Wednesday, October 1, 2008

Solve x^2 - 10000x - 10000 = 0 without a calculator! A Precalculus Investigation

No, there's no mistake in the constant term in that equation. Imagine giving this to your Precalculus/Math Analysis/Adv Math students! Actually, 'solving' it with the TI-84 requires some effort using Solver since one needs to make an approximate guess or adjust the lower bound so that the positive root is obtained. Using the graph is no 'walk in the park' either! The TI-89 or Mathematica would have much less difficulty in displaying the exact radical form or a suitable decimal approximation but they may not be within reach. Perhaps an important issue here is that sometimes technology gives us unexpected or even inaccurate results. That's when students need some understanding of theory to recognize the limitations of the technology and adjust accordingly.

Here's the point of all this. The given quadratic is not factorable over the integers, however we can replace it with a 'nicer' quadratic that is. The roots of the desired quadratic can be shown to be approximately the same as the 'nice' quadratic and we can show that the absolute error is less than two ten-thousanths (and a much much smaller relative or % error)! Does this 'numerical analysis' have any practical value? Why approximate roots when powerful technology can produce exact answers? Do professionals who need to apply mathematics to the solution of 'real' problems ever use such approximation techniques? Could it be that theory actually provides practical application!

(1) Show that the roots of the x2-10000x-10001 = 0 are 10001 and -1 by factoring.

(2) Show that the roots of x2-10000x-10000=0 can be approximated by 10001 and -1 with an error of less than 0.0001.
(a) By direct calculation: Using the quadratic formula and, yes, you may use the calculator!

(b) (Challenging) By comparing, in general,
(*) the roots of x
2-bx-(b+1)=0 and
(**) the roots of
x2-bx-b=0.Here we are assuming that b > 0.
(i) First show by factoring that the roots of equation (*) are b+1 and -1.
(ii) Then use the quadratic formula to express the roots of (**) in terms of b.
(iii) Compare the positive roots of these equations by subtracting them and (after algebraic manipulation and simplication), show that the absolute value of the difference is less than 1/(b+1).
Note: For b=10000, this error is therefore less than 0.0001.

(c) Explain intuitively why the roots of the original equation and the 'approximating' equation are virtually the 'same' for 'large' values of b. One possibility here is to consider how the graphs of the associated quadratic functions are related. What do they have in common? How are they different?
Note: Subtle point here for students. Even though the difference of the function values (i.e., y-values) is always 1, this is not true of the difference between their zeros! This may be the essence of the numerical analysis in this investigation.

Ok, now "solve" x2 - (googol)x - googol = 0 without a calculator.

Without your calculator show that √(10001) - √(10000) is less than 0.005.
Does this provide us with an effective method of approximating the square root of some large numbers or is it limited and impractical?

For Calculus students: How does this compare to using linearization to approximate the square root?

For more advanced calculus students: Newton's Method? The Binomial Formula (using fractional exponents)? A Taylor Polynomial approximation? All equivalent?


Kate said...

Nice...thanks for sharing. Have you used this? How much do you give away when you introduce the problem? Do you try to get them to figure something out, or let them flail for a while and then tell them about the "close" equation? How do you tie it to other stuff you're doing?

Dave Marain said...

You symbolize for me the reason for my continuing to write this blog. You immediately asked the central question (I'm paraphrasing or intuiting what some others might be thinking):

"Nice problem, but how exactly do I implement this in an already packed curriculum? What does your lesson look like?"

I certainly do not believe that one could lift the problem directly off the screen and drop it on a group of youngsters. Parts may be too sophisticated for your group and would require more 'scaffolding.' Further, considerable revision may be needed. The post is merely a framework. However, here are a few thoughts:

Content Tie-In: Possibly the chapter on Polynomial Equations/Functions; also, this can be used for review of solving equations and appropriate uses of technology

Suggested Implementation in the Classroom:
Set aside at least a 1/2 period for this investigation or one might think of it as performance assessment with students working in pairs.

A Possble Way to Introduce:
Look at the two equations on the board (handout, PowerPoint, etc.).

Which one do you think can be solved without a calculator?

Note: If this is an assignment or assessment, these questions would appear on the handout. Otherwise, you are asking questions for discussion purposes.

Now solve both equations using the graphing calculator (or Mathematica as needed. Express roots to the nearest whole number for now. Which equation has integer roots? What does this imply about the quadratic?
[Ans: It's factorable over the integers].

