Can anyone solve these problems? (1 Viewer)

HAX0R · Oct 4, 2013

Hello everyone.

Can anyone solve http://www.maths.unsw.edu.au/sites/default/files/final_junior_51.pdf with solutions? It'll be much appreciated.

Thank you very much and have a nice day,
With regards,
HAX0R

HeroicPandas · Oct 4, 2013

Question 1 beautiful

$c = \sqrt{1+2\sqrt{1+2\sqrt{\dots}}} \\ \\ \frac{c^2-1}{2} = c \\ \\ c^2 - 2c - 1 = 0$

By completing the square or applying the quadratic formula, solve for c

HAX0R · Oct 5, 2013

How did you get to step 2?

HeroicPandas · Oct 5, 2013

$c = \sqrt{1+2\sqrt{1+2\sqrt{\dots}}} \\ \\ c^2 - 1 = 2\sqrt{1+2\sqrt{1+2\sqrt{\dots}}} \\ \\ \frac{c^2 - 1}{2} = \sqrt{1+2\sqrt{1+2\sqrt{\dots}}} \\ \\ \frac{c^2 - 1}{2} = c$

I hope you can follow this

HeroicPandas · Oct 6, 2013

$c = \sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{\dots}}}}$

Square both sides, noting that $(\sqrt{a})^2 = a$

$c^2 = 1+2\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{\dots}}}}$

Subtract '1' from both sides,

$c^2 \bold{-1} = 1+2\left ( \sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{\dots}}}} \right ) \bold{-1}$

$c^2 - 1 = 2\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{\dots}}}}$

Divide both sides by 2,

$\frac{c^2 - 1}{\bold{2}} = \frac{2\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{\dots}}}}}{\bold{2}}$

$\frac{c^2 - 1}{2} = \sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{\dots}}}}$

Do you notice that on the right-hand side, it is equal to 'c'?

$\frac{c^2 - 1}{2} = c$

Here HAX0R, I've guided you with every step! Hope you now understand it

As practice, from $\frac{c^2 - 1}{2} = c$ , get to $c^2 - 2c -1 = 0$ in one step (doing steps in ur head)

Can anyone solve these problems? (1 Viewer)

HAX0R

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HeroicPandas

Heroic!

HAX0R

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HeroicPandas

Heroic!

HeroicPandas

Heroic!

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