/sci/ - Science & Math

[Anonymous No.10987403]

Hello /sci/ I am interning at a middle school as a sub teacher and I want to explain 7th graders how to derive the quadratic formula.
What do you think? Is this clear enough for them to get the main idea? (It is an AP algebra class). I made this derivation myself since a typical derivation you find on the web is too cryptic and doesn't explain the main idea well.

Problem: derive the solution of the quadratic equation [math]ax^2 + bx + c.[/math]
Our derivation is based on the identity: [math](a+b)^2 = a^2 + 2ab + b^2 \qquad (1)[/math]
Any quadratic equation can be transformed to the following form:
[math](x + m)^2 = d \qquad (2)[/math].
Why is it important? Because m and d are constants and we have to solve for just one uknown variable x by taking the square root on both sides:
[math]x + m = \pm \sqrt{d}[/math]
[math]x = -m \pm \sqrt{d}[/math]
Let's consider a simple example [math]x^2 + 6x + 5 = 0[/math]
How dow we convert it to the identity (1)? Well we can first re-write it as
[math]x^2 + 2\cdot x\cdot 3 + 5= 0[/math]
Notice the second term looks like the second term in (1) so b=3. But obviously [math]b^2[/math] then needs to be 9. So we can add 4 to both sides of the equation:
[math]x^2 + 2\cdot x\cdot 3 + 5 + 4 = 4[/math]
[math]x^2 + 2\cdot x\cdot 3 + 3^2 = 4[/math]
[math](x + 3)^2 = 4[/math]
This is the same as form (2). We can now solve it:
[math]x +3 = \pm \sqrt{4}[/math]
[math]x = -3 \pm 2[/math]
[math]x=-1, x=-5[/math]
Now we can apply the same technique to solving the general equation [math]ax^2 + bx +c = 0[/math]. We first divide by a:
[math]x^2 + \dfrac{bx}{a} + \dfrac{c}{a} = 0[/math]
Now we take [math]\dot{b} = \dfrac{b}{a}, \dot{c} = \dfrac{c}{a}; \qquad[/math] And the equation becomes:
[math]x^2 + \dot{b}x + \dot{c} = 0[/math]
Let's add a certain unknown number [math]d[/math] to both sides:
[math]x^2 + \dot{b}x + \dot{c} + d = d[/math]
[cont'd]

[Anonymous No.10987404]

Let's rewrite it as
[math]x^2 + 2x(\dfrac{\dot{b}}{2}) + \dot{c} + d = d[/math]
To convert it to identity (1), we need to make sure
[math]\dot{c} + d = \left(\dfrac{\dot{b}}{2}\right)^2 \qquad(3)[/math]
Then we can rewrite our equation as
[math]x^2 + 2x\left(\dfrac{\dot{b}}{2}\right) + \left(\dfrac{\dot{b}}{2}\right)^2 = d[/math]
[math]\left(x + \dfrac{\dot{b}}{2}\right)^2 = d[/math]
[math]x = - \dfrac{\dot{b}}{2} \pm \sqrt{d}[/math]
Now we get [math]d[/math] from (3):
[math]d = \left(\dfrac{\dot{b}}{2}\right)^2 -\dot{c} = \dfrac{\dot{b}^2}{4}-\dot{c}[/math]
[math]x = - \dfrac{\dot{b}}{2} \pm \sqrt{\dfrac{\dot{b}^2}{4}-\dot{c}}[/math]
And unwinding our substitutions:
[math]x = - \dfrac{b}{2a} \pm \sqrt{\dfrac{b^2}{4a^2}-\dfrac{c}{a}}[/math]
[math]x = - \dfrac{b}{2a} \pm \sqrt{\dfrac{b^2}{4a^2}-\dfrac{4ac}{4a^2}}[/math]
[math]x = - \dfrac{b \pm \sqrt{b^2 - 4ac}}{2a}[/math]

QED

[Anonymous No.10987408]

too much for a typical 7th grader

[Anonymous No.10987421]

>>10987408
I know but this is an AP class and some kids seem to be really motivated...

