1
$\begingroup$

The following is an abbreviation of a common derivation of the quadratic formula: $$ ax^2+bx+c=0\\ \vdots\\ \sqrt{(x+\frac{b}{2a})^2}=\sqrt{\frac{b^2-4ac}{4a^2}}\\ x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}\\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ $$ Following the square root extraction, the equation should have three plus-minus signs: $$ \pm(x+\frac{b}{2a})=\frac{\pm\sqrt{b^2-4ac}}{\pm2a}\\ $$ But instead there's only one. Is it because the signs cancel each other out, or are moved from one expression to the other (through multiplication in $-1$ or $\frac{-1}{-1}$) in six of the eight permutations that the three plus-minus signs allow; or is it a matter of choosing a principal root for some arcane reason beyond the scope of elementary algebra (eg. branch points)?

$\endgroup$
  • 1
    $\begingroup$ A simpler way of looking at it is if $u^2 = v^2$, then $u = v$ or $u = -v$. $\endgroup$ – N. F. Taussig Feb 14 '16 at 15:37
  • 1
    $\begingroup$ One avoids at least half the problem by initially not dividing by $a$, but multiplying by $4a$. $\endgroup$ – André Nicolas Feb 14 '16 at 15:59
  • $\begingroup$ "Multiplying by $4a$": math.stackexchange.com/questions/49229/… $\endgroup$ – David K Feb 14 '16 at 16:20
3
$\begingroup$

There shouldn't be $2$ plus-minus signs in the fraction:

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

Once we put the plus-minus sign up, we simplify the radicals without worrying about plus-minus signs.

Why don't we have the plus-minus sign on the left side?

Well, if we have (as N.F.Taussig mentions) $$u^2=v^2$$, then it simplifies down to $u=\pm v$ or $\mp u=v$.

We just decided to put it on the right because we were solving for $x$.

You don't want two plus-minus signs because they sort of merge into one plus-minus sign:

$$\pm u=\pm v$$

$$u=\pm(\pm v)=+v,-v$$

It can only come out to be positive or negative, and so we just merge it together.

$\endgroup$
0
$\begingroup$

Cancellation and "moving" signs (as you put it) are one explanation for why only one $\pm$ sign is required. We know that \begin{gather} \frac{-p}{-q} = \frac{p}{q}, \\[.5ex] -\left(\frac{p}{q}\right) = \frac{-p}{q} = \frac{p}{-q}, \\[.5ex] p = -q \quad\text{if and only if}\quad -p = q. \end{gather} Using these facts we can show that the next four equations are all equivalent, and hence by stating any one of them we also imply that each of the others is true: \begin{align} x+\frac{b}{2a} &= \frac{\sqrt{b^2-4ac}}{2a}, & x+\frac{b}{2a} &= \frac{-\sqrt{b^2-4ac}}{-2a}, \\ -\left(x+\frac{b}{2a}\right) &= \frac{-\sqrt{b^2-4ac}}{2a}, & -\left(x+\frac{b}{2a}\right) &= \frac{\sqrt{b^2-4ac}}{-2a}. \end{align}

The following four equations also are all equivalent to each other: \begin{align} x+\frac{b}{2a} &= \frac{-\sqrt{b^2-4ac}}{2a}, & x+\frac{b}{2a} &= \frac{\sqrt{b^2-4ac}}{-2a}, \\ -\left(x+\frac{b}{2a}\right) &= \frac{\sqrt{b^2-4ac}}{2a}, & -\left(x+\frac{b}{2a}\right) &= \frac{-\sqrt{b^2-4ac}}{-2a}. \end{align}

By asserting the equation $$\pm\left(x+\frac{b}{2a}\right)=\frac{\pm\sqrt{b^2-4ac}}{\pm2a},$$ then, assuming we can choose each of the $+$ or $-$ signs independently, we assert that at least one of the previous eight equations is true. So either one of the first four of those equations is true, in which case $x+\frac{b}{2a} = \frac{\sqrt{b^2-4ac}}{2a}$ (as implied by any of those four equations), or one of the other four of those equations is true, in which case $x+\frac{b}{2a} = \frac{-\sqrt{b^2-4ac}}{2a}$. That is, the equation with three $\pm$ signs is simply asserting that $$ x+\frac{b}{2a} = \frac{\sqrt{b^2-4ac}}{2a} \quad\text{or}\quad x+\frac{b}{2a} = \frac{-\sqrt{b^2-4ac}}{2a}, $$ which is just what we mean when we write $$x+\frac{b}{2a} = \frac{\pm\sqrt{b^2-4ac}}{2a}.$$

It is simpler, however, if we apply the rules of sign cancellation, etc., earlier in the procedure, as detailed in the previous comments and answer, so that we never end up with three $\pm$ signs in the first place.

$\endgroup$
0
$\begingroup$

There are 2 roots and 4 possibilities. Which two will you eliminate? Answer is in the selection criterion.

$\endgroup$
0
$\begingroup$

Actually, there're cyclic properties for higher order equations.

For cubic $x^{3}+3px=2q$,

$$x_{k}=e^{\frac{2i\pi k}{3}} \sqrt[3]{q+\sqrt{p^{3}+q^{2}}}+ e^{\frac{4i\pi k}{3}} \sqrt[3]{q-\sqrt{p^{3}+q^{2}}}$$ for $k=0,1,2$.

We have same set of roots by writting $x_{k}$ as $$e^{\frac{4i\pi k}{3}} \sqrt[3]{q+\sqrt{p^{3}+q^{2}}}+ e^{\frac{2i\pi k}{3}} \sqrt[3]{q-\sqrt{p^{3}+q^{2}}}$$

For biquadratic $x^{4}-2ax^{2}+b=0$, $$x=\pm \sqrt{a \pm \sqrt{a^{2}-b}}= \pm \sqrt{\frac{a+\sqrt{b}}{2}} \pm \sqrt{\frac{a-\sqrt{b}}{2}}$$ Four combinations of $\pm$ signs giving four distinct roots.

Also de Moivre's quintic $x^{5}+5ax^{3}+5a^{2}x=2b$, $$x_{k}=e^{\frac{2i\pi k}{5}} \sqrt[5]{b+\sqrt{a^{5}+b^{2}}}+ e^{\frac{8i\pi k}{5}} \sqrt[5]{b-\sqrt{a^{5}+b^{2}}}$$ for $k=0,1,2,3,4$.

It also gives same set of roots by writing $x_{k}$ as $$x_{k}=e^{\frac{4i\pi k}{5}} \sqrt[5]{b+\sqrt{a^{5}+b^{2}}}+ e^{\frac{6i\pi k}{5}} \sqrt[5]{b-\sqrt{a^{5}+b^{2}}}$$

Last but not least, there are $n$ roots for $n$ degree polynomial equation. Redundancy of root (don't mess up with multiple roots) means you've done an extra cycle, but make sure you don't miss any.

$\endgroup$
0
$\begingroup$

The solutions of $Y^2=(b^2-4 a c)/4 a^2\;$ are $\;Y=+\sqrt {b^2-4 a c}/2 a\;$ and $\;Y=-\sqrt {b^2-4 a c}/2 a.$

If $(x+b/ 2 a )^2=Y^2$ then $x+b/2 a$ must be equal to one of these solutions or to the other.

So $x+b/2 a=\pm \sqrt {b^2-4 a c}/2 a$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.