This one goes up to Infinity!

Algebra Level 1

What is $\sqrt { 6+\sqrt { 6+\sqrt { 6+\sqrt { 6..... } } } }$ up to $\infty$ ?

The answer is 3.

3 solutions

Anish Puthuraya
Apr 2, 2014

$S = \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6\ldots}}}}$

$S = \sqrt{6+S}$

$S^2 = 6+S$

$S^2-S-6=0$

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

But, $\displaystyle S$ cannot be negative, so therefore,

$S = \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6\ldots}}}} = \boxed{3}$

Why can't S be negative??

Satvik Golechha - 7 years, 2 months ago

The principle square root of any positive real number is taken as positive.

Anish Puthuraya - 7 years, 2 months ago

Why is it taken as positive, then?

Satvik Golechha - 7 years, 2 months ago

@Satvik Golechha – It is a widely accepted and logical assumption. For eg. Just find the square root of $16$ in your calculator, you will see that it displays " $+4$ " instead of " $-4$ ".

Moreover, there can only be one answer for the problem, so the first choice is ought to be $+3$ .

If you want $y=\sqrt{x}$ to be a function, then you have to assume $y>0$ or $y<0$ , otherwise, by definition, it would not be a function at all, since any function has to give only one output for any input.

So, if you assume $y=\pm\sqrt{x}$ , then you see that it provides two values of $y$ for any positive value of $x$ . This must not be the case if you want this relation to be a function. Hence, we take it to be $y=+\sqrt{x}$

Anish Puthuraya - 7 years, 2 months ago

that's why didn't get into iit........... ;)

Aarambh Sanjay - 7 years, 2 months ago

How could you say that the s in sqrt (6 +s) is equal to s in which you already deducted 1 from an infinity?

Marx Sargent Mercurio - 7 years, 2 months ago

Than what will be the value? If we would have 4 in place of 6...

Vinod Chhallany - 6 years, 11 months ago

Saurabh Mallik
Jun 14, 2014

Let:

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

$x=\sqrt{6+x}$

Squaring both sides:

$x^{2}=6+x$

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

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

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

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

So, the values of x $=3$ and $-2$

But, $x$ cannot be negative, since square root of negative integers is not possible.

Thus, the answer is: $x=\boxed{3}$

But the square root of negative numbers can be expressed in the form of iota, e.g.-- Square root of (-2) can be written as 0.414i(i=iota). Therefore, the answers can be 3 as well as (-2).

Sourabh shukla - 6 years, 10 months ago

Ninad Jadkar
May 19, 2014

First square both sides

where are both sides ,buddy ?

Rishabh Jain - 7 years ago

This one goes up to Infinity!

The answer is 3.

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