[Anonymous No.10987423]

Looks fine to me. You might have to spend some time answering questions though. You might want to make the steps explicit when you put it on the board in case their algebra is weak.

[Anonymous No.10987432]

>>10987423
Yes, absolutely, I thought about that. I would have to go back and forth and point at different parts explicitly to show how they are related to make sure they understand the main ideas, since simply reading this line by line may look confusing.

[Anonymous No.10987445]

>>10987432
Double it up with the graphical explanation. The one where you're using the discriminant to find the distance of the points from the axis of reflection. Maybe derive the discriminant and the axis of reflection too.

[Anonymous No.10987618]

[math]
\begin{align*}
ax^2 + bx +c &= 0 \\
ax^2 + bx &= -c \; \; \; \; | \; \cdot 4a \\
4a^2x^2 + 4abx &= -4ac \; \; \; \; | \; +b^2 \\
4a^2x^2 + 4abx +b^2 &= b^2 -4ac \\
(2ax + b)^2 &= b^2 -4ac \\
2ax + b &= \pm \sqrt{b^2 - 4ac} \\
2ax &= -b \pm \sqrt{b^2 - 4ac} \\
\displaystyle
x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\end{align*}
[/math]

[Anonymous No.10987642]

complete the square graphically

[Anonymous No.10987649]

>>10987403
>>10987404
This is filled with a million too many steps, bloated

[Anonymous No.10987671]

not needed

[Anonymous No.10987685]

the quadratic formula is literally isolating x after completing the square.

why are students such brainlets that they need a teacher to teach them, worse, a teacher has to consider if they can even understand the proof.

[Anonymous No.10987688]

why not just use the complete the square method?

[Anonymous No.10987702]

in what fucking country do 12 year olds know quadratic equations? why is the education system in North America so shite?! We learned this shit in 10th grade.

[Anonymous No.10988490]

>>10987642
This, then just get the quadratic formula from that

[Anonymous No.10988539]

>>10987618

How the fuck did you enter the 4a in your second step???

[Anonymous No.10988562]

>>10987702
>10th grade
bitch what country?

[Anonymous No.10988563]

>>10988539
He can do that because he's doing to both sides

It's like x+a=y
Then x+a+1=y+1

Or x+a=y
4a(x+a)=4ay

[Anonymous No.10988569]

>>10988539
4 is the smallest integer
with its square root also integer.
Gives nicer numbers down the line

[Anonymous No.10988574]

>>10987403
>>10987404

Crystal clear, OP. I recommend putting a dabbing Filthy Frank instead of 'QED' at the end, and replacing the variable names with dancing fortnite characters.

[Anonymous No.10988743]

>>10987421
>AP
>7th grade
OP is confirmed middle schooler who needs help with his algebra homework.

[Anonymous No.10988747]

>>10988562
Canada

[Anonymous No.10988872]

>>10987618

What mathematical symbol is ' | ' ?
What does this symbol mean, it is not being used as a division symbol, so what is it?

[Anonymous No.10988878]

>>10988872
it's just a separator

[Anonymous No.10988879]

>>10987403
Just do a change of coordinates to put the parabola in standard position with respect to some uv-plane, and then compute the new coordinates of the axis of the xy-place

[Anonymous No.10988890]

>>10987702
Japan/SKorea/China/Taiwan/India/etc I learned it when I was in middle school. When I moved to US, I realized how retarded US students were with regards to math.

[Anonymous No.10988910]

>>10987403
>derive
what?
>quadratic equation
why is it quadric if it's only a squared number?
>identity
what's an identity?
>Why is it important?
will this come on the exam.....?
>m and d
what's m and d?
>±
it cant be both plus and minus at the same time lmao
> we convert it to the identity (1)
what's an identity?
>re-write it as
why?

I'm bored. can we have class outside?

[Anonymous No.10988928]

>>10987618

No. Second step involves multiplying both sides by 4a, but if a = 0 this makes no sense.

