# Problem based on square root

• Mar 30th 2009, 08:50 PM
siddscool19
Problem based on square root
If $\sqrt{6+\sqrt{6+\sqrt{6+......}}}$, then

a) x=3
b) x= $\infty$
c) x=6
d) x=9

• Mar 30th 2009, 08:59 PM
mollymcf2009
Quote:

Originally Posted by siddscool19
If $\sqrt{6+\sqrt{6+\sqrt{6+......}}}$, then

a) x=3
b) x= $\infty$
c) x=6
d) x=9

The answer is a. As you take the square root of 6 infinitely and continuously add it to itself, you will slowly approach 1/2 of 6 which is three. You will get closer and closer to three, but the repeating square root prohibits is from getting there.
• Mar 30th 2009, 09:26 PM
o_O
Let: $x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$

Now, note that if you square it and move +6 to the LHS: $x^2 = 6 + \sqrt{6+\sqrt{6+ \cdots}} \ \ \Leftrightarrow \ \ x^2 - 6 = \sqrt{6+\sqrt{6+ \cdots}}$

But the RHS is $x$ again!

Thus, we see that $x$ is the positive solution to the quadratic: $x^2 - x - 6 = 0$

which is simply 3.
• Mar 30th 2009, 09:39 PM
Gaurav
Quote:

Originally Posted by o_O
Let: $x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$

Now, note that if you square it and move +6 to the LHS: $x^2 = 6 + \sqrt{6+\sqrt{6+ \cdots}} \ \ \Leftrightarrow \ \ x^2 - 6 = \sqrt{6+\sqrt{6+ \cdots}}$

But the RHS is $x$ again!

Thus, we see that $x$ is the positive solution to the quadratic: $x^2 - x - 6 = 0$

which is simply 3.

$x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$
$x=\sqrt{6+x}$
$x^2=6+x$
x=3 or -2

ans=(a)3