To the nearest integer, you should now have discovered that the 2 equations have the "same" roots. Since the equations are not equivalent, how is this possible? In looking for an explanation, you might consider the graphs of the corresponding quadratic functions.

Sometimes, I would just let it flow and avoid overly scripting. Spontaneity is needed because you may experience some very unexpected results! I've never actually implemented this problem in my classes. I just found the idea of "approximating" roots to be of value, particularly the way it ties together so many different concepts and deepens student understanding of appropriate uses of technology.

I'd be interested in how this plays out in the classroom, Kate. Actually, solving both equations using the graphing calculator can be very instructive. We would hope that some would recognize the factorability of one of the equations, but we know this may not happen unless we tell them this ahead of time!

Kate said...

Thanks for the clarification. I'm thinking of including something like it in a problem set for Honors Precalculus. I like your idea of showing the two together, leading them through how close the solutions are except that one is factorable, and then asking a follow-up question.

Eric Jablow said...

I have a couple of comments while I avoid watching the VP debate.

1. Once one sees that the two roots are near -1 and 10001, a student who knows some calculus can see that the slope of the function near -1 is about 20000. This means that the function will change by 1 unit in a space of about 1/20000. So, the smaller root of the second equation is within 1/20000 of -1.

2. I first saw this trick in Forman Acton's book, "Numerical Methods that (Usually) Work". Rewrite the second equation as

x^2 -10000 = 10000x, and divide by 10000:

x = -1 + x^2/10000.

Now, guess at x, say x = 0, and try the iteration x->-1 + x^2/10000. Convergence is really fast.

3. You never want to compute √(10001) - √(10000) in the obvious manner. As an experiment, try working on blackboard with a mythical 6-digit decimal floating-point calculator. Show how one loses 3 digits of precision of the 6. Show how converting to 1/[√(10001) + √(10000)] solves the problem.

Dave Marain said...

I do not want to impose my style when it comes to actual implementation of these investigations. All I am comfortable doing is sharing my experiences and what I've learned.

I really like your slope argument to explain the insignificant change in x for unit change in y. However, I was also suggesting how precalculus students might compare two different functions: f(x) and f(x)-1. It might seem paradoxical at first ("but the function values always differ by 1 for the same x"). They need to grasp that we're looking at the inverse situation here: Comparing how close the x's are when y=0 for both functions. Your slope argument provides strong theoretical and visual support for this.
I've read several references to Acton's book. i think I would enjoy reading it.

Eric Jablow said...

I had the sign of the slope wrong in my last comment. More importantly, I didn't need to refer to differentiation at all.

For x near -1, say between -2 and 0, write

x² - 10000x - 10001 = (x - 10000) x - 10001,

and realize that x - 10000 is bounded between -10002 and -10000. So, the expression is bounded between two lines with slopes -10002 and -10000.

Realize for the second equation that

x² - 10000x - 10000 = 0 is the same as

x² - 10000x - 10001 = -1.

Then, write the equation of the two lines, see how close the solutions are to -1, and see what this implies for the solution of the quadratic.

One thing a teacher should realize is that high school and lower-level college students are not comfortable with axes and coordinate systems other than the standard Cartesian system. Showing them an out-of-scale pair of axes that might allow them to understand the geometry may only confuse them.I never learned how to accustom them to scaling. I would guess that engineering teachers have similar problems accustoming students to log or log-log paper.

San said...

hi .
thanks for sharing.
i read it but i have a problem .
you said :"the absolute value of the difference is less than 2/(b+1)"
how did you find it?
i find positive root of x^2-bx-b=0 and i subtract the positive roots .
but i dont know why its less than 2/(b+1).
help me please.
thanks again .

Dave Marain said...

The algebra for the "2/(b+1)" is messy and I need time to either do it in TeXiFy or make a video or in Word using Eqn Editor (then saving it as an image). I'm still trying to get the Image Capture technology to work but there's never enough time in my life to do it. Pls be patient with me and I will post the solution. The manipulations are really nice and worthwhile for students to try.

Dave Marain said...

San and others--
I've been rechecking my error bound and I'm thinking the "2" in the numerator can be improved to "1", i.e., the absolute error is less than 1/(b+1). I haven't located my original calculations but, in re-doing them, I think I was able to prove this newer version. This needs to be independently verified of course! Thus, for b = 10000, one can show that the error in using 10001 as the root will be less than 0.0001. Let me know if you agree with this. I will revise the post to reflect this improved error bound if someone confirms it.

San said...

i agree.
i 'm wating for your revise.
thanks a lot.