What's with the crazy roots!!?

www.imomath.com Its a nice website for good questions.Try it once.

Indeed, Dinesh's recommended website is good, but it's a little "too hard" for you right now.

Check out this huge list and also this specific olympiad , along with one of the best websites that you could ever find something to study .

EDIT: Lorenc, add me up on Facebook.

Lorenc, just call me Guilherme . :D

Solutions:

a : We can see that $(\sqrt{3} \pm \sqrt{2})^4 = 49 \pm 20\sqrt{6}$ . This means $\sqrt{a} = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} \Rightarrow \boxed{a =12.}$ This problem was created by the famous Indian math genius Ramanujan .

b : Squaring both sides, we get $2+\sqrt{2+\sqrt{2+\sqrt{\cdots}}} = b^2$ . But $\sqrt{2+\sqrt{2+\sqrt{\cdots}}} = b$ , so $\; 2 + b = b^2$ . This results in $\boxed{b=2.}$

c : This one can be solved by using functions. Let $c = \sqrt{\alpha x+(n+\alpha)^2+x\sqrt{\alpha (x+n)+(n+\alpha)^2+(x+n)\sqrt{\cdots}}}$ . Squaring both sides leads us to $c^2 = \alpha x+(n+\alpha)^2+x\sqrt{\alpha (x+n)+(n+\alpha)^2+(x+n)\sqrt{\cdots}}$ . Allowing $c = F(x)$ , we have $F(x)^2 = \alpha x + (n+\alpha)^2 + x \cdot F(x+n)$ , and it can be proved that $F(x) = x + n + \alpha$ . Letting $\alpha = 0, n= 1, x= 2$ , we have $c = 3$ . This problem also came from Ramanujan. Check out a study about it in this presentation , from pages $8$ to $14$ .

d : Cube both sides to get $d^3 = 6 + \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{\cdots}}}$ . But $\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{\cdots}}} = d$ , so $d^3 = 6 +d$ . It is clear that $\boxed{d = 2.}$ .

Lorenc, NEVER GIVE UP STUDYING !

EDIT: Lorenc, add me up on Facebook.