# Coefficients inquiry

# The prompt

**Mathematical inquiry processes: **Test different cases; generalise and prove. **Conceptual field of inquiry: **Factorisation of quadratic expressions; quadratic formula; the discriminant.

The coefficients prompt is designed to be used after students have learnt about solving quadratic equations by factorising (**solving quadratic equations prompt**).

It provides an opportunity for the teacher to introduce the quadratic formula and the discriminant, particularly when students discover that they cannot solve an equation with the method they have learnt previously. Indeed, students might select a **regulatory card** to request an explanation of an alternative method.

In the orientation phase of the inquiry, students aim to understand the prompt. Firstly, in the general form of a quadratic equation *ax*^{2} + *bx* + *c* = 0 the coefficients (*a* and *b*) and constant (*c*) are all odd. Secondly, if *a, b *and *c* are odd, then there are no integer solutions.

Another way of saying the second point is that you cannot factorise the quadratic expression on the left-hand side of the equation. If you could, the solutions would be integers. When *a* = 1, the factorisation would be in the form (*x *+ *p*)(*x* + *q*) where *p* and *q* are integers, giving -*p* and -*q *as the two solutions. When *a* > 1 the factors are (*mx* +* p*) and (*nx* + *q*). In this case, the solutions (-*p*/*m* and -*q*/*n*) might or might not be integers. In the classroom, it has been useful to explain this idea carefully.

At the start of the inquiry, students pose questions and make observations:

A quadratic equation has an

*x*^{2}.Is the quadratic equation equal to zero?

A quadratic equation could be 2

*x*^{2}+ 4*x*+ 7A quadratic equation you can factorise is

*x*^{2}- 5*x*- 24*x*- 8)(*x*+ 3) = 0.Why are they all odd? Do they have to be odd?

What does an 'integer solution' mean?

If there isn't an integer solution, could we have decimals?

The solution is the value of

*x*that makes the quadratic equation work.

# Reasoning through inquiry

**Savneet Bhadare***, a teacher of mathematics in west London (UK), used the coefficients prompt in sessions designed to introduce students to post-16 study. The prompt bridges between concepts at GCSE and A-level. Savneet describes how the inquiry developed: *

Students learn how to solve quadratic equations with the formula at GCSE. I wanted to draw on this knowledge, but also extend it in the sessions. Using the discriminant to decide how many roots an equation has would be new learning.

The sessions started with students trying to understand the prompt. I encouraged contributions with probing questions:

What words do you recognise in the prompt?

Can you give me an example of a quadratic equation?

How would you solve it?

Students then made up numerical examples and explored. I taught them about the discriminant. If it showed an equation had roots, students used the formula to find them.

I gave them the example sets (below). In each set, the discriminants for the first and third equations are less than zero and they do not have roots. The second and fourth equations have solutions, but they are not integers.

Students in all the sessions began to conjecture that the prompt is true.

Reasoning

Pardeep, one of the students in a session, reasoned that when the constant is odd, the only way to factorise the equation is with odd numbers: (*x* + odd)(*x* + odd).

That's because if the product of two numbers is odd, both numbers must be odd. But then *b* will be even because:

(odd)*x* + (odd)*x* = (even)*x *and (odd)*x* - (odd)*x *= (even)*x*

Pardeep used an example to illustrate his reasoning. If *c *= 21 (and *a *= 1), the four ways to factorise all lead to an even value for *b*:

(*x* + 3)(*x* + 7), *b* = 10

(*x* + 3)(*x* - 7), *b* = -4

(*x* - 3)(*x* + 7), *b* = 4

(*x* - 3)(*x* - 7), *b* = -10

Proof

Another student, Marisa, tried to prove the prompt is always true. She used an approach she had learned at GCSE. In her proof, 2*n* - 1 stands for any negative number.

The working out should end 2*n* = 1, in which case *n* = 0.5 and 2*n* - 1 = 0. Despite the impressive algebra, Marisa proved that 0*x*^{2} - 0*x* - 0* *= 0 rather than construct a general proof.

The sessions ended with a proof by contradiction based on Pardeep's reasoning (see below). Again, the prompt acted as a bridge from algebraic forms of proof at GCSE to a new form of proof at A-level.

August 2022

# Proof by contradiction

The proof of the statement in the prompt is accessible to students in upper secondary school. It starts with two assumptions - that *a*, *b* and *c *are odd and that there are real roots - and then contradicts the assumptions.

As the chain of reasoning leads to a contradiction, the assumptions must be false and there are no integer solutions. The prompt is true for all cases in which *a*, *b* and *c *are odd.