[Anonymous No.10988947]

>>10988928
If a=0 it's not a quadratic equation kek

[Anonymous No.10988950]

>>10988872
Here, it's used to indicate that the operation will be performed to both sides on the next line

[Anonymous No.10988951]

>>10988928
made my day. kek

[Anonymous No.10989138]

>>10988563
>He can do that because he's doing to both sides


This is the shit that you have to explain to those that are a little slower or are prone to questioning things. WHY can you just do this to both sides? WHY did you choose to add 4a? Why not 3a? WHY does this make things better in the long run.

Remember, you have significantly more knowledge than they do. Skipping propositions is only going to hinder the way they understand maths.

I always missed this in my own mathematical schooling, the "why". The teachers always just "did", but never explained why. "It works" or "you're just allowed to do it" aren't sufficient answers.

[Anonymous No.10989171]

>>10989138
>the "why"
they don't teach you that in American, because they expect you to be an engineer or some other form of smooth brained loser.

[Anonymous No.10989212]

>>10989138
i agree in general, but what kind of explanation do you expect for things like adding the same value on both sides of an expression? Is your question why do we need to do it? Or why does it work?

[Anonymous No.10989640]

>>10989212
It's clear THAT it works by following the algebra, but a perfectly reasonable question a student might ask is, "what purpose does it serve to multiply both sides by 4a?" to which a competent teacher should answer, "to neatly complete the square with x on line 5, allowing us to take the square root and finish solving for x."

[Anonymous No.10990065]

>>10988562
Canada

[Anonymous No.10990291]

>>10988928
The solution has 2a in the denominator, thus proving your quadratic accusation has no solution.

[Anonymous No.10991131]

>>10987403
Draw colored squares and label them with the variables.

[Anonymous No.10991139]

>>10989138
This. Professors writings proofs that took centuries to find like it's trivial are the most infuriating thing.

[Anonymous No.10991657]

>>10987421
Pre-Ap?

[Anonymous No.10991708]

>>10987403
>AP Algebra
Fuck off

[Anonymous No.10991779]

>>10988879
Like, let [math]ax^2 + bx + c[/math] be the given parabola. Then we want another coordinate system where the parabola has the simplest possible expression.
So we put [math]u = x + k[/math] and [math]v = y + h[/math].
Then out parabola becomes [math] a(u-k)^2 + b(u-k) + c = v - h \implies a(u^2-2uk+k^2) + b(u-k) + c = v-k \implies au^2 + u(b-2ak) + ak^2 - bk + c = v- h[/math]. So chosing [math]k = b/(2a)[/math] the second term becomes zero and we have [math]au^2 + b^2/(4a) - b^2/(2a) + c = y-h \implies au^2 - b^2/(4a) + c = y - h[/math]. We want to make the constant term disappear, so pick [math]h= b^2/(4a)-c[/math] we get [math]au^2 = v[/math].
So we have [math]u = x + b/(2a)[/math] and [math]v = y + (b^2-4ac)/(4a)[/math].
Now, this coordinate transform clearly sends [math]x = 0\mapsto x = -b/(2a)[/math] and [math]y = 0 \mapsto y = -(b^2-4ac)/(4a)[/math]. See pic related where the green axis are the original ones, the red ones the new axis, A is the original origin and B is the new origin.
Finally, to recover the quadratic formula note that [math]v=u^2\implies y + (b^2-4ac)/(4a) = a(x+b/(2a)) \implies (b^2-4ac)/(4a^2) = (x+b/2a)^2 \implies \pm\sqrt(b^2-4ac)/(2a) = x + b/(2a) \implies x = \frac{-b\pm\sqrt(b^2-4ac)}{2a}[/math].
Where we put y=0 because we want a zero of the polynomial.

[Anonymous No.10991785]

>>10991779
The fuck happened to the last part?

[Anonymous No.10992088]

>>10991785
you really need to use {} not () for sqrt but that shouldn't result in an error. did you check it in the preview window?

