How Do You Solve Nested Square Root Expressions?

[Jameson]

Evaluate [math]\left( \sqrt{11+\sqrt{11+\sqrt{11+\cdots}}} \right) - \left( \sqrt{7+\sqrt{7+\sqrt{7+\cdots}}} \right)[/math]

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[Jameson]

Congratulations to the following members for their correct solutions:

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

Solution (from soroban):

[sp]Evaluate: .[/color]$\left(\sqrt{11 + \sqrt{11 + \sqrt{11 + \cdots}}}\right) - \left(\sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots}}}\right) $Let $$x \:=\:\sqrt{11 + \sqrt{11 + \sqrt{11 + \cdots}}} \quad\Rightarrow\quad x^2 \:=\:11 + \sqrt{11 + \sqrt{11 + \cdots}}$$

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

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

Hence: .[/color]$$x \:=\:\frac{1 + 3\sqrt{5}}{2}$$Let $$y \:=\:\sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots }}} \quad\Rightarrow\quad y^2 \:=\:7 + \sqrt{7 + \sqrt{7 + \cdots }}$$

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

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

Hence: .[/color]$$y \:=\:\frac{1 + \sqrt{29}}{2}$$Therefore: .[/color]$$x - y \;=\;\left(\frac{1+3\sqrt{5}}{2}\right) - \left(\frac{1+\sqrt{29}}{2}\right) \;=\;\frac{3\sqrt{5} -\sqrt{29}}{2}$$
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General solution to this form (from BAdhi):

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

[math]\sqrt{a+\sqrt{a+\sqrt{a+\dots }}}=x[/math]

by squaring the both sides,

[math]a+\underbrace{\sqrt{a+\sqrt{a+\sqrt{a+\dots }}}}_{x}=x^2[/math]

[math]a+x=x^2[/math]

[math]x^2-x-a=0[/math]

since this is a quadric equation the answer is,

[math]x=\frac{-(-1)\pm \sqrt{(-1)^2-4\times 1\times (-a)}}{2\times 1}[/math]

[math]x=\frac{1\pm \sqrt{1+4a}}{2}[/math][/sp]