1. ## Various Surds Questions

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?

2. Hello prettiestfriend
Originally Posted by prettiestfriend
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?

(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?

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# Surds hard questions

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