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I was wondering if it's erroneous to use the quadratic formula on a quadratic equation where there is no constant term. What I figured I'd try was to just assome the constant term is +0.

I was doing a trigonometric equation, which looks like this:

  • $2\cos^2 x - 3\sqrt 3 \cos x = 0$

I did like this:

  • $\cos x = y$

    $2y^2 - 3\sqrt 3y = 0$

    $$y = \dfrac{3\sqrt 3 \pm \sqrt{27 - 4(2)(0)} }4$$

and ended up down the line with y equaling 0 (when confined to a domain between -1 and 1). When I try doing the equation on my calculator with x equaling 90 or 270 (as it has to be for cos x to equal 0), it does end up being 0. So I did arrive at the right answer doing it this way, but for all I know that might just be a lucky coincidence. So I'm wondering if it's right to use the quadratic formula in this case, or if I just lucked out on having it work?

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There is a constant term, and you found it correctly. This approach is fine, though if you know the constant is zero there are easier ways. –  Ross Millikan Dec 25 '13 at 15:31
    
If there is no constant term, just factor (and do not forget that y=0 is one of the two solutions). –  Claude Leibovici Dec 25 '13 at 15:51

4 Answers 4

up vote 10 down vote accepted

That is correct, you can use the quadratic formula for $c = 0$. And your work is all fine.

But note, you can save yourself time by simply factoring your equaton, and noting that when once has factors $a, b$, then $$a b = 0 \;\text{ if and only if } \;a= 0 \;\text{ or }\;b = 0$$


$$\begin{align} 2y^2 - 3\sqrt 3y = 0 & \iff y\,(2y - 3\sqrt 3) = 0\\ \\ &\iff y = 0\;\text{ or }\; \left(2y - 3\sqrt 3 = 0 \iff \;y = \frac{3\sqrt 3}2\right)\end{align}$$

Don't forget that $y = \cos x$, so we need to solve for $x$ given $$\cos x =y = 0\text{ or } \cos x = y = \dfrac {3\sqrt 3}{2}$$ and you are correct that we can throw out the possibility that $\cos x = \frac{3\sqrt 3}2 > |1|$.

So, for $x\in [0, 2\pi) = [0^\circ, 360^\circ) $ then indeed, $x = \pi/2 = 90^\circ, $ or $\,x = 3\pi/2 = 270^\circ$

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Ah, now that you show it like that I see I don't even need to change cos x into y, I can just factor it as cosx(2cosx - 3√3) and solve for both inside and outside the parentheses. Thanks! –  Threethumb Dec 25 '13 at 15:36
    
You're welcome! –  amWhy Dec 25 '13 at 15:37
    
Maybe you forgot to write $y=0$ at the second sentence after the line. –  mathlove Dec 25 '13 at 15:38
    
@mathlove Second statement after the line: $$ \iff {\bf y = 0}\;\text{ or }\; (2y - 3\sqrt 3 = 0 \iff \;y = \frac{3\sqrt 3}2)$$ bold face added (also just added parentheses to clarify the disjunction). –  amWhy Dec 25 '13 at 15:41

Yes, you can mechanically apply the quadratic formula when $\,c=0,\,$ so the trinomial $\,ax^2+bx+c\,$ degenerates to the binomial $\,ax^2+bx.\,$ But that is not a very inefficient way to proceed because binomials are easily solved by factoring, viz. $\,ax^2+bx = (ax+b)x$.

Why does the quadratic formula work in this degenerate case? If you examine the standard proof of the quadratic formula obtained by completing the square, you will see that it proceeds as follows

$$\, f(x)=0 \iff f_1(x) = f_2(x)\iff f_3(x) = f_4(x) \iff \cdots \iff x =\, \ldots$$

For example the first step subtracts $\,c\,$ from both sides. This yields an equivalent equation since $\,r = s\!\iff\! r-c = s-c.\,$ However, later steps multiply by $\,4a$ and divide by $2a$, and this yields an equivalent equation iff $\,a\ne 0,\,$ since the direction $(\Leftarrow)$ of $\ r = s\! \iff\! ar = as\ $ fails when $\,a=0,\, r\ne s.\,$ So the proof breaks down when $\,a=0,\,$ as is apparent in the obtained formula, which involves division by $a$. But the proof works fine if $\,b=0\,$ or $\,c=0\,$ since all steps in the proof remain equivalences for all values of $\,b,c\,$ (in particular, the proof never scales an equation by $\,b,c\,$ or their inverses, as it does for $\,a).\,$

But there is a way of making the quadratic formula work for $\,a = 0.\,$ Namely, use the quadratic formula for $X = 1/x$. It is a root of the reversed polynomial $\,c X^2 + b X + a,\, $ to which we can now apply the quadratic formula, and invert the result to obtain $x = 1/X\,$ (here we assume $c\ne 0$, else $\,c=0=a\,$ so the equation is trivial: $\, bx = 0).$ This solution by "inversion symmetry" is equivalent to "rationalizing the numerator" of the quadratic formula. See this answer for further discussion.

In any case, as we see above, understanding the genesis of the quadratic formula allows use to better understand when it works, and how to tweak it to work in cases when it doesn't immediately apply. This knowledge will also come in handy later when you try to apply the quadratic formula to more general number systems, where the quadratic formula can also break down in other ways, e.g. $\, x^2 \equiv 1\,$ has $\,4\,$ roots $\,x \equiv \pm1, \pm3\,$ in $\,\Bbb Z/8 = $ integers mod $8$. Intuition usually trumps rote mechanical calculation. It is almost always worthwhile to analyze a problem first, before diving head-first into mechanical, rule-based calculations.

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+1, I have never thought of inverting the polynomial, although this is a simple idea. –  Ian Mateus Dec 25 '13 at 17:24
    
+ 1: "Intuition usually trumps rote mechanical calculation...." Hear, hear! (And a nice "blurb" for one's profile!) –  amWhy Dec 26 '13 at 13:01

If the constant term is $0$, the following is better :

$$y(2y-3\sqrt 3)=0.$$ Then?

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y = 0 / (2y-3√3) | y = 0? –  Threethumb Dec 25 '13 at 15:29
    
NO! $y=0\ \text{or} \ y=\frac{3\sqrt 3}{2}.$ –  mathlove Dec 25 '13 at 15:29
    
Ah, yes, you're right. Thanks! –  Threethumb Dec 25 '13 at 15:40
    
You are welcome! –  mathlove Dec 25 '13 at 15:41

$$\text{As the roots of }ax^2+bx+c=0$$ $$\text{are }\frac{-b\pm\sqrt{b^2-4ca}}{2a}$$

If $c=0,$ the roots become $\displaystyle0,-\frac ba$

If $b=0,$ the roots become $\displaystyle\pm\frac{\sqrt{-4ca}}a$

If $b,c$ both are zero, $x=0$

I leave the following as exercise, but with a hint :

If $a=0,$ set $x=\frac1y$

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That way of handling $\,a=0\,$ is explained further in this answer. It can also be done, equivalently, by "rationalizing the numerator", see e.g. Arturo's answer in the same thread. –  Bill Dubuque Dec 25 '13 at 16:07

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