# Explaining the concept of identities in Math to a grade 8 student

I was tutoring a student in the quadratic formula today, and wanted them to see different ways of solving $$8x^2 - 64$$.

Suppose we want to solve $$8x^2 - 64 = 0$$ using the quadratic formula. I then told her that to apply the quadratic formula, we need the values of $$a, b, c$$ in $$ax^2 + bx + c = 0$$.

She identified $$a = 8$$ correctly, but was not able to deduce $$b$$ or $$c$$. She said she knows that $$bx + c = -64$$, but doesn't know how to simplify further. How do I explain to her that $$b = 0, c = -64$$ because we are comparing coefficients in an identity, not solving an equation? She is in grade 8.

• "I was tutoring a student in the quadratic formula today, and wanted them to see different ways of solving $8x^2-64$." Stop right there. One can solve an equation, but $8x^2-64$ isn't an equation – it has no equal sign. Please, if you're teaching someone, use mathematical terms correctly. May 31 at 2:29
• Yeah Gerry, you would never say roots instead of solutions or vice versa. May 31 at 2:39
• C’mon, we all know what was intended here. Don’t be difficult. May 31 at 2:41
• Personally, I don't think I would be introducing the quadratic formula, if the student cannot yet find $a,b$ and $c.$ It sounds like a level of abstraction that the student is not yet ready to tackle. I would build up to this by first looking at the multiplication of binomials, then factoring, and eventually the quadratic formula. May 31 at 6:07

Try from the other way. Start with $$ax^2 + bx + c = 0$$, and say you want to get down to $$8x^2 - 64 = 0$$. So how do we get $$8x^2 - 64 = 0$$ from $$ax^2 + bx + c = 0$$? Well, what we can do is choose values for $$a,b,c$$. So what values do we need to choose for $$a,b,c$$ so that $$ax^2 + bx + c = 0$$ becomes $$8x^2 - 64 = 0$$?

Based on her prior response, she'll probably choose $$a$$ correctly. Next is to guide her to choose $$b,c$$. Make her aware that there is no $$x$$ term in $$8x^2 - 64 = 0$$, and that for almost any $$b$$ she choose, there will be an $$x$$ term left. So what $$b$$ must she choose? Then $$c$$ should be easy to finish off.

IMO she has all right to be confused, because this identification-process is actually not as trivial as we perceive it. Maths education doesn't really explain what's really going on there, and perhaps for good reasons (learning by doing is likely the most effective). But to actually get down to the principles, on the risk of making it way more complicated: what we're dealing with here is a problem of inverse modelling.

The general formulation is this:

You are given two functions, \begin{align} & f : \mathbb{R} \to \mathbb{R} \\ &f(x) = 8x^2 - 64 \end{align} and \begin{align} & g : \mathbb{R}^3\times\mathbb{R} \to \mathbb{R} \\ & g_{(a,b,c)}(x) = ax^2 + bx + c. \end{align} Now your task is to find values for $$a$$, $$b$$ and $$c$$ that make $$g_{(a,b,c)}(x)$$ as similar as possible to $$f(x)$$. In general this would require probing with concrete values and some kind of optimisation procedure (and first of all a notion what we mean by "similar"). We're lucky because we have closed forms for both $$f$$ and $$g$$ and can actually make them equal by mere pattern matching. Specifically, she's already determined that $$a$$ must be $$8$$. This takes care of the quadratic terms, simplifying the problem to finding $$b,c$$ such that $$\tilde f(x) = -64$$ and $$\tilde g_{(b,c)}(x) = bx + c$$ are equal.

