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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.

 Sep 9, 2018
 #1
avatar+5172 
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so you need to find g such that that infinitely nested square root is equal to 2

 

I showed you somthing similar earlier.

 

let the nested square root be x then

 

\(x = \sqrt{g +x} \\ x^2 =g+x \\ x^2 - x- g = 0 \\ x = \dfrac{1\pm \sqrt{1+4g}}{2} = 2 \\ 1 \pm \sqrt{1+4g}=4 \\ \pm \sqrt{1+4g}=3 \\ \text{we need only consider the + case} \\ 1+4g = 9 \\ 4g = 8 \\ g=2\)

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 Sep 9, 2018
 #2
avatar+1177 
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Thanks! I was a bti confused about the \(\bowtie\) sign.

Lightning  Sep 9, 2018

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