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Let \(a \bowtie b = a+\sqrt{b+\sqrt{b+\sqrt{b+...}}}\). If \(7\bowtie g = 9\), find the value of g.

 May 6, 2019
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\(\text{let }\sqrt{b+\sqrt{b+\sqrt{b+\dots}}} = \beta\\ \sqrt{b+\beta}=\beta\\ b+\beta=\beta^2\\ \beta^2-\beta-b=0\\ \beta=\dfrac{1\pm \sqrt{1+4b}}{2}\\ \beta>0 \text{ as it's a square root so }\\ \beta = \dfrac{1+\sqrt{1+b^2}}{2}\\ a\bowtie b = a + \dfrac 1 2 (1+\sqrt{1+b^2})\)

 

\(7\bowtie 9 = 9\\ 7+\dfrac 1 2 (1+\sqrt{1+g^2})=9\\ \sqrt{1+g^2}=3\\ 1+g^2=9\\ g=\pm\sqrt{8} = \pm 2\sqrt{2}\)

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 May 6, 2019

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