Advanced Quadratic and Polynomial Equations
One important realization is precisely how useful it is to recognize function composition when we see. We will examine a series of problems that are the composition of a polynomial, $p(x)$ and another function, $f(x)$. Ordinarily, solving transcendental equations is very difficult, but if we can bring the techniques we have learned from polynomials to bear, we will find that we can solve many more equations that we initially thought we could.
1. Solve $ln(x^2) = (ln(x))^2$.
It may not be immediately obvious, but the preceding equation is quadratic in $\ln(x)$ once we apply some log properties. We proceed to solve this like any quadratic equation from this point. Some people prefer to make the substitution $y = ln(x)$ after the second step and proceed that way. That is fine as well.
$ln(x^2) = (ln(x))^2$
$2ln(x) = (ln(x))^2$
$0 = (ln(x))^2-2ln(x)$
$0 = ((ln(x)-2)ln(x)$
$ln(x) = 0 \rightarrow x = 1$ or $ln(x) – 2 = 0 \rightarrow x = e^2$
Now, we have to be careful when dealing with solutions of transcendental equations and check that they work.
$\begin{equation} \begin{array}
\ln\ln(1^2) \stackrel{?}{=} (\ln(1))^2 \\
\ln(1) \stackrel{?}{=} (0)^2 \\
0 = 0 \\ \\
\ln((e^2)^2) \stackrel{?}{=} (\ln(e^2))^2 \\
\ln(e^4) \stackrel{?}{=} (2\ln(e))^2 \\
4\ln(e) \stackrel{?}{=} (2)^2 \\
4 = 4
\end{array} \end{equation}$
2. Solve $8cos^3(\theta) + 4cos^2(\theta) – 4cos(\theta) – 2 = 0$.
Recall the rational roots theorem. We can use this to extract a rational root. We find that $-\frac{1}{2}$ is a root and we write this as the factor $2cos(\theta)+1$. We can use polynomial long division then to obtain
$(2cos(\theta) + 1)(4cos^2(\theta) – 2) = 0$
The next factor we see is the difference of perfect squares and we can rewrite this as:
$(2cos(\theta) + 1)(2cos(\theta) + \sqrt{2})(2cos(\theta) – \sqrt{2}) = 0$
Now, we have to be aware of properties of trigonometric functions. These functions have many zeroes and we’d like to list them all. We start with
$2cos(\theta) + 1 = 0 \rightarrow cos(\theta) = -\frac{1}{2} \rightarrow \theta = \frac{2\pi}{3} + 2\pi n, \frac{4\pi}{3} + 2\pi n, n \in \mathbb{Z}$
$2cos(\theta) + \sqrt{2} = 0 \rightarrow cos(\theta) = -\frac{\sqrt{2}}{2} \rightarrow \theta = \frac{3\pi}{4} + 2\pi n, \frac{5\pi}{4} + 2\pi n, n \in \mathbb{Z}$
$2cos(\theta) – \sqrt{2} = 0 \rightarrow cos(\theta) = \frac{\sqrt{2}}{2} \rightarrow \theta = \frac{\pi}{4} + 2\pi n, \frac{7\pi}{4} + 2\pi n, n \in \mathbb{Z}$
3. Solve $27e^{3x} – 54e^{2x} + 36e^x – 8 = 0$.
The ease with which you do this problem will depend upon your familiar with cubic polynomials. You may start by employing the rational roots theorem here and that is fine. You might notice that this is a cube of the form $(ae^{x} + b)^3$. Then you obtain
$(3e^x – 2)^3 = 0$ which implies
$3e^x – 2 = 0 \rightarrow e^x = \frac{2}{3}$ which further simplifies to
$x = ln(\frac{2}{3})$.
4. Solve $3sin(x) – 3cos(x) -ln(x)sin(x) +ln(x)cos(x) = 0$.
$3sin(x) – 3cos(x) -ln(x)sin(x) +ln(x)cos(x) = 0$
$3sin(x) – 3cos(x) -ln(x)(sin(x) – cos(x)) = 0$
$3(sin(x) – cos(x)) -ln(x)(sin(x) – cos(x)) = 0$
$(sin(x) – cos(x))(3-ln(x)) = 0$
$ln(x) = 3 \rightarrow x = e^3$ or
$sin(x) = cos(x) \rightarrow tan(x) = 1 \rightarrow x = \frac{\pi}{4} + 2\pi n, \frac{5\pi}{4} + 2\pi n, n\in \mathbb{Z}$.