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