+1245 Let \(a\bowtie b = a+\sqrt{b+\sqrt{b+\sqrt{b+...}}}\). If \(7\bowtie g = 9\), find the value of g.
+6251 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
\(x = \sqrt{g +x} \\ x^2 =g+x \\ x^2 - x- g = 0 \\ x = \dfrac{1\pm \sqrt{1+4g}}{2} = 2 \\ 1 \pm \sqrt{1+4g}=4 \\ \pm \sqrt{1+4g}=3 \\ \text{we need only consider the + case} \\ 1+4g = 9 \\ 4g = 8 \\ g=2\)
