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?
Best Answer
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 $\,x = 3\pi/2 = 270^\circ$