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# Help!

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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
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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
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Thanks! I was a bti confused about the $$\bowtie$$ sign.

Lightning  Sep 9, 2018