These are the more difficult questions at the end of my chapter and I REALLY have no idea how to even approach them.
Would any of you mind giving me a few hints for them and nudging me
in the right direction?
Hello prettiestfriend(5) $\displaystyle \sqrt{\frac{3+2\sqrt2}{3-2\sqrt2}} =\sqrt{\frac{(3+2\sqrt2)^2}{(3-2\sqrt2)(3+2\sqrt2)}}$
$\displaystyle = \frac{3+2\sqrt2}{3^2 - (2\sqrt2)^2}$
$\displaystyle =3 + 2\sqrt2$
(6) $\displaystyle \frac{1}{\sqrt3}-\frac{1}{\sqrt2}=\frac{\sqrt2-\sqrt3}{\sqrt2\sqrt3}=p$
$\displaystyle \Rightarrow \left(\frac{\sqrt2-\sqrt3}{\sqrt2\sqrt3}\right)^2=p^2$
$\displaystyle \Rightarrow \frac{2 - 2\sqrt2\sqrt3 + 3}{6} = p^2$
Can you complete this now?
(7) $\displaystyle x = \sqrt5-\sqrt3$
$\displaystyle \Rightarrow x^2 = 5 - 2\sqrt5\sqrt3 + 3 = 8 - 2\sqrt{15}$
$\displaystyle \Rightarrow x^4 = 64 - 32\sqrt{15} + 4\times 15 = 124 - 32\sqrt{15}$
Can you complete it?
(8) For $\displaystyle n=1: u_1 = \frac{1}{\sqrt5}\left[\left(\frac{1+\sqrt5}{2}\right)^1-\left(\frac{1-\sqrt5}{2}\right)^1\right]$
$\displaystyle = \frac{1}{\sqrt5}\left[\frac{1+\sqrt5 - 1 +\sqrt5}{2}\right]=\frac{1}{\sqrt5}\times \sqrt5=1$
For $\displaystyle n = 2$, note that $\displaystyle \left(\frac{1+\sqrt5}{2}\right)^2 = \frac{1 + 2\sqrt5+ 5}{4}= \frac{3+\sqrt5}{2}$. Use a similar expression for $\displaystyle \left(\frac{1-\sqrt5}{2}\right)^2$, and simplify the result.
For $\displaystyle n = 3$, note that $\displaystyle \left(\frac{1+\sqrt5}{2}\right)^3= \left(\frac{1+\sqrt5}{2}\right)^2\times\left(\frac {1+\sqrt5}{2}\right)$
$\displaystyle =\left(\frac{3+\sqrt5}{2}\right)\times\left(\frac{ 1+\sqrt5}{2}\right)$
Multiply out the brackets, and simplify. Can you do this, and $\displaystyle n = 4$ now?
Grandad