$\color{white}{\require{cancel}{3}}$
So I was looking on Youtube for math equations that I thought that I could probably solve when I came across this video by the channel Maths and many more. The question in this video was$$\text{If }a^2-a-1=0\text{, then what is the value of }a^6\text{, where }a\gt0$$which I wanted to try and solve on my own. Here are the steps that I took to solve the equation:$$\text{Put it into the quadratic formula since it is unfactorable with rational numbers}$$$$\frac{1\pm\sqrt{1-4(1)(-1)}}{2}$$$$=\frac{1\pm\sqrt{5}}{2}$$however, since the value of $a$ is greater than $0$, it is actually $$a=\frac{1+\sqrt{5}}{2}$$$$\text{Then, all you have to do is square it}$$$$\left(\frac{1+\sqrt{5}}{2}\right)^2$$$$\implies\frac{1}{4}(1+\sqrt{5})^2\text{, and simplifying that gets}$$$$\frac{1}{4}(6+2\sqrt{5})$$$$\iff\frac{6+2\sqrt{5}}{4}$$$$\iff1.5+\frac{\sqrt{5}}{2}=a^2$$$$\implies(1.5+\frac{\sqrt{5}}{2})^2=a^4$$
| $\color{white}{.}$ | $1.5$ | $\frac{\sqrt{5}}{2}$ |
|---|---|---|
| $1.5$ | $2.25$ | $\frac{3\sqrt{5}}{4}$ |
| $\frac{\sqrt{5}}{2}$ | $\frac{3\sqrt{5}}{4}$ | $1.25$ |
$$2.25+2\left(\frac{3\sqrt{5}}{4}\right)+1.25$$$$\iff3.5+\frac{3\sqrt{5}}{2}=a^4$$
| $\color{white}{.}$ | $3.5$ | $\frac{3\sqrt{5}}{2}$ |
|---|---|---|
| $1.5$ | $5.25$ | $2.25\sqrt{5}$ |
| $\frac{\sqrt{5}}{2}$ | $1.75\sqrt{5}$ | $3.75$ |
$$5.25+(2.25+1.75)\sqrt{5}+3.75=a^6$$$$\implies\therefore9+4\sqrt{5}=a^6$$$$\text{And,when we plug it into Wolfram Alpha:}$$
$$\text{We get the same thing! :)}$$
My question
Is my solution correct, and if not, what could I do to attain the correct solution or what could I do to attain it more easily?
To clarify
- Sorry if this seems like a trivial/short question
- If you want to edit this question to improve it, I am so sorry about the compressed formatting, it's just easier for me to type it like this for it to be faster.
- Sorry if the tags aren't correct, they most likely are but still.
- Sorry if I used any math functions incorrectly.
Best Answer
Your solution is fine, but there are easier ways to do this problem. Specifically, notice that
$$a^2=a+1$$
implies
$$a^6=(a^2)^3=(a+1)^3=a^3+3a^2+3a+1$$ $$=a(a+1)+3(a+1)+3a+1=a^2+a+6a+4$$
$$=(a+1)+7a+4=8a+5$$
Then using the $a$ you found the answer is
$$a^6=4+4\sqrt{5}+5=9+4\sqrt{5}$$