[Anonymous No.10992111]

>>10992088
I was phoneposting, so no, I didn't check.

Does this work?
[math]v=u^2 \implies y + (b^2-4ac)/(4a) = a(x+b/(2a))^2 \implies (b^2-4ac)/(4a^2) = (x+b/(2a))^2 \implies \pm \sqrt{b^2-4ac}/(2a) = x + b/(2a) \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]

[Anonymous No.10992122]

>>10992111
Weird, it worked on the preview visualizator

[Anonymous No.10992142]

>>10992111
>>10992122
yeah this should work . must be a phone posting thing

[math]v=u^2 \implies y + (b^2-4ac)/(4a) = a(x+b/(2a))^2 \implies (b^2-4ac)/(4a^2) = (x+b/(2a))^2 \implies \pm \sqrt{b^2-4ac}/(2a) = x + b/(2a) \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]

[Anonymous No.10992145]

>>10992142
[math]v=u^2 \implies y + (b^2-4ac)/(4a) = a(x+b/(2a))^2 \implies (b^2-4ac)/(4a^2) = (x+b/(2a))^2 \implies \pm \sqrt{b^2-4ac}/(2a) = x + b/(2a) \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]

[Anonymous No.10992146]

>>10992145
bug?

[Anonymous No.10992208]

Just do it using geometry.

First of all, prove that the maximum or minimum is located at b^2/2a.
Since you know the quadratic is symmetrical, if you shift the "center" to the middle you know that you can factor it like this (x+a)(x-a).

You can solve it using geometry and algebra, although calculus is another way to do it

[Anonymous No.10992284]

>>10992208
to show you how this works out:
[math]y = g((x-m) + p)((x-m) - p)[/math]
[math]y = g(x^2 - x[(m-p)+(p+m)] - (p-m)(p+m))[/math]
[math]y = gx^2 - 2gmx + g(m^2-p^2)[/math]
from this we discern
[math]g = a[/math]
[math]-2gm = b [/math]
[math]g(m^2-p^2) = c[/math]
solving, we find:
[math]g = a[/math]
[math]m = -\frac{b}{2a}[/math]
[math]p = \pm\sqrt{(\frac{b}{2a})^2 - \frac{c}(a)}[/math]

plug these in to our initial equation, we find

y = a((x + \frac{b}{2a}) \pm \sqrt{(\frac{b}{2a})^2 - \frac{c}(a)})(x + \frac{b}{2a}) \mp \sqrt{(\frac{b}{2a})^2 - \frac{c}(a)}))

thus, we know that when y = 0, either the first or second term must also equal zero

so for the first term
[math]0 = x + \frac{b}{2a}) \pm \sqrt{(\frac{b}{2a})^2 - \frac{c}(a)}[/math]
and the second term
[math]0 = x + \frac{b}{2a}) \mp \sqrt{(\frac{b}{2a})^2 - \frac{c}(a)}[/math]

thus both equations yield
[math]x = -\frac{b}{2a}) \pm \sqrt{(\frac{b}{2a})^2 - \frac{c}(a)}[/math]
which can be factored like this:
[math]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/math]

[Anonymous No.10992295]

[math]y = g((x-m) + p)((x-m) - p)[/math]
[math]y = g(x^2 - x[(m-p)+(p+m)] - (p-m)(p+m))[/math]
[math]y = gx^2 - 2gmx + g(m^2-p^2)[/math]
from this we discern
[math]g = a[/math]
[math]-2gm = b [/math]
[math]g(m^2-p^2) = c[/math]
solving, we find:
[math]g = a[/math]
[math]m = -\frac{b}{2a}[/math]
[math]p = \pm\sqrt{(\frac{b}{2a})^2 - \frac{c}{a}[/math]

plug these in to our initial equation, we find

[math]y = a((x + \frac{b}{2a}) \pm \sqrt{(\frac{b}{2a})^2 - \frac{c}{a})(x + \frac{b}{2a}) \mp \sqrt{(\frac{b}{2a})^2 - \frac{c}{a}))[/math]

