Question

If $x=\sqrt{6+\sqrt{6+\sqrt{6+...00}}},$ then

• A. $x$ is an irrational number
• B. $2<x<3$
• C. $x=3$
• D. none of these

Answer 1

The correct option is C $x=3$

$\begin{array}{c}x=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+.....}}}}\\ \Rightarrow x=\sqrt{6+x}\\ \Rightarrow x^2-x-6=0\\ \Rightarrow x^2-3x+2x-6=0\\ \Rightarrow x\left(x-3\right)+2\left(x-3\right)=0\\ \Rightarrow \left(x+2\right)\left(x-3\right)=0\\ \Rightarrow x=3\\ Hence,\\ optionCiscorrectanswer.\end{array}$