Question

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

Answer 1

Answer: g = 2

================================================

Work Shown:

$a \bowtie b = a + \sqrt{b+\sqrt{b+\sqrt{b+...}}}\\\\7 \bowtie g = 7 + \sqrt{g+\sqrt{g+\sqrt{g+...}}} = 9\\\\\sqrt{g+\sqrt{g+\sqrt{g+...}}} = 2$

After subtracting 7 from both sides.

Note how because we have an infinite sequence of nested radicals, we can let $x = \sqrt{g+\sqrt{g+...}}$ which means x is equal to 2 as well.

This lets us say

$\sqrt{g+\sqrt{g+\sqrt{g+...}}} = \sqrt{g+x} = x$

Solve the equation $\sqrt{g+x}= x$ for x to get

$\sqrt{g+x} = x\\\\g+x = x^2\\\\x^2-x-g = 0\\\\x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-g)}}{2(1)}\\\\x = \frac{1\pm\sqrt{1+4g}}{2}\\\\x = \frac{1+\sqrt{1+4g}}{2} \ \text{ or } \ x = \frac{1-\sqrt{1+4g}}{2}$

Since x is positive, this means we only focus on the first equation in the last line above.

Earlier we let x be equal to the infinite nested radicals involving g, but x is also equal to 2. So plug in x = 2 and use that to find g.

$x = \frac{1+\sqrt{1+4g}}{2}\\\\2 = \frac{1+\sqrt{1+4g}}{2}\\\\4 = 1+\sqrt{1+4g}\\\\3 = \sqrt{1+4g}\\\\9 = 1+4g\\\\1+4g = 9\\\\4g = 8\\\\g = 2$

As a check, we can do the following

$7 + \sqrt{2+\sqrt{2}} \approx 8.84775906502258\\\\7 + \sqrt{2+\sqrt{2+\sqrt{2}}} \approx 8.96157056080647\\\\7 + \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} \approx 8.9903694533444$

we're slowly approaching 9