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

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Jameson

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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) Sudharaka
2) BAdhi
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 [tex]x \:=\:\sqrt{11 + \sqrt{11 + \sqrt{11 + \cdots}}} \quad\Rightarrow\quad x^2 \:=\:11 + \sqrt{11 + \sqrt{11 + \cdots}} [/tex]

. . [tex]x^2 - 11 \:=\:\sqrt{11 + \sqrt{11 + \cdots}} \quad\Rightarrow\quad x^2 - 11 \:=\:x[/tex]

. . [tex]x^2 - x - 11 \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{1\pm\sqrt{45}}{2}[/tex]

Hence: .[tex]x \:=\:\frac{1 + 3\sqrt{5}}{2}[/tex]


Let [tex]y \:=\:\sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots }}} \quad\Rightarrow\quad y^2 \:=\:7 + \sqrt{7 + \sqrt{7 + \cdots }} [/tex]

. . [tex]y^2 - 7 \:=\:\sqrt{7 + \sqrt{7 + \cdots }} \quad\Rightarrow\quad y^2 - 7 \:=\:y [/tex]

. . [tex]y^2 - y - 7 \:=\:0 \quad \Rightarrow \quad y \:=\:\frac{1 \pm \sqrt{29}}{2}[/tex]

Hence: .[tex]y \:=\:\frac{1 + \sqrt{29}}{2}[/tex]


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


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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