It has been interesting watching my son learn a bit more algebra this year. He seems to have made quite a bit of progress studying linear equations, but he’s still at a point where the quadratic formula is the first thing he thinks about with any sort of non-linear equations (which, given that he’s just learning algebra are mostly quadratic equations).

Today he ran across a problem that was probably designed quite specifically to help kids see beyond the quadratic formula. The problem is #20 from the 2007 AMC 10 a

Here’s the problem:

Suppose that the number $a$ satisfies the equation = . What is the value of ?

This problem gave him some difficulty and I asked him to explain his original approach using the quadratic formula first:

Next we talked about how to approach this problem without solving for first. We had briefly talked through this approach in the morning but this was his first time trying to explain it.

Next we went to Wolfram Alpha to see that the solution he’d found with the quadratic formula actually produced the answer of 194. After that we talked about the graph of . It was a little hard for him to see that the minimum value on the graph occurred at , but zooming in a little helped him see it.

Finally, I wrapped up by showing him one way that we could use the quadratic formula to help us see where that minimum occurred. I took this approach to help him see that even though the quadratic formula wasn’t so helpful in solving the original problem, it still could be helpful as a way to learn a little bit about x + 1/x.

So, a nice little problem that provides a good example of a situation where the quadratic formula isn’t so helpful. Hopefully examples like this one will help him see that there are lots of to approach non-linear equations, and the quadratic formula is just one of them.

5 thoughts on “Moving beyond the quadratic formula”

There were two moments in these videos that I’ve been thinking about since I saw them last night. The first one was when your son said that the equation was true if x was an integer, and you referred to the graph, noting that x could be any number. He agreed but I wasn’t so convinced (I’m not sure he was either!), because it just intuitively didn’t seem right. I paused and plugged in 1/2 and quickly realized what was happening within the dynamics of the equation, then (and this is the second thing this video brought to mind) I realized that this equation is what I think of as “one of those” equations that math minds recognize: There was a moment for me,years ago, when someone told me that as soon as I see (x^2 – 25) I should know what the factors are without having to stop and figure them out. At that moment I realized that math is full of these recognizable relationships that, at a certain level “everyone knows.” Grasping this concept of recognizable relationships that didn’t have to be worked out fresh each time really opened up things for me.

There’s another direction using the quadratic equation that is very nice and leads to some interesting thinking:
a^2 – 4a +1 =0 can be reinterpreted as a^2 = 4a -1

Using that, we can reduce any polynomial in a to a linear polynomial. Working up the powers, a^3 = 4a^2 – a = 4(4a-1) – a = 15a – 4
a^4 = 15a^2 – 4a = 15(4a – 1) – 4a = 56a-15

Reasoning based on symmetry, we see that the same thing holds for the reciprocal of a, that is a^(-4) = 56/a – 15

This thought process connects to two cool areas of more advanced algebra. First, residue rings. Say f(x) is a polynomial with coefficients in a base ring R, leading coefficient 1, then we can think of f as giving us a way to reduce powers of x in R[x]/(f(x)). In particular, if f is quadratic, then all the elements of R[x]/(f(x)) are nicely represented by linear terms in x (of the form cx + d).

This is exactly what we did before, where we recast all higher powers of a as a linear expression involving a.

Second, but closely related, is the observation that any polynomial expression that is symmetric in a and 1/a, with integer coefficients will have to be an integer. Similarly, if you have rational coefficients. This seems pretty amazing that all the square roots drop out.

Another example is the closed form expression for terms of the Fibonacci involving powers of the golden ratio where the fact that we always get integers appears absolutely magical (to me, at least).

The power reduction observation above helps show why this is true (the reduction steps preserve the symmetry, we end up with a linear expression involving a and 1/a, symmetry means that we can factor all of the a terms out out with a single (a + 1/a), etc). This is a nice example of a key insight from Galois Theory.

Following up on the Fibonacci connection in Joshua’s reply, since the number a satisfies the quadratic equation x^2 = 4 x – 1, one can check that the sequence of powers 1, a, a^2, a^3,… satisfies the linear recurrence x_n = 4 x_{n-1} – x_{n-2}. The same statement holds for the sequence of powers of the other zero 1/a. Since the recurrence is linear, adding these two sequences term-by-term will yield a new sequence which also satisfies this recurrence. If we write:

x_n = a^n + (1/a)^n

for the nth term of this new sequence, then we can compute x_4 using the recurrence and the initial values x_0 = 1 + 1 = 2 and x_1 = a + 1/a = 4.