thus, we know that when y = 0, either the first or second term must also equal zero

so for the first term
[math]0 = x + \frac{b}{2a}) \pm \sqrt{(\frac{b}{2a})^2 - \frac{c}{a}[/math]
and the second term
[math]0 = x + \frac{b}{2a}) \mp \sqrt{(\frac{b}{2a})^2 - \frac{c}{a}[/math]

thus both equations yield
[math]x = -\frac{b}{2a}) \pm \sqrt{(\frac{b}{2a})^2 - \frac{c}{a}[/math]
which can be factored like this:
[math]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/math]

[Anonymous No.10992314]

>>10992208
So like this:
[math]y = g((x-m) + p)((x-m) - p)[/math]
[math]y = g(x^2 - x[(m-p)+(p+m)] - (p-m)(p+m))[/math]
[math]y = gx^2 - 2gmx + g(m^2-p^2)[/math]
from this we discern
[math]g = a[/math]
[math]-2gm = b [/math]
[math]g(m^2-p^2) = c[/math]
solving, we find:
[math]g = a[/math]
[math]m = -\frac{b}{2a}[/math]
[math]p = \pm\sqrt{(\frac{b}{2a})^2 - \frac{c}{a}}[/math]

plug these in to our initial equation, we find

[math]y = a((x + \frac{b}{2a}) \pm \sqrt{(\frac{b}{2a})^2 - \frac{c}{a}})((x + \frac{b}{2a}) \mp \sqrt{(\frac{b}{2a})^2 - \frac{c}{a}})[/math]

thus, we know that when y = 0, either the first or second term must also equal zero

so for the first term
[math]0 = x + \frac{b}{2a} \pm \sqrt{(\frac{b}{2a})^2 - \frac{c}{a}}[/math]
and the second term
[math]0 = x + \frac{b}{2a} \mp \sqrt{(\frac{b}{2a})^2 - \frac{c}{a}}[/math]

thus both equations yield
[math]x = -\frac{b}{2a} \pm \sqrt{(\frac{b}{2a})^2 - \frac{c}{a}}[/math]
which can be factored like this:
[math]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/math]

[Anonymous No.10992605]

>>10987403
Like others said, I was taught to do it by completing the square. >>10987642
but, algebraically.
If you teach them completing square first, that is moving c over and adding (b/2)^2 to each side to get perfect square on left, and then have them to that to the general form, you'll get the quadratic.

[Anonymous No.10993231]

>>10987702
I was taugh that in middle school in Mexico too.

[Anonymous No.10993264]

>>10987702
I Taught them in 6th grade
in the 10th grade we study trigonometric functions, derivatives and number theory
Pyccкий

[Anonymous No.10993270]

>>10988947
This lol why the fuck would they use the quadratic equation to solve bx + c = 0

[Anonymous No.10993290]

>>10987421
Just go for it. You'll do more of a service engaging the few at the top than pandering to the normies who whine, "when will I ever use this in real life?".

[Anonymous No.10993292]

>>10987445
That seems like itll complicate it and make it more confusing

[Anonymous No.10993343]

>>10988928
[math]x = lim_{a \rightarrow 0} \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/math]

for the top, we find the expansion
[math]\sqrt{b^2 - 4ac} = b\sqrt{1 - 4\frac{ac}{b^2}} \approx b\left 1 - 2\frac{ac}{b^2} - \frac{1}{2} \left \frac{ac}{b^2} \right^2 .......\right[/math]
therefore, approximating
[math]x = lim_{a \rightarrow 0} \frac{-b + b - 2\frac{ac}{b}}{2a} = lim_{a \rightarrow 0} -\frac{c}{b}[/math]

So
[math]x = -\frac{c}{b}[/math]

Which is the correct result

[Anonymous No.10993358]

