Question

The value of the expression $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$ upto $\infty$ is

• A. 2
• B. 3
• C. 30
• D. 5

Answer 1

The correct option is B 3

$Letx=\sqrt{6+\sqrt{6+\sqrt{6+-----}}}\therefore x^2=6+x\therefore x^2-x-6=0\therefore (x-3)(x+2)=0\therefore x=3,-2.Werejectthe-2valueasitisimpossibleatthegivensituation.\therefore x=3\left(Ans\right)$