Respuesta :
Answer: g = 2
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Work Shown:
[tex]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[/tex]
After subtracting 7 from both sides.
Note how because we have an infinite sequence of nested radicals, we can let [tex]x = \sqrt{g+\sqrt{g+...}}[/tex] which means x is equal to 2 as well.
This lets us say
[tex]\sqrt{g+\sqrt{g+\sqrt{g+...}}} = \sqrt{g+x} = x[/tex]
Solve the equation [tex]\sqrt{g+x}= x[/tex] for x to get
[tex]\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}[/tex]
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.
[tex]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\\\\[/tex]
As a check, we can do the following
[tex]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\\\\[/tex]
we're slowly approaching 9
Answer: [tex]g = \boxed{2}[/tex]
Step-by-step explanation:
[tex]\sqrt{g+\sqrt{g+\sqrt{g+...}}}=2[/tex]
implies that
[tex]\sqrt{g+\sqrt{g+\sqrt{g+...}}}=\sqrt{g+2}=2[/tex].
Squaring both sides of this new equality, we have [tex]g + 2 = 4 \implies g = \boxed{2}[/tex]
