When deriving the quadratic equation as shown in the Wikipedia article about the quadratic equation (current revision) the main proof contains the step:
$$
\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}
$$
the square root is taken from both sides, so why is
$$\sqrt{4a^2} = 2a$$
in the denominator and not
$$ \sqrt{4a^2} = 2\left |a \right | $$
Could somebody explain this to me? Thank you very much
Algebra – Why Quadratic Equation Denominator Lacks 2|a|
algebra-precalculusquadratics
Best Answer
One could take the square root as $2|a|$ instead, which would lead to:
$$ x+\frac {b}{2a} = \pm{\frac {\sqrt{b^{2}-4ac}}{2|a|}} \quad\iff\quad x = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{2|a|}} \tag{1} $$
However, given that $\,|a|\,$ is either $\,a\,$ or $\,-a\,$ it follows that $\,\pm|a|=\pm a\,$, so the formula simplifies to:
$$ x = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{2|a|}} = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{\color{red}{2a}}} = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \tag{2} $$
$(1)\,$ and $\,(2)\,$ are entirely equivalent, but $\,(2)\,$ is more convenient to use.