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


Try This!

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.

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


Go to Watch (Steps to Derive the Quadratic Formula)