Question

Brain teaser .
what is the value of X.

Answer 1

$Let    \sqrt{x+\sqrt{x+ \sqrt{x....}  }}= \sqrt{x\sqrt{x \sqrt{x....}  }}=z\\\\z^{2} =(\sqrt{x+\sqrt{x+ \sqrt{x....}  }})^{2} = (\sqrt{x\sqrt{x \sqrt{x....}  }})^{2} \\\z^{2} =x+\sqrt{x+\sqrt{x+ \sqrt{x....}  }}=x\sqrt{x\sqrt{x \sqrt{x....}  }}\\z^{2} =x+z=xz\\z^{2} =xz\\x=z                 [dividing by z on both sides]\\x+z=xz\\x+x=x*x\\        [As x=z]2x=x^{2} \\x^{2} -2x=0\\x(x-2)=0\\Either x=2 or x=0\\\\Hope you understand :)\\Thanks for the wonderful question$