Let $a \bowtie b = a+\sqrt{b+\sqrt{b+\sqrt{b+...}}}$. If $7\bowtie g = 9$, find the value of $g$.
I notice that the expression $\sqrt{b+\sqrt{b+\sqrt{b+...}}}$ is an infinite nested radical, which can sometimes be simplified by finding a fixed point. That is, I want to find a value of $x$ such that $\sqrt{b+x}=x$. Squaring both sides, I get $b+x=x^2$, or $x^2-x-b=0$. Using the quadratic formula, I find that $x=\frac{1\pm\sqrt{1+4b}}{2}$. However, since $x$ has to be positive, I only consider the positive root, $x=\frac{1+\sqrt{1+4b}}{2}$. Therefore, I can rewrite $\sqrt{b+\sqrt{b+\sqrt{b+...}}}$ as $\frac{1+\sqrt{1+4b}}{2}$. Now, I can simplify the original equation, $7\bowtie g = 9$, by substituting this expression for the infinite nested radical. I get $7+\frac{1+\sqrt{1+4g}}{2}=9$. Subtracting $7$ from both sides, I get $\frac{1+\sqrt{1+4g}}{2}=2$. Multiplying both sides by $2$, I get $1+\sqrt{1+4g}=4$. Subtracting $1$ from both sides, I get $\sqrt{1+4g}=3$. Squaring both sides, I get $1+4g=9$. Subtracting $1$ from both sides, I get $4g=8$. Dividing both sides by $4$, I get $g=2$. 