money were withdrawn $1 at a time, no one in their right mind would expect the “balance of” column to be anything like the amount deposited. It would be 49+48+47+46…….+2+1 which is way over $50 (I think it’s 1225).

No one balances their checkbook by adding up the daily balances.


There’s no reason the remaining sums should add up to the original balance. Imagine if he’d withdrawn the money $1 at a time; the right-hand column would be $49 + $48 + $47 + $46… etc., and would total $1225.