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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

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6 Answers 6

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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.

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When taking the square root we put a $\pm$ on the right hand side to account for the two roots, so it is unnecessary to strip off the sign of $a$, as we will put it back anyways.

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The two square roots of $a^2$ are $a$ and $-a$, sometimes written together as $\pm a$.

For real numbers $\pm a$ is equivalent to $\pm |a|$ but this is not true for complex numbers. So putting the absolute value operation in would make the proof less general.

We could write the proof as

$$ \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}} $$

$$ \pm\left(x+{\frac {b}{2a}}\right)={\frac {\pm\sqrt{b^{2}-4ac}}{\pm2a}} $$

$$ x+{\frac {b}{2a}}={\frac {\pm\sqrt{b^{2}-4ac}}{2a}} $$

But generally it is considered sufficient to put in just a single $\pm$ from the start rather than putting in one for each square root and then removing the redundant ones.

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If you put $x =\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} $ into $ax^2+bx+c$, since $x^2 =\dfrac{b^2\mp2b\sqrt{b^2-4ac}+(b^2-4ac)}{4a^2} =\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a^2} $ you get

$\begin{array}\\ ax^2+bx+c &a\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a^2} +b\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}+c\\ &=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a} +\dfrac{-2b^2\pm 2b\sqrt{b^2-4ac}}{4a}+c\\ &=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}-2b^2\pm 2b\sqrt{b^2-4ac}+4ac}{4a}\\ &=0\\ \end{array} $

If you use $|a|$, it won't work since you can't combine the terms.

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    $\begingroup$ The OP is only putting the absolute value when taking the square root, not putting it on $-b/2a$. This is correct, but unnecessary, see dxiv's answer. $\endgroup$
    – Fan Zheng
    Jun 25, 2018 at 18:29
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my preference for remembering and using the quadratic formula (and electrical engineers seem to do that often) is to remember the root quadratic equations as:

$$ x^2 \ + \ b\,x \ + \ c \ = \ 0 $$

which has solution:

$$ x \ = \ \begin{cases} -\tfrac{b}{2} \pm \sqrt{\left(\tfrac{b}{2}\right)^2 - c} \qquad & \text{for }\left(\tfrac{b}{2} \right)^2 > c \\ \\ -\tfrac{b}{2} \qquad & \text{for }\left(\tfrac{b}{2} \right)^2 = c \\ \\ -\tfrac{b}{2} \pm i \sqrt{c - \left(\tfrac{b}{2}\right)^2} \qquad & \text{for }\left(\tfrac{b}{2} \right)^2 < c \\ \end{cases}$$

normalizing out the "$a$" does not make the quadratic equation less general. the only degrees of freedom are $b$ and $c$, so that means normally (except for a double root), there are two independent solutions.

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Alternatively, noting $a\ne 0$: $$\begin{align}\left(x+{\frac {b}{2a}}\right)^{2}&={\frac {b^{2}-4ac}{4a^{2}}} \iff \\ 4a^2\left(x+{\frac {b}{2a}}\right)^{2}&=b^{2}-4ac \iff \\ \left(2ax+b\right)^{2}&=b^{2}-4ac \iff \\ 2ax+b&=\pm \sqrt{b^2-4ac} \iff \\ x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.\end{align}$$

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