Question 132

The value of the following is: $$\sqrt{12+\sqrt{12+\sqrt{12+.....}}}$$

Solution

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

=> $$x=\sqrt{12+x}$$

Squaring both sides, we get :

=> $$x^2=x+12$$

=> $$x^2-x-12=0$$

=> $$x^2-4x+3x-12=0$$

=> $$x(x-4)+3(x-4)=0$$

=> $$(x-4)(x+3)=0$$

=> $$x=4,-3$$

$$\because x$$ cannot be negative, => $$x=4$$

=> Ans - (D)

Create a FREE account and get: