1. Jan 18, 2012

### Bacle2

Hi, Everyone: I have a job interview tomorrow where I must give a 10-min presentation on the quadratic formula for an intro class , where we are assuming students know both how to factor and how to complete the square. Please comment:

1. The problem statement, all variables and given/known data

O.K. We are given an equation ax2+bx+c=0 , where a,b,c are real numbers, and we want to find the values of x that satisfy the equation -- if these values exist *, i.e., we want to find values of x , so that when we plug them in, in the formula, we get a 0. As an example, if we have x2+2x+1 , we sub-in , say, x=1 , and get:

12+2.1 +1= 4 ≠ 0 , so 1 does not work. We can see also that values like x=0 or x=1 don't work either.

We will see there is a formula that helps us find these values, and we will derive the formula. We will then see some examples of how to use it

* We're not allowing complex numbers yet.

2. Relevant equations

The formula we will use is: -b/2a ±[$\sqrt{b2-4ac}$]/2a

3. The attempt at a solution

Let's see how we get the formula:

ax2+bx+c=0 , and assume a≠ 0 .

We then divide thru by a , to get:

x2+(b/a)x+c/a=0

We then go on to complete the square , by adding and subtracting (b/2a)2

to get:

x2+(b/a)x+ (b/2a)2- (b/2a)2 +c/a=0 , so:

(x+(b/2a))2= (b2/4a2)-c/a=

[b2-4ac]/4a2.

Now we solve for x, by first taking the square root:

x=-b/2a±$\sqrt{b2-4ac}$/2a

And this does it.

Let's see some examples:

First , let's apply it to our original formula x2-2x+1:

here we have :

a=1 , b=-2 , c=1 , so we sub-in the formular to get:

x= -(-2)/2(1) ±√ [ (-2)2-4(1)(1)]/2(1)

=1± √0/2 , so we get that x is a double root .

2. Jan 18, 2012

### ArcanaNoir

You have visuals or plan to draw on the board, right?

3. Jan 18, 2012

### Dick

Well, for one thing you stated an original 'formula' (I'd call it an equation and add =0) of x^2+2x+1 and said that 1 is NOT a root. Then at the end you said "let's apply it to our original formula x^2-2x+1" and showed 1 IS a root. But you've changed the original formula! Are you just teasing them with that maneuver? It's definitely confusing.

Last edited: Jan 18, 2012
4. Jan 18, 2012

### micromass

Staff Emeritus
You might want to talk about what happens if $b^2-4ac$ is >0, =0 or <0.

You also might want to introduce parabola's and tell them that is where it crosses the X-axis.

I would give examples of the quadratic formula first, and only then derive it in the end.

5. Jan 18, 2012

### Bacle2

.

No, my bad; I just wasn't careful enough; I should have written x^2-2x+1 the first time around

ArcanaNoir wrote:

"You have visuals or plan to draw on the board, right? "

Yes, I will be drawing on a board. Good point; this is still difficult, even after a few years of teaching.

Micromass Wrote:

"You might want to talk about what happens if b2−4ac is >0, =0 or <0.

You also might want to introduce parabola's and tell them that is where it crosses the X-axis.

I would give examples of the quadratic formula first, and only then derive it in the end. "

You're right about describing what happens if b2-4ac< ,> or =0 , but the problem is I only have 10 minutes for the whole presentation. Introducing the idea of parabolas may take a good chunk of time, but, given that they already know how to complete squares , this may be easier under the time constraint. I may assign the parabola approach as an extra-credit. And introducing complex numbers may take too much time.

Re giving examples at first, that is a close call: it helps keep their attention, but they may end up staring at the new formula and not paying attention to the rest. A speech prof. commented on this, and I am kind of in the fence here.

6. Jan 18, 2012

### Mentallic

I don't think he meant for you to introduce complex numbers, but instead that you should give examples of parabolas in each case. What does it mean to have $\pm a, a>0$ in the quadratic formula? What about $\pm 0$ and even $\pm \sqrt{-a}$? Don't mention complex numbers, but rather show how the parabolas behave.

Take a few seconds to break down the formula? I always liked that a parabola is symmetric, so you can maybe draw some parabola, then x=-b/2a is the line of symmetry and the $\pm$ tells us how far left and right to go to find where it cuts the x-axis. This would probably come before explaining the discriminant.

7. Jan 19, 2012

### vela

Staff Emeritus

As micromass suggested, you should definitely talk about the discriminant and what happens in the different cases. You might want to use a few examples where you can factor by hand, since the students supposedly know how to do that, and show you get the same result.

8. Jan 19, 2012

### Bacle2

Thanks all Again. I ended up using only what I was told students knew: factoring and completing the square. I gave an example of an expression ax^2+bx+c ; actually using x^2+x+1 (to make clear what the quadratic was for) , showing that x=1, x=0 did not work. Then I stated that one could always find the values for which the quadratic formula works , if these values existed (i.e., are real), and gave the quick proof of the quadratic in my first post.
Then I stated that if one could factor a quadratic (using the second part that students know to factor) into an expression (x-r1)(x-r2) , then r1 and r2 satisfied ax^2+bx+c , but that if the factoring was not obvious, we could use the square.

I just tried to stick to the two things I knew the students had covered, as vela suggested.

Anyway, I did get the job; they must be pretty desperate! :) . Thanks to all again.

9. Jan 19, 2012

### Bacle2

In case someone is looking for a position like this -- at intro level-- I was told by the interviewer that he was looking for a presentation suited to the material that the students were said to already know, which is what vela suggested. I don't know how general this advice is, but just as data for anyone else in a similar position. So no more \$0.99 pizza slice dinners.

10. Jan 19, 2012

### vela

Staff Emeritus
Congrats on the new job!

11. Jan 19, 2012

### micromass

Staff Emeritus
I'm glad you got the job!!

12. Jan 21, 2012

### Bacle2

Thanks again for the good wishes and the comments .