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Not the most challenging contest problem but something to give to your students to develop logical careful thinking and some "basic skills." There's a slight 'twist' but nothing that will faze our math experts out there.
SOLVE
(x2 - 6x + 9)(x2 - 4x + 3) = 1
Posted by
Dave Marain
at
8:31 AM
Labels: algebra 2, exponents, math contest problems, warmup
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14 comments:
Very nice. I actually caught the trick pretty early, then it was just a matter of making sure I applied it correctly. I got three answers.
Glad you enjoyed it Sean, and, yes, I think you nailed it! I always enjoy these kinds of problems personally even though they appear to have little "practical real-world application"! I do believe it develops student thinking while reviewing some nice skills and concepts.
I used calculus (at least limits and L'hopital's rule) to rule out one of the answers. Is there an algebra 2 way to do that?
I get only two solutions. There is a third that does appear, but checking it gives an impossible result.
Calculus Dave, I don't think any limits are necessary are they? Just plain factorisation and substitution for checking?
Well, at x = 3 you get 0^0 which is undefined. So, you have to do limit as x -> 3 and see what it approaches. In this case I got that it does not go to zero as I wanted for this example (via L'hopital's and logarithms).
Ok, that should teach me to check things VERY carefully before posting. I did find 3 solutions, not 2. But again, only using factorisation (and a graphical calculator)
I am so going to give this one to my IB HL students :)
Oops, sorry, I meant the limit shows that at x = 3, the function is e^4
Nice discussion ongoing for such an 'innocent' little problem!
Another discussion point for students might be to ask the significance of choosing a perfect square trinomial for the base. An easily overlooked set of solutions for these types is determining where the base is -1 and the exponent is an even integer. This could not happen in this instance but I think it is a point worth mentioning since we normally restrict all bases to be positive.
Another afterthought...
I expect someone to point out that I also did not restrict the domain to reals, so we could even allow roots of unity and other complex solutions! Okay, enough already -- assume real values please!
Ok, I took lunch break to work out my issues with x=3 on this problem and I think I figured it out. But what is the rationale to rule it out for algebra 2 students?
To Calculus Dave: But isn't it unnecessary to check the limit as x tends to 3? We see that the left hand side is undefined when x = 3, and hence x = 3 couldn't possibly be a root.
Anon,
I'm also not sure why CalcDave is considering the limit, since, as you indicated, the function is undefined at x = 3. As i calculated it, the function has a removable discontinuity at x=3, since the limit there is 1.
At this point, I'm guessing most have obtained the 3 solutions 1,2, and 4.
I do think that the issue of the base being a perfect square is important here. If we choose a different base we might need to consider all solutions of the form (-1)^(even integer), which would make for an interesting analysis!
Yeah, well, I was doing the limit just to see what happened. But I was curious if we teach alg 2 students about indeterminant forms like 0^0.
I solved it by taking natural log of both sides and going from there. Is that how the students are meant to do it or just think about how the base should be 1 and/or the exponent should be 0?
So I came late, which is what I have been doing, found 3 solutions, but got stuck on the next two.
With 3+2i, 3-2i I find -1 to the -4+4i and -4 -20i powers respectively, but do not know how to evaluate them. Any clues?
Jonathan
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