>>10988928
x = lim_{a \rightarrow 0} \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

for the top, we find the expansion
[math] \sqrt{b^2 - 4ac} = b\sqrt{1 - 4\frac{ac}{b^2}} \approx b\left 1 - 2\frac{ac}{b^2} - \frac{1}{2} \left \frac{ac}{b^2} \right^{2} .......\right [/math]
therefore, approximating
[math]x = lim_{a \rightarrow 0} \frac{-b + b - 2\frac{ac}{b}}{2a} = lim_{a \rightarrow 0} -\frac{c}{b}[/math]

So
[math]x = -\frac{c}{b}[/math]
Which is the correct result

[Anonymous No.10993362]

>>10988928
[math]x = lim_{a \rightarrow 0} \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/math]

for the top, we find the expansion
[math] \sqrt{b^2 - 4ac} = b \sqrt{1 - 4\frac{ac}{b^2}} [/math] [math] \approx b \cdot \left 1 - 2\frac{ac}{b^2} - \frac{1}{2} \left \frac{ac}{b^2} \right^{2} .......\right [/math]
therefore, approximating
[math]x = lim_{a \rightarrow 0} \frac{-b + b - 2\frac{ac}{b}}{2a} = lim_{a \rightarrow 0} -\frac{c}{b}[/math]

So
[math]x = -\frac{c}{b}[/math]
Which is the correct result

[Anonymous No.10993370]

>>10988928
It does make sense. Let's take the limit
[math]x = lim_{a \rightarrow 0} \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/math]

Now since "a" is small, we can expand the radical using a taylor approximation
[math]\sqrt{b^2 - 4ac} = b \sqrt{1 - 4\frac{ac}{b^2}} [/math] [math] \approx b \cdot \left( 1 - 2\frac{ac}{b^2} - \frac{1}{2} \left( \frac{ac}{b^2} \right)^{2} .......\right)[/math]

Putting this into our equation
[math]x = lim_{a \rightarrow 0} \frac{-b + b - 2\frac{ac}{b}}{2a} = lim_{a \rightarrow 0} -\frac{c}{b}[/math]

then taking the limit
[math]x = -\frac{c}{b}[/math]

this is the correct result

[Anonymous No.10993385]

>>10987403
>AP algebra
“Education” companies really have no fucking shame do they. This shit is no doubt a money grab that has no real benefit for society.

[Anonymous No.10993390]

>>10993385
I wish every class in my first two years of uni had an AP equivalent in high school. Too bad they either make you stay back with the herd or if you want IB, have no life. Self paced high school works for certain students.

[Anonymous No.10993586]

OP here. Yes it is a pre-AP (for some reason it's just called AP here, I guess to motivate the kids).
https://pre-ap.collegeboard.org/
I was told some of them have been enrolled since as early as 4th grade. It is common in charter schools around here. And no, the quadratic equations is NOT the standard curriculum in the 7th grade. It is not typically taught until high school, and even then it can be an AP level in the 9th grade. Consider it sort of an experimental class for those willing to go an extra mile. However 7th graders ARE expected to know about the exponents and radicals. The point is to try to explain something they are NOT familiar with based on some common building blocks: equations/exponents/radicals/identities etc.
That said, I talked with my cooperating teachers and they have doubts. They 'recommended' that I just explain the (a+b)^2 identity and stop there haha. According to them, even a lot of high school kids (non-AP) struggle with basic algebra, especially fractions/negative powers etc. But like this anon said: >>10993290
this is my goal, not expecting everyone to get this. But there is a lot of red tape here and they keep rescheduling things so this may never happen. HOWEVER this discussion is incredibly useful for me. Thanks all for your input.
Oh and those who commented on "completing the square", yes, that's what it is. With just a little bit of additional details that are better explained verbally.

[Anonymous No.10993657]

>>10993264
Same. I am from ukraine and qe are studied in 7th-8th grade. At 11th : trigonometric functions, limits, derivatives and a bit of analysis with formalism.

