Nested!

Algebra Level 2

If $x > 0$ , what is

$\sqrt{x \sqrt{x \sqrt{x.. ...... \infty}}} ?$

x+ \text{constant}

x-\text{constant}

none of the rest

x

1 solution

A Former Brilliant Member
Feb 28, 2015

$\sqrt{x\sqrt{x\sqrt{x\cdots}}} =Y$ $\sqrt{xY} = Y$ Squaring both sides , $\Rightarrow xY = Y^2$ $\Rightarrow {x = Y}$
Therefore it is equal to x only
$\sqrt{x\sqrt{x\sqrt{x\cdots}}} =x$

You can view a Tricky similar version of this problem here

A Former Brilliant Member - 6 years, 3 months ago

Wait! How can you cancel $Y$ on both the sides?? How do you know that it is not equal to zero???

A Former Brilliant Member - 6 years, 3 months ago

Oh yes you are right .

i didn't thought of that.

I think it should be mentioned in the problem that x must be greater than 0.

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – Yea...Well, you can edit it...hehe

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – hehe.......

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – Since 0 is also a valid solution, you can solve it this way: instead of dividing by Y, subtract the term to the other side:

$Y^2 - xY = 0$

which yields: $Y(Y-x) = 0$

So either Y = 0 or Y-x = 0.

Jay Nabonne - 6 years, 1 month ago

Nested!

1 solution

0 pending reports