Hey, I have a problem: solve for exact value (simplest radical form) $-3\sqrt{27}$ , the result is $-9 \sqrt3$ . I'm in 8th grade studying for a Math placement test to take trigonometry as a freshman next year. This doesn't seem to be covered in my textbook. Can anyone explain to me what's going on here? Thanks!
[Math] Simplifying as an exact value (Simplest Radical Form)
algebra-precalculus
Related Solutions
If you suspect that $\sqrt{97+56\sqrt3}$ (or similar) equals something like $a+\sqrt b$ with $a$, $b$ rational you must have $$97+56\sqrt 3=a^2+b+2a\sqrt b.$$ To match up the surds you'll have to have $a^2+b=97$ and $2a\sqrt b=56\sqrt3$. Thus $a^2b=28^2\times 3=2352$. Therefore $a^2$ and $b$ are the solutions of the quadratic $$x^2-97x+2352=0.$$ The solutions are $x=48$ and $x=49$. Then $a^2=49$ as that's the one that's a square, and $b^2=48$. So $$\sqrt{97+56\sqrt3}=\sqrt{49}+\sqrt{48}=7+4\sqrt3.$$
Expanding on my comments under OP's question, suppose we do not know about the characteristic polynomial, and do not have the foresight to try to find solutions in the form of a geometric progression. Then the $2^{nd}$ order recurrence can still be solved by reducing it to the known case of a $1^{st}$ order recurrence $g_n = \mu g_{n-1}$ which is easily solved by telescoping as $g_n = \mu g_{n-1}=\mu^2g_{n-2}=\dots=\mu^ng_0\,$.
Let the $2^{nd}$ order homogeneous recurrence be $f_n=a f_{n-1}-bf_{n-2}$. We can assume WLOG that $b\ne0$, otherwise it would not be a $2^{nd}$ order recurrence.
The idea is to try to rewrite the relation $f_n=a f_{n-1}-bf_{n-2}$ in the form $g_n = \mu g_{n-1}$ where $g_n=f_n-\lambda f_{n-1}$ is a linear combination between consecutive terms of the original sequence:
$$ g_n = \mu g_{n-1} \;\iff\; f_n - \lambda f_{n-1} = \mu(f_{n-1}-\lambda f_{n-2}) \;\iff\; f_n = \color{red}{(\lambda+\mu)}\,f_{n-1}- \color{green}{\lambda\mu}\, f_{n-2} $$
The last equality matches $f_n=\color{red}{a} f_{n-1}-\color{green}{b}f_{n-2}$ if we choose $\lambda,\mu$ such that $\lambda+\mu=a$ and $\lambda\mu = b$ $\;\left(\dagger\right)\;$ with $\lambda,\mu \ne 0$ because $b \ne 0$.
Then $g_n=\mu^{n-1} g_1\,$, and $f_n$ can be determined by adding the following and telescoping again:
$$ \begin{cases} f_n - \lambda f_{n-1} &= \mu^{n-1}\,g_1 && \mid \,\cdot\,1 \\ f_{n-1} - \lambda f_{n-2} &= \mu^{n-2}\,g_1 && \mid \,\cdot\, \lambda \\ f_{n-2} - \lambda f_{n-3} &= \mu^{n-3}\,g_1 && \mid \,\cdot\, \lambda^2 \\ \dots \\ f_2 - \lambda f_1 &= \mu \,g_1 && \mid \,\cdot\, \lambda^{n-2} \\ f_1 - \lambda f_0 &= \;\;\,g_1 && \mid \,\cdot\, \lambda^{n-1} \end{cases} $$
$$ \implies\;\;\;\;f_n = \lambda^n f_0 + \underbrace{\left(\lambda^{n-1}+\lambda^{n-2}\mu+\dots+\mu^{n-1}\right)}_{=\;\begin{cases}\begin{align} (\lambda^n-\mu^n)/(\lambda-\mu) && \text{if}\;\; \lambda \ne \mu \\ n\,\lambda^{n-1} && \text{if}\;\; \lambda = \mu\end{align}\end{cases}}\,\left(f_1-\lambda f_0\right) $$
$\;\left(\dagger\right)$ Note that, by Vieta's relations, these $\lambda, \mu$ are the roots of the quadratic $t^2 - at + b = 0$, which "happens" to be the characteristic polynomial associated with the original recurrence.
Best Answer
Presumably you are familiar with the rule that (for positive $a$ and $b$) $$\sqrt{a\times b}=\sqrt{a}\,\,\times\sqrt{b}$$ Can you see how to break up 27 into $a\times b$ for the right $a$ and $b$?