[Anonymous No.10993845]

>>10987403
>>10987649
This. There are too many steps. Smart students will get it, but dumber ones will simply think the derivation is hard and just memorize the formula without giving a fuck about understanding why it works.

You could keep this derivation for exposition, but highlight the general steps. For example, you could do something like below:

If you forgot the formula, you can find it by remembering that:
[math](x+m)^2 = n \Rightarrow x-m\pm\sqrt{n} \\
x^2 + 2mx + (m2-n) = 0[/math]
So we can compare it to the general formula:
[math]ax2+bx+c = 0 \Rightarrow x2 + \frac{b}{a}x + \frac{c}{a}= 0[/math]
Which leads us to the system:
[math]2m = \frac{b}{a} \\ m2-n = \frac{c}{a}[/math]
Then you just solve the system:
[math]2m=\frac{b}{2a} \Rightarrow m=\frac{b}{2a}[/math]
Substituting this result in the second equation:
[math]n = m2-\frac{c}{a} \Rightarrow n = (\frac{b}{2a})^2-\frac{c}{a} \\
n= \frac{(b^2 - 4ac)^2}{(2a)^2}[/math]
Therefore,
[math]x = -\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{(2a)^2}} \\
x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\
x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}[/math]

Now they only need to memorize 3 steps: the perfect square form, general equation and the system that it leads to.

[Anonymous No.10993865]

>>10987702
learned them in 6th grade in america. sorry, you're just dumb.

[Anonymous No.10993873]

>>10993370

Brilliant answer! Good job.

[Anonymous No.10993880]

>>10987403 (OP)
>>10987649
This. There are too many steps. Smart students will get it, but dumber ones will simply think the derivation is hard and just memorize the formula without giving a fuck about understanding why it works.

You could keep this derivation for exposition, but highlight the general steps. For example, you could do something like below:

If you forgot the formula, you can find it by remembering that:
[math](x+m)^2 = n \Rightarrow x = -m \pm \sqrt{n} \\
x^2 + 2mx +(m^2 - n) = 0 [/math]
So we can compare it to the general formula:
[math]ax^2 + bx + c = 0 \Rightarrow x^2 + \frac{b}{a}+\frac{c}{a} = 0[/math]
Which leads us to a system:
[math]2m = \frac{b}{a} \\
m^2 - n = \frac{c}{a}[/math]
Notice that:
[math]2m = \frac{b}{a} \Rightarrow m = \frac{b}{2a}[/math]
Substituting this result in the second equation:
[math]n = m^2 - \frac{c}{a} \Rightarrow n = (\frac{b}{2a})^2 - \frac{c}{a}\\ n = \frac{b^2 - 4ac}{(2a)^2}[/math]
Therefore,
[math]x = - \frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{(2a)^2}}\\
x = - \frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/math]

Now they only need to memorize 3 steps: the perfect square form, general equation and the system that it leads to.

[Anonymous No.10993897]

>>10993880
Btw, obviously assume [math]a \neq 0[/math] and [math]n \geq 0[/math].

[Anonymous No.10994453]

>>10987403
Too deep
Just outline the major steps, then go back and explain how you magically completed the square
Use ordinary language so you don't lose half the class before the third sentence
Your goal is to teach, not to filter

[Anonymous No.10994712]

>>10987403
ax^2 + bx + c = 0\qquad (1)
x^2 + b/a x = -c/a\qquad (2)
but we want to get to something like
(x+d)^2 = x^2 + 2dx + d^2 = e\qquad (3)
So we do
x^2 + 2b/2a x + (b/(2a))2 = (b/(2a))2 -c/a\qquad (4)
and we get
(x+ b/2a)^2 = (b^2-4ac)/4a^2\qquad (5)
x + b/2a = \pm \sqrt{(b^2-4ac)/4a^2} (6)
and finally
x = (-b \pm \sqrt{(b^2-4ac)})/(2a) (7)

This is objectively the correct pedagogy. You need to remind them of what the goal is right after you make the first few intuitive steps.

Once you get to the important point, they'll understand that the rest is rearranging.