Problem of the week #16 - July 16th, 2012

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Jameson

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Congratulations to the following members for their correct solutions:

1) Sudharaka
3) Reckoner
4) soroban
5) veronica1999

Solution (from soroban):

Evaluate: .$\left(\sqrt{11 + \sqrt{11 + \sqrt{11 + \cdots}}}\right) - \left(\sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots}}}\right)$

Let $$x \:=\:\sqrt{11 + \sqrt{11 + \sqrt{11 + \cdots}}} \quad\Rightarrow\quad x^2 \:=\:11 + \sqrt{11 + \sqrt{11 + \cdots}}$$

. . $$x^2 - 11 \:=\:\sqrt{11 + \sqrt{11 + \cdots}} \quad\Rightarrow\quad x^2 - 11 \:=\:x$$

. . $$x^2 - x - 11 \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{1\pm\sqrt{45}}{2}$$

Hence: .$$x \:=\:\frac{1 + 3\sqrt{5}}{2}$$

Let $$y \:=\:\sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots }}} \quad\Rightarrow\quad y^2 \:=\:7 + \sqrt{7 + \sqrt{7 + \cdots }}$$

. . $$y^2 - 7 \:=\:\sqrt{7 + \sqrt{7 + \cdots }} \quad\Rightarrow\quad y^2 - 7 \:=\:y$$

. . $$y^2 - y - 7 \:=\:0 \quad \Rightarrow \quad y \:=\:\frac{1 \pm \sqrt{29}}{2}$$

Hence: .$$y \:=\:\frac{1 + \sqrt{29}}{2}$$

Therefore: .$$x - y \;=\;\left(\frac{1+3\sqrt{5}}{2}\right) - \left(\frac{1+\sqrt{29}}{2}\right) \;=\;\frac{3\sqrt{5} -\sqrt{29}}{2}$$

General solution to this form (from BAdhi):

let's consider a general expression for the statement in the brackets of the given problem and take it as 'x'

$$\displaystyle \sqrt{a+\sqrt{a+\sqrt{a+\dots }}}=x$$

by squaring the both sides,

$$\displaystyle a+\underbrace{\sqrt{a+\sqrt{a+\sqrt{a+\dots }}}}_{x}=x^2$$

$$\displaystyle a+x=x^2$$

$$\displaystyle x^2-x-a=0$$

since this is a quadric equation the answer is,

$$\displaystyle x=\frac{-(-1)\pm \sqrt{(-1)^2-4\times 1\times (-a)}}{2\times 1}$$

$$\displaystyle x=\frac{1\pm \sqrt{1+4a}}{2}$$

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