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I'd like to know if the approach I took below (solving quadratic formula for function $u$) was valid.

Solve for $\theta$:

$2\cos \theta = 3\cos^2 \theta - 1$

Let $u = \cos \theta$:

$0 = 3u^2 - 2u - 1$

Apply quadratic formula:

$u = \frac{1\pm 2}{3}$

$\cos \theta_1 = 1, \cos \theta_2 = -\frac{1}{3}$

$\theta_1 = \arccos(1), \theta_2 = \arccos(-\frac{1}{3})$

I only wanted $\theta$ in $(0..\pi)$, which is $\theta_2$, which I happen to know is indeed the correct answer.

Again, what I'd like to know is if my approach was valid. Specifically, substituting a variable for a function, solving the quadratic equation for that variable, then solving for the function equal to that variable. It felt wrong, but gave the correct answer. Could I do this for any $u$, or was I just fortunate that $\cos \theta$ met some condition? I tried searching the web but everything I found talking about the quadratic formula only showed examples for $x$.

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This is absolutely valid. In fact, there was no need to use the formula. We have $3\cos^2\theta-2\cos\theta-1\equiv (3\cos\theta+1)(\cos\theta-1)$. –  Fly by Night Oct 9 '13 at 19:21

2 Answers 2

up vote 2 down vote accepted

This is absolutely valid. In fact, the quadratic formula was unnecessary. Just as factorsies to give $$3u^2-2u-1 \equiv (3u+1)(u-1),$$ we have the exactly analogous expression $$3\cos^2\theta-2\cos\theta-1 \equiv (3\cos\theta+1)(\cos\theta-1)$$ Solutions to $3\cos^2\theta-2\cos\theta-1=0$ are then given by $3\cos\theta+1=0$ or $\cos\theta-1=0$.

This works for all kinds of functions. For example $$\operatorname{e}^{2x}-3\!\operatorname{e}^x+2 \equiv \left(\operatorname{e}^x-1\right)\left(\operatorname{e}^x-2\right)$$ and so $\operatorname{e}^{2x}-3\!\operatorname{e}^x+2=0$ over the reals if, amd only if, $\operatorname{e}^x=1$ or $\operatorname{e}^x=2$, i.e. $x=0$ or $x = \ln 2$.

~~~ EDIT ~~~

In general, you may apply the formula. For example $$\cos^2\theta + 4\cos\theta + 1=0 \iff \cos\theta = -2\pm\sqrt{3}$$ $$2\ln^2x+3\ln x-4=0 \iff \ln x= -\tfrac{3}{4} \pm \tfrac{1}{4}\sqrt{31}$$

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thanks. if i remember correctly, one way to derive the quadratic formula is to try to factorize the generic problem. i was mainly just concerned about substituting an arbitrary function for x then applying the quadratic formula –  Kip Oct 9 '13 at 19:42
    
@Kip The standard way to derive the formula is to "complete the square", e.g. to write $x^2+4x+1=0$ as $(x+2)^2-3=0$. Then $(x+2)^2=3$ and so $x=-2\pm\sqrt{3}$. This is different to factorising. To factorise, you would need to spot that $x^2+4x+1 \equiv (x+2-\sqrt{3})(x+2+\sqrt{3})$, i.e. spot the solution straight away. –  Fly by Night Oct 9 '13 at 19:48

Your approach is absolutely correct, as long as the quantity $x$ (where $x$ is a variable, function, power of a variable, etc.) exactly matches at all terms.

This method is also applicable for solving an equation in multiple variables in terms of one of those variables.

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