My solution: Let $x=\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+\cdots}}}}$ $\Longrightarrow x^2=a-b\sqrt{a+bx}\Longrightarrow (b\sqrt{a+bx})^2=(a-x^2)^2$ $x^4-2ax^2-b^3x+(a^2-ab^2)=0$ $\Longrightarrow x^4-2ax^2-b^3x+(a^2-ab^2)=0$ $\Longrightarrow (x^2-bx-a)(x^2+bx+b^2-a)=0$ Solving the quadratics and using $a>b^2$, we find that the positive values for $x$ are $\sqrt{a+\frac{b^2}{4}}+\frac{b}{2}$ and $\sqrt{a-\frac{3b^2}{4}}-\frac{b}{2}$. If it is the former, we equate it to the big expression and square it to get ${\sqrt{a-b\sqrt{a+\cdots}}}=\sqrt{a+\frac{b^2}{4}}+\frac{b}{2}$ Again squaring and rearranging, we get $\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+\cdots}}}}=-\left(\sqrt{a+\frac{b^2}{4}}+\frac{b}{2}\right)$ And it must mean $\sqrt{a+\frac{b^2}{4}}+\frac{b}{2}=0$ which is not possible. This root along with others are extraneous which means the value of that expression is $\sqrt{a-\frac{3b^2}{4}}-\frac{b}{2}$ as required.

gouthamphilomath wrote: My solution: Let $x=\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+\cdots}}}}$ In-fact you must prove, that such $x$ exist (i.e. sequence $x_n=\underbrace{\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+\cdots}}}} }_{n-times}$) has limit. But all prove is quite nice

Absolutely, one can not manipulate infinite sums before being assured they exist (by the convergence of the process used in their formation). Otherwise we get things like: let $S = 1 - 1 + 1 - 1 + \cdots$; we have $S = 1 - S$, hence $S = 1/2$ (none other than Leibniz thought that was valid!).

Does anybody know any other solution to this task other than this standard, reccurence-based one? I'm mostly thinking of geometrical solution to this problem- do You know one?

Please, verify my initial observation: this looks somewhat as an orthogonalization (as in Gramm-Schmitt), but where $b$ is the length and $a$ is already a field (length times length). The "but" part of the sentence makes it imho harder to find the geometrical tantamount task. Why is $ \sqrt{a-\frac{3b^{2}}{4}}-\frac{b}{2} $ not like $ \sqrt{a^{2}-\frac{3b^{2}}{4}}-\frac{b}{2}$- ie. do You know how this task was created?

So who stops us to take $a = \alpha^2$, with $\alpha > |b|$?

I know we can do substitutions, but Im wondering as how to find tantamount task within geometry. Unless You are claiming "turning fields into linear values in geometry", I don't want to use substitutions of this kind.

Imagine two rectangles: A and B(x). Denote area of A P(A)=a=$a_{1}a_{2}$. Choose such length x so that P(B(x))=bx. Add those two areas (as fields) and try to find the square of the same area. Find the length of its side (m). Then, construct another rectangle where sides are: b and m, and substract it from the area of A so that the remaining area is a square of side x. Find such x. Why would anyone want to to that? http://dreamcubic.com/dream_of_prime/im_poker_donkey.jpg Transform the task so that the answer explains the connection between 30 degrees (or 60 degrees) and 1/2. Why would ie. 30 degrees would give the solution? (how is that possible?)

Should I lock this ?