Livio begins by describing the history of polynomial equations. First, we have linear equations, which appear extensively in the ancient Egyptian Rhind Papyrus. Second, we have quadratic equations, and a form of the Quadratic Formula was used by the ancient Babylonians in order to relationship between the perimeter and area of a rectangular plot of land.
We all know the Quadratic Formula from Algebra I. Here I state both the general quadratic equation and the Quadratic Formula in order to demonstrate how I will render polynomials in ASCII in order to discuss them on the blog.
Quadratic Equation: ax^2 + bx + c = 0
Quadratic Formula: (-b +- sqrt(b^2 - 4ac))/2a
The Quadratic Formula is notoriously difficult to remember, and so a common way to get students to remember it is to make it into a song, such as "Pop Goes the Weasel":
I remember back when I was first learning about the Quadratic Formula, I often wondered whether there was such a thing as a Cubic Formula -- a formula that could solve any cubic equation of the form ax^3 + bx^2 + cx + d = 0. As it turns out, mathematicians in early Renaissance Italy also sought out a Cubic Formula.
Livio devotes much of Chapter 3 to the story of four Italian mathematicians who worked on solving the cubic equation. As we've seen before on this blog, the stories behind mathematicians can be highly intriguing. (Last week, I didn't get to sub in any math courses but I was in a eighth grade physical science class, and they were studying buoyancy, including the Principle of Archimedes. So of course I couldn't resist telling the story of how Archimedes discovered his Principle while taking a bath -- as I mentioned in my April Fool's post.)
Livio credits Scipione dal Ferro with discovering the solution of a simple case of the cubic equation, namely ax^3 + bx = c, in the year 1515. He did not publish his formula, but instead told it directly to his student Antonio Maria Fiore. Now Fiore kept the formula a secret as well, but he wanted to prove to everyone how smart he was, so he challenged the mathematician Niccolo Tartaglia -- which means Niccolo the Stammerer -- to a duel, a mathematical duel. The two of them would give each other 30 cubic equations to solve. Unfortunately for Fiore, not only had Tartaglia independently discovered dal Ferro's formula, but he had also found out how to generalize it to all cubics.
In between all of this, another mathematician, Ludovico Ferrari, not only generalized all of these results on cubic equations, but also discovered a Quartic Formula -- one that would solve equations of degree four, ax^4 + bx^3 + cx^2 + dx + e = 0. Livio tells the story much more dramatically than I have here on the blog, but by the end of the 16th century, the Quadratic, Cubic, and Quartic Formulas were all known. So this raises the question -- why do we expect high school students to learn only the Quadratic Formulas and not the Cubic or Quartic Formulas? After all, if the Babylonians developed the Quadratic Formula to solve perimeter and area problems, then why can't high school students learn the Cubic Formula to solve volume problems?
Well, let's take a look at this Cubic Formula. Livio states the dal Ferro formula in Appendix 5:
Cubic Equation: x^3 + px = q
Cubic Formula: cbrt(q/2 + sqrt(p^3/27 + q^2/4)) + cbrt(q^2 - sqrt(p^3/27 + q^2/4))
So we see that even for this simpler form of the cubic ax^3 + bx^2 + cx + d = 0, where a = 1, b = 0, c = p, and d = -q, this is already more complicated than the full Quadratic Formula. Since the Quadratic Formula involves square roots, it's not surprising that the Cubic Formula uses cube roots, abbreviated "cbrt" above.
Many Algebra II texts claim that imaginary numbers were created because mathematicians wanted to find solutions for quadratic equations such as x^2 + 1 = 0. But this is false -- mathematicians around fifteenth century Italy would have simply declared x^2 + 1 = 0 to have no solution. Complex numbers weren't created to solve quadratic equations -- they were created to solve cubic equations. The cubic equation mentioned earlier, x^3 - 15x = 4, requires imaginary numbers that end up canceling out to produce a real solution. As it turns out, if a cubic has just one real root, the dal Ferro formula gives the root without imaginary numbers appearing, but if it has three real roots, then the formula doesn't give any of the roots without requiring imaginary parts that eventually cancel out.
Finally, we notice that we have solved the full cubic ax^3 + bx^2 + cx + d = 0 yet. Well, we can easily eliminate the a coefficient simply by dividing by a, just as we do when completing the square. As it turns out, we can eliminate the term bx^2 by doing something like "completing the cube" -- we ultimately perform the transformation x' = x - b/3a. Notice that this transformation is, in fact, a horizontal translation. So we see that solving cubic equations is much more complicated than solving quadratic equations.
But then we could ask the opposite question. If the Cubic Formula is deemed too hard for high school students to learn, then why do we make them learn the Quadratic Formula? After all, quadratics can be solved using numerical methods on a computer instead as well. Any argument against teaching the Cubic Formula can be used to argue against teaching the Quadratic Formula too.
The status quo in high school math is that students should learn all of the units that help prepare them for calculus, and within each unit, learn as much as they reasonably can at that stage. If the Common Core Standards required students to learn the Cubic Formula, then opponents would point out that the Cubic Formula is too difficult and would prevent students from succeeding and making it to Calculus, but if the Common Core Standards didn't require students to learn the Quadratic Formula, then opponents would say that the Common Core is watering down Algebra I. Only the current path, where students learn the Quadratic but not the Cubic Formula, would be acceptable to most traditionalists.
And so I return to Geometry. Many people discuss the Core's emphasis on transformations and wonder whether students should have to learn translations, reflections, and rotations. Are the Common Core's transformations like the Quadratic Formula -- not completely necessary, but still within the range of student understanding? Or are they like the Cubic Formula -- completing frustrating to students and would be a huge barrier to advancement in math.