This is where it pays off to write it explicitly as functions, because $$\tilde f$$ is still more than just a number literal: it is a function of $$x$$, whose value just happens to not depend on $$x$$. This highlights that what we need to solve is not just $$-64 = bx + c$$ (which would be indeterminate as such), but specifically

$$\forall x\in\mathbb{R} : \quad -64 = bx + c.$$

To have any chance of being true for all numbers, it should certainly be true on any examples, so let's start with the simplest conceivable examples: $$x=0$$.

$$-64 = b\cdot0 + c \quad\Rightarrow\quad c = -64,$$

and then $$x = 1$$,

$$-64 = b + c = b - 64 \quad\Rightarrow\quad b = 0.$$

So at this point we have determined that $$b=0$$ and $$c=-64$$ are the only possible candidate values. Now we still need to show that the expressions are equal for all $$x$$, but that's just a matter of confirming that the $$b=0$$ cancels any variations that $$x$$ could introduce.

Of course, once the student has grokked the principle, she should avoid actually doing it this awkward and slow way and instead just notice that $$f$$ can be rewritten, adding a term that's equal to zero but pattern-matchable:

$$f(x) = 8x^2 + 0x - 64$$

(in German, this is called a "nahrhafte Null", i.e. "nutricious zero").

This is comparable to the difficulty I had in my early stages of learning English: It was difficult for me to understand that "Thank you" is a sentence. Later I was told that it is a grammatically incorrect short form of the grammatically correct "I thank you".

Now you can say that we use such sloppy shortenings in mathematics also which are never stated/taught explicitly. So the quadratic equation "$$3x^2+1x +5$$" is shortened as "$$3x^2+x+5$$" (suppress the coeff 1, except when it is the constant term)

"$$2x^2 + -5x +7$$" is shortened as "$$2x^2-5x +7$$" ( suppress the connecting plus sign when it is followed by a term with negative number as the coefficient)

"$$4x^2 +3x+0$$" is shortened as "$$4x^2+3x$$" (suppress the whole term and the connecting plus sign for it if the coefficient is zero).

Actually one learns these rules subconsciously the way a (pre-school) child learns grammar of its mother tongue.

I was forced to learn these rules explicitly when I tried to write a computer program that converts a list of coefficients into a polynomial. 

This isn’t really a situation where you need to equate coefficients and solve. $$a, b$$ and $$c$$ are simply names given to the coefficients when they are all moved to one side.

Furthermore, have you tried just explaining to her that an identity is true for all x whereas an equation is true for some specific $$x$$ which you need to solve for? Questions like this make one wonder if you even understand the difference between an equation and an identity.

• The OP wishes the student to get comfortable with applying the quadratic formula for solving quadratic equations, one pre-requisite being to identify the correct values of the parameters to use in the quadratic formula. I believe the difficulty is that she doesn't know how to choose $b$ and $c$ for the formula, since the specific example is different form the standard form. May 31 at 2:45
• Yeah and I’m saying that that’s not equating coefficients, or even really about identities. It’s just that $a, b$ and $c$ are names given to the coefficients. May 31 at 2:51
• Yeah, Malady, you would never say identity instead of equation, or vice versa. May 31 at 9:02
• That’s not even what I’m saying Gerry. C’mon bud. May 31 at 10:55

You could ask her to write any quadratic equation first in terms of the general formula $$ax^2 + bx + c$$ in the sense that she looks for the coefficients.

An example would be:

if the equation is $$8x^2 - 64$$

ask her to look for $$x$$ terms and then write it first like $$8x^2 + 0x - 64$$

To explain the concept of coefficients, you could give her some linear equations like $$8x-64=64$$ and show her that even when $$bx-c=c$$, the value of $$b = 0$$

This would lead to the understanding that when there isn't any one coefficient $$a, b, c$$, that value equals $$0$$

John, this is all you need to do: present every quadratic expression to your student in the $$ax^2+bx+c$$ format, using $$b=0$$ or $$c=0$$ if necessary, until she feels confident applying the quadratic formula to any quadratic expression. Once that happens you can start phasing it out, as she will have learned where to place implied coefficients of $$0$$ and that these do not impact her ability to use the quadratic formula.