The following is a derivation of the quadratic formula. Assume \(a, b, c \in \mathbb{R}\)
\( ax^2 + bx + c = 0 \)
The first step will be to complete the square. Factor out \(a\).
\( a \left( x^2 + \frac{b}{a}x + \frac{c}{a} \right) =0 \)
Now add and subtract the square of half of the coefficient of the \(x\) term.
\( a \left( x^2 + \frac{b}{a}x + \frac{c}{a} + \left( \frac{b}{2a} \right)^2 – \left( \frac{b}{2a} \right)^2 \right) = 0 \)
Now complete the square.
\( a \left[ \left( x + \frac{b}{2a} \right)^2 + \frac{c}{a} – \left( \frac{b}{2a} \right)^2 \right] = 0 \)
Now solve for \(x\).
Either \( a=0 \) or \( \left( x + \frac{b}{2a} \right)^2 + \frac{c}{a} – \left( \frac{b}{2a} \right)^2 = 0 \)
Isolate \(x\)
\( \left( x + \frac{b}{2a} \right)^2 = \frac{b^2}{4a^2} – \frac{c}{a} \)
Take the square root of both sides
\( \left| x + \frac{b}{2a} \right| = \frac{ \sqrt {b^2 – 4ac}}{2a} \)
Apply the definition of absolute value to determine intervals.
\( \left| x + \frac{b}{2a} \right| = \begin{cases} \ \ x + \frac{b}{2a}, & x + \frac{b}{2a} \ge 0 \\ -\left( x + \frac{b}{2a} \right), & x + \frac{b}{2a} < 0 \end{cases} \)
For \(x \ge -\frac{b}{2a}\)
\( x + \frac{b}{2a} = \frac{ \sqrt {b^2 – 4ac}}{2a}\)
\( x = – \frac{b}{2a} + \frac{ \sqrt {b^2 – 4ac}}{2a}\)
For \(x < -\frac{b}{2a}\)
\( -\left( x + \frac{b}{2a} \right) = \frac{ \sqrt {b^2 – 4ac}}{2a} \)
\( x = – \frac{b}{2a} – \frac{ \sqrt {b^2 – 4ac}}{2a}\)
Therefore, the formula for the quadratic formula is written as
\( x = \frac{ -b \pm \sqrt {b^2 – 4ac}}{2a}\)
Review the specific example below
\( 3x^2 + 8x + 2 = 0 \)
\( 3 \left( x^2 + \frac{8}{3}x + \frac{2}{3} \right) =0 \)
\( 3 \left( x^2 + \frac{8}{3}x + \frac{2}{3} + \left( \frac{8}{2\cdot3} \right)^2 – \left( \frac{8}{2\cdot3} \right)^2 \right) = 0 \)
\( 3 \left[ \left( x + \frac{8}{6} \right)^2 + \frac{2}{3} – \left( \frac{8}{6} \right)^2 \right] = 0 \)
\( a \neq 0 \) we only need to consider \( \left( x + \frac{4}{3} \right)^2 + \frac{3}{2} – \left( \frac{4}{3} \right)^2 = 0 \)
\( \left( x + \frac{4}{3} \right)^2 = \frac{4^2}{3^2} – \frac{2}{3} \)
\( \left| x + \frac{4}{3} \right| = \frac{ \sqrt {64 – 24}}{6} \)
\( \left| x + \frac{4}{3} \right| = \frac{ \sqrt {40}}{6} \)
\( \left| x + \frac{4}{3} \right| = \frac{ \sqrt{4} \sqrt {10}}{6} \)
\( \left| x + \frac{4}{3} \right| = \frac{ 2 \sqrt {10}}{6} \)
\( \left| x + \frac{4}{3} \right| = \frac{ \sqrt {10}}{3} \)
\( \left| x + \frac{4}{3} \right| = \begin{cases} \ \ x + \frac{4}{3}, & x + \frac{4}{3} \ge 0 \\ -\left( x + \frac{4}{3} \right), & x + \frac{4}{3} < 0 \end{cases} \)
\( x + \frac{4}{3} = \frac{ \sqrt {10}}{3}\)
\( x = \frac{ – 4 + \sqrt {10}}{3}\)
\( -\left( x + \frac{4}{3} \right) = \frac{ \sqrt {10}}{3}\)
\( x = \frac{ -4 – \sqrt {10}}{3}\)