POLQ 2 | Lesson 4 | Try This! (Deriving the Quadratic Formula)

Try This!

Deriving the Quadratic Formula

The standard form of the quadratic equation is \(ax^2+bx+c=0\). Use your understanding of completing the square to put the steps below in the correct order to derive the quadratic formula.

\( \left( x + \dfrac{b}{2a} \right) = \pm \sqrt{ \dfrac{b^2-4ac}{4a^2}}\)
\(x^2 + \dfrac{b}{a} x + \bigg( \dfrac{b}{2a} \bigg)^2 = – \dfrac{c}{a}+\bigg( \dfrac{b}{2a}\bigg)^2 \)
\( \sqrt{\bigg( x + \dfrac{b}{2a}\bigg)^2} = \pm \sqrt{ \dfrac{b^2-4ac}{4a^2}}\)
\(x = \dfrac {-b \pm \sqrt{b^2-4ac} } {2a}\)
\( \bigg( x + \dfrac{b}{2a} \bigg)^2 = -\dfrac{c}{a} \cdot \dfrac{4a}{4a}  + \dfrac{b^2}{4a^2}\)
\( x + \dfrac{b}{2a} = \pm  \dfrac{\sqrt{ b^2-4ac}}{2a} \)
\(x^2 + \dfrac{b}{a}  x + \dfrac{c}{a} = 0\),  \(a \neq 0 \)
\( x = -\dfrac{b}{2a}  \pm  \dfrac{\sqrt{b^2-4ac}}{2a}\)
\( \bigg( x + \dfrac{b}{2a}\bigg)^2 = -\dfrac{c}{a}+\bigg( \dfrac{b}{2a}\bigg)^2 \)
\( x + \dfrac{b}{2a}  = \pm \dfrac{\sqrt{b^2-4ac}}{ \sqrt{4a^2}}\)
\(\bigg( x + \dfrac{b}{2a}\bigg)^2 = -\dfrac{c}{a}  + \dfrac{b^2}{4a^2}\)
\(x^2 + \dfrac{b}{a} x = \ – \dfrac{c}{a} \)
\(\bigg( x + \dfrac{b}{2a}\bigg)^2 = \dfrac{b^2-4ac}{4a^2}\)


Go to Watch (Steps to Derive the Quadratic Formula)