+781 If \(\sqrt{x+\!\sqrt{x+\!\sqrt{x+\!\sqrt{x+\cdots}}}}=9\), find x.
I know that if this equation were switched around, where the x is 9 and 9 is x instead, we'd solve it like this:
\(\sqrt{9+\!\sqrt{9+\!\sqrt{9+\!\sqrt{9+\cdots}}}}=x\\ x^2=9+\sqrt{9+\!\sqrt{9+\!\sqrt{9+\cdots}}}\\ x^2=9+x\\ x^2-x-9=0\\ x=\frac{1+\sqrt{37}}{2} \)
I think I should be applying the same logic, but it's kind of difficult because of the x and it's making me confused. Could anyone help?
+23246 Your instinct is correct!
Call that whose expression on the left side of the equal sign "expression" so you have
[expression] = 9
Square both sides -- you'll get:
x + [expression] = 81
Since [expression] = 9 -- by substituting '9' for [expression] -- you'll now have:
x + 9 = 81
Therefore, x = 72
