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How the Quadratic Formula is derived:
We start with any quadratic equation in the form
$ax^{2} + bx + c = 0$
where $a$, $b$, and $c$ are numbers and $a \neq 0$[1]
We solve this general equation for $x$ by completing the square and using the Square Root Property, which says:
If $Z^{2} = Q$, then $Z = \pm \sqrt{Q}$
OK, here we go: to solve $ax^{2} + bx + c = 0$
move the constant term $c$ to the other side:
$ax^{2} + bx = -c$
Now divide both sides by $a$ so that the leading coefficient will be 1.
$\frac{ax^{2}}{a} + \frac{bx}{a} = \frac{-c}{a}$
Simplify:
$x^{2} + \frac{b}{a}x= -\frac{c}{a}$
Now we complete the square on the left-hand side. To find what needs to be added to complete the square, compare:
$x^{2} + 2Ax + A^{2} = (x+A)^{2}$
$x^{2} + \frac{b}{a}x + ?$
This tells us that the coefficient of $x$, which is $2A$ in the first line, must equal the coefficient of $x$ in the second line, which is $\frac{b}{a}$
So $2A = \frac{b}{a}$
$\implies A = \frac{1}{2}\left(\frac{b}{a}\right) = \frac{b}{2a}$[2]
The number we have to add to complete the square is
$A^{2} = \left(\frac{b}{2a}\right)^{2} = \frac{b^{2}}{4a^{2}}$
Add this to both sides of the equation and we get:
$x^{2} + \frac{b}{a}x + \frac{b^{2}}{4a^{2}} = -\frac{c}{a} + \frac{b^{2}}{4a^{2}}$
We’ll simplify the right-hand side by combining over a common denominator $4a^{2}$[3].
To change $-\frac{c}{a}$ so that it has the common denominator, we need to multiply the top and bottom by $4a$:
$-\frac{c}{a}\cdot \frac{4a}{4a} = -\frac{4ac}{4a^{2}}$
So the right-hand side of the equation becomes
$-\frac{c}{a} + \frac{b^{2}}{4a^{2}} = -\frac{4ac}{4a^{2}} + \frac{b^{2}}{4a^{2}} = \frac{-4ac + b^{2}}{4a^{2}} = \frac{b^{2} – 4ac}{4a^{2}}$
Now our equation looks like this:
$x^{2} + \frac{b}{a}x + \frac{b^{2}}{4a^{2}} = \frac{b^{2} – 4ac}{4a^{2}}$
Write the left-hand side in its factored form so we can use the Square Root Property.
$\left(x + \frac{b}{2a}\right)^{2} = \frac{b^{2} – 4ac}{4a^{2}}$
The Square Root Property tells us that
If $\left(x + \frac{b}{2a}\right)^{2} = \frac{b^{2} – 4ac}{4a^{2}}$
then $x + \frac{b}{2a} = \pm \sqrt{ \frac{b^{2} – 4ac}{4a^{2}}}$
That square root on the right-hand side can be simplified:
$\sqrt{ \frac{b^{2} – 4ac}{4a^{2}}} = \frac{\sqrt{b^{2} – 4ac}}{\sqrt{4a^{2}}} = \frac{\sqrt{b^{2} – 4ac}}{2a}$ provided that $a > 0$. We’ll return to that last bit later.
So now our equation looks like
$x + \frac{b}{2a} = \pm\frac{\sqrt{b^{2} – 4ac}}{2a}$
We’re almost finished. We just have to subtract $ \frac{b}{2a}$ from both sides and simplify a bit.
$x = -\frac{b}{2a} \pm\frac{\sqrt{b^{2} – 4ac}}{2a}$
$x = \frac{-b \pm\sqrt{b^{2} – 4ac}}{2a}$
[1] Why do we require that $a \neq 0$ at the beginning?
[2] Dividing by $2$ is the same as multiplying by $\frac{1}{2}$, which is easier to do since we are working with an algebraic fraction here.
[3] We have not yet worked with algebraic fractions (rational expressions) in this course, but it is coming up, so just realize that it is basically the same as working with regular numerical fractions, and do the best you can with this for now.