My answer is that only time will tell. We have to wait and see how well students are able to learn about transformations on the Common Core tests. If it can be shown that students don't learn about congruence using transformations well, but succeeded when SSS, SAS, and ASA were all considered postulates, then transformations should be dropped in favor of the old path.
With that, we look at today's PARCC Question. Question 12 on the PARCC Practice Test is on tangents to circles:
The figure shows two semicircles with centers P and M. The semicircles are tangent to each other at point B, and Ray DE is tangent to both semicircles at F and E.
(In the figure, Ray DA passes through P and M, with
If PB = BC = 6, what is ED?
(A) 6
(B) sqrt(48)
(C) 8
(D) sqrt(72)
Here's how we solve this problem. Section 13-5 of the U of Chicago text tells us that the tangent of a circle is perpendicular to the radius, so both DFP and DEM are right angles. Both Triangles DFP and DEM have angle D in common, so by AA Similarity (Section 12-9) these are similar triangles.
To find the scale factor, we notice that PF is a radius of circle P, so its length must be equal to that of PB, or 6. This side corresponds to ME, a radius of circle M. A diameter of circle M is BC, or 6, so the radius ME must be half of this, or 3. So the scale factor must be 1/2.
So DM must be half of DP -- and notice that MP must be the other half of DP. Now MP is the sum of the radii MB and PB, so it must be 3 + 6 = 9. So DP = 18 and DM = 9. Now DM is the hypotenuse of right triangle DEM, with known leg ME = 3, and we want to find the other leg ED. So all that remains is to use the Pythagorean Theorem (Section 8-7):
ME^2 + ED^2 = DM^2
3^2 + ED^2 = 9^2
9 + ED^2 = 81
ED^2 = 72
ED = sqrt(72), which is choice (D).
This is a very tricky problem. As we can see, we needed information from three Sections, 13-5, 12-9, and 8-7, in order to get an answer. Students might know the Pythagorean Theorem, but it's not obvious that Pythagoras must be used especially since no triangles at all -- much less right triangles -- appear in the drawing.
This is just like those similarity problems where one ends up needing to use the Quadratic Formula in order to find the desired length. Students may remember the Quadratic Formula from the previous year, but not realize that they need to use it in the problem.
Some people believe that students need to be challenged. They need to know how to use the Pythagorean Theorem and Quadratic Formula even in situations where it's not obvious that those are the tools to be used. But in questions like this one, it may end up frustrating students, who are likely to get this question wrong, especially with the correct answer being the last choice (D).
Here's an analogy -- suppose a third-grade teacher, being a strong traditionalist, wanted to test the students' knowledge of the multiplication tables from 0 * 0 to 9 * 9, but she doesn't want to grade 100 questions for every student. So instead, she decides to have the students multiply two 10-digit pandigital numbers -- that is, numbers containing all of the digits:
4019782536
* 4092578361
Since the numbers are pandigital, students must know all of 0 * 0 to 9 * 9 to get the answer, and as the correct answer has 20 digits, it's easier for the teacher to grade than 100 problems. If all 20 digits are correct, the student gets a score of 100%, if one digit is wrong, the student gets 95%, and so on.
Obviously, this is a lousy test. It's very easy for the students to make a mistake, since there are so many digits to keep track of and so much carrying and adding. Notice that if students use the much maligned lattice method, the lattice will contain a complete multiplication table before the students have to add and carry -- where they will probably make a mistake!
If we decide that the best questions are content-based skill questions, then notice that a question that gives a right triangle or a quadratic equation and asks for students to use the Pythagorean Theorem or Quadratic Formula to solve it is indeed a skill-based question. Notice that there's a question in the Chapter 13 SPUR section similar to this PARCC question -- and it's a Properties question, not a Skills question. After all, it's a property of the tangent line that produces the right angle needed for the Pythagorean Theorem. Yesterday's question was more obviously a Properties question since it asks for the properties of rotations, but today's question is no less a Properties question.
To finish, let me point out that after Ferrari discovered the Quartic Formula so quickly after finding the Cubic Formula, the race was on to find the Quintic Formula -- a formula that would solve the fifth-degree equation ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0. But the solution to this quintic wasn't discovered as quickly as the Italian mathematicians had hoped. According to Livio, it wasn't until the end of the 18th century when Paolo Ruffini suspected that the reason that no one had discovered the Quintic Formula yet was because there is no Quintic Formula. In other words, the quintic equation is, as mentioned in the title of the book, The Equation That Couldn't Be Solved.
PARCC Practice EOY Exam Question 12
U of Chicago Correspondence: Section 13-5, Tangents to Circles and Spheres
Key Theorem: Radius-Tangent Theorem
A line is tangent to a circle if and only if it is perpendicular to a radius at the radius's endpoint on the circle.
A line is tangent to a circle if and only if it is perpendicular to a radius at the radius's endpoint on the circle.
Common Core Standard:
Commentary: There is a question in this section that looks much like the PARCC problem, namely Question 7. This question is written below:
CCSS.MATH.CONTENT.HSG.C.A.2
Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.
Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.
7. Line CA below is a common tangent to circles P and Q at points A and B. If CB = 5, AB = 10 and the radius of circle Q is 3, what is the radius of circle P?
Notice that this question does not require the Pythagorean Theorem to solve. A question that would be more analogous to the PARCC problem would be to ask for PQ, as this would require the Pythagorean Theorem. Also, PQ cannot be determined simply by adding the radii since, unlike the PARCC Question, the circles are not tangent to each other. In fact, we calculate that PA = 9, yet PQ = 8. The drawing is not to scale as point Q should actually be inside circle P. We see that there are more questions like this in the SPUR section, under Properties -- Questions 24 and 25 under Objective F.
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