Question

Solution set of equation $x=\sqrt{12+\sqrt{12+\sqrt{12+....\mathrm{upto}\infty }}}$

• A. $4$
• B. $-4$
• C. $3$
• D. $-3$

Answer 1

The correct option is A

$4$

Explanation for the correct answer:

Determining the solution set for the equation.

The given Equation is:

$x=\sqrt{12+\sqrt{12+\sqrt{12+....\infty }}}$

therefore, substitute the terms with $x$.

thus, $x=\sqrt{12+x}$

Squaring on both the sides, we get

$$\begin{gathered} \Rightarrow x^2={\left(\sqrt{12+x}\right)}^2 \\ \Rightarrow x^2=12+x \\ \Rightarrow x^2-12-x=0 \end{gathered}$$

Simplify the above Equation as

$$\begin{gathered} \Rightarrow x^2-x-12=0 \\ \Rightarrow x^2-4x+3x-12=0 \\ \Rightarrow x(x-4)+3(x-4)=0 \\ \Rightarrow (x-4)(x+3)=0 \end{gathered}$$

$x=4,x=–3$ but since $x>0$, $x=-3$ is rejected.

Therefore, $x=4$

Thus, the correct answer is option (A).