I want a very elementary derivation of the quadratic formula. Something accessible to a middle-schooler.
Thanks for the help.
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2This seems elementary enough. – Sahiba Arora Jul 29 '17 at 16:52
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Have a look at this-https://math.stackexchange.com/questions/1586919/derive-quadratic-formula – John_dydx Jul 29 '17 at 17:00
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@SahibaArora +1, I don't think I can make an answer simpler than that, (although 1 most likely exists). – Tobi Alafin Jul 29 '17 at 17:09
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How's this? $$ax^2+bx+c=0$$ $$x^2+\frac{b}{a}x+\frac{c}{a}=0$$ $$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}$$ $$\bigg(x+\frac{b}{2a}\bigg)^2=\frac{b^2}{4a^2}-\frac{c}{a}$$ $$\bigg(x+\frac{b}{2a}\bigg)^2=\frac{b^2-4ac}{4a^2}$$ $$\bigg|x+\frac{b}{2a}\bigg|=\frac{\sqrt{b^2-4ac}}{2a}$$ $$x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$$ $$x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$$ $$\color{red}{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$$
Franklin Pezzuti Dyer
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Every quadratic equation can be expressed in the form: $ax^2 +bx + c$, where:
$a$ is the coefficient of $x^2$.
$b$ is the coefficient of $x$.
$c$ is the constant.
Using completing the square method, we'll derive the quadratic formula.
$(x + k)^2 = x^2 + 2kx + k^2$ is an algebraic identity, and the basis of completing the square method.
First express the quadratic equation in the form: $ax^2 +bx + c$
$$ax^2 + bx + c = 0$$
Divide through by $a$:
$$x^2 + \frac{bx}{a} + \frac{c}{a} = 0$$
Complete the square:
$$x^2 + \frac{bx}{a} + \left(\frac{b}{2a}\right)^2 -\left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0$$
$$\left(x^2 + \frac{bx}{a} + \left(\frac{b}{2a}\right)^2\right) + \frac{c}{a} -\left(\frac{b}{2a}\right)^2 = 0$$
Apply the afore mentioned quadratic identity:
$$\left(x + \frac{b}{2a}\right)^2 + \frac{c}{a} - \left(\frac{b}{2a}\right)^2 = 0$$
$$\left(x + \frac{b}{2a}\right)^2 = \left(\frac{b^2}{4a^2}\right) - \frac{c}{a}$$
Make the denominator common in the RHS above:
$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$$
Take the square root of both sides:
$$\left(x + \frac{b}{2a}\right) = \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
Recall that $\sqrt{x^2} = -x \text{ or } +x \, \forall x \in \mathbb{R}$
$$\left(x + \frac{b}{2a}\right) = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
Make $x$ the subject of the formula:
$$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
$$\therefore x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Q.E.D
Tobi Alafin
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