The value of

Question:

The value of $\sqrt{6+\sqrt{6+\sqrt{6+}} \ldots .}$ is

(a) 4
(b) 3
(c) −2
(d) 3.5

Solution:

Let $x=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+} \ldots . .}}}$

Squaring both sides we get

$x^{2}=6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\ldots . . .}}}}$

$x^{2}=6+x$

$x^{2}-x-6=0$

$x^{2}-3 x+2 x-6=0$

$x(x-3)+2(x-3)=0$

 

$(x-3)(x+2)=0$

The value of x cannot be negative.

Thus, the value of x = 3

Therefore, the correct answer is $(b)$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now