Quadratic proof

So the quadratic formula and completing the square give the same answer so

\[ax^2 + bx + c \Rightarrow \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}\] Right?

So making a formula for completing the square should lead me to the quadratic formula

  1. \(ax^ + bx + c \Rightarrow x^2 + \frac {bx + c}{a}\)
  2. \(x^2 = \frac {bx}{a} + \frac {c}{a} \Rightarrow (x + \frac {b}{2a})^2 - \frac {b^2}{4a^2} + \frac {c}{a}\)
  3. \((x + \frac {b}{2a})^2 - \frac {b^2}{4a^2} + \frac {c}{a} \Rightarrow (x + \frac {b}{2a})^2 - \frac {b^2 + 4ac}{4a^2}\)

If we assume that \(ax^2 + bx + c = 0\) Then

  1. \((x + \frac {b}{2a})^2 = \frac {b^2 + 4ac}{4a^2}\)
  2. \(x + \frac {b}{2a} = \pm \sqrt {\frac {b^2 + 4ac}{4a^2}} \Rightarrow \frac {\pm \sqrt {b^2 + 4ac}}{\pm \sqrt {4a^2}}\)
  3. \(x + \frac {b}{2a} = \frac {\pm \sqrt {b^2 - 4ac}}{2a}\)

So this then leads to

\[x = \frac {-b \pm \sqrt {b^2 -4ac}}{2a}\]

Note by Jack Rawlin
3 years, 6 months ago

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