+0  
 
+1
81
2
avatar+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?

 Jul 2, 2020
 #1
avatar+21953 
+1

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

 Jul 2, 2020
 #2
avatar+781 
+1

ah ok, I get it now. Thanks.

 Jul 2, 2020

69 Online Users

avatar
avatar
avatar
avatar
avatar