This is not the most efficient solution (the “repeated squaring” process wins there) but I’ve always found the fact that a sequence formed by powering the roots of a polynomial satisfies a linear recurrence directly determined by that polynomial to be very beautiful! 🙂

There were two moments in these videos that I’ve been thinking about since I saw them last night. The first one was when your son said that the equation was true if x was an integer, and you referred to the graph, noting that x could be any number. He agreed but I wasn’t so convinced (I’m not sure he was either!), because it just intuitively didn’t seem right. I paused and plugged in 1/2 and quickly realized what was happening within the dynamics of the equation, then (and this is the second thing this video brought to mind) I realized that this equation is what I think of as “one of those” equations that math minds recognize: There was a moment for me,years ago, when someone told me that as soon as I see (x^2 – 25) I should know what the factors are without having to stop and figure them out. At that moment I realized that math is full of these recognizable relationships that, at a certain level “everyone knows.” Grasping this concept of recognizable relationships that didn’t have to be worked out fresh each time really opened up things for me.

There’s another direction using the quadratic equation that is very nice and leads to some interesting thinking:

a^2 – 4a +1 =0 can be reinterpreted as a^2 = 4a -1

Using that, we can reduce any polynomial in a to a linear polynomial. Working up the powers, a^3 = 4a^2 – a = 4(4a-1) – a = 15a – 4

a^4 = 15a^2 – 4a = 15(4a – 1) – 4a = 56a-15

Reasoning based on symmetry, we see that the same thing holds for the reciprocal of a, that is a^(-4) = 56/a – 15

Thus, a^4 + a^(-4) = 56a -15 + 56/a -15 = 56 (a + 1/a) -30 = 56 * 4 -30 = 194

This thought process connects to two cool areas of more advanced algebra. First, residue rings. Say f(x) is a polynomial with coefficients in a base ring R, leading coefficient 1, then we can think of f as giving us a way to reduce powers of x in R[x]/(f(x)). In particular, if f is quadratic, then all the elements of R[x]/(f(x)) are nicely represented by linear terms in x (of the form cx + d).

This is exactly what we did before, where we recast all higher powers of a as a linear expression involving a.

Second, but closely related, is the observation that any polynomial expression that is symmetric in a and 1/a, with integer coefficients will have to be an integer. Similarly, if you have rational coefficients. This seems pretty amazing that all the square roots drop out.

Another example is the closed form expression for terms of the Fibonacci involving powers of the golden ratio where the fact that we always get integers appears absolutely magical (to me, at least).

The power reduction observation above helps show why this is true (the reduction steps preserve the symmetry, we end up with a linear expression involving a and 1/a, symmetry means that we can factor all of the a terms out out with a single (a + 1/a), etc). This is a nice example of a key insight from Galois Theory.

I actually thought a little bit about how to approach talking about these sorts of relations with the boys just a few days later:

https://mikesmathpage.wordpress.com/2015/12/25/relations-from-i-to-geometry/

Following up on the Fibonacci connection in Joshua’s reply, since the number a satisfies the quadratic equation x^2 = 4 x – 1, one can check that the sequence of powers 1, a, a^2, a^3,… satisfies the linear recurrence x_n = 4 x_{n-1} – x_{n-2}. The same statement holds for the sequence of powers of the other zero 1/a. Since the recurrence is linear, adding these two sequences term-by-term will yield a new sequence which also satisfies this recurrence. If we write:

x_n = a^n + (1/a)^n

for the nth term of this new sequence, then we can compute x_4 using the recurrence and the initial values x_0 = 1 + 1 = 2 and x_1 = a + 1/a = 4.

We have: 2, 4, 16 – 2 = 14, 56 – 4 = 52 and finally x_4 = 208 – 14 = 194.

This is not the most efficient solution (the “repeated squaring” process wins there) but I’ve always found the fact that a sequence formed by powering the roots of a polynomial satisfies a linear recurrence directly determined by that polynomial to be very beautiful! 🙂

Just noticed question 12 on Patrick Vennebush’s recent post: 16 problems for 2016