If $\displaystyle \sqrt{6+\sqrt{6+\sqrt{6+......}}}$, then

a) x=3

b) x= $\displaystyle \infty$

c) x=6

d) x=9

Please tell the answer with complete explaination :)

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- Mar 30th 2009, 07:50 PMsiddscool19Problem based on square root
If $\displaystyle \sqrt{6+\sqrt{6+\sqrt{6+......}}}$, then

a) x=3

b) x= $\displaystyle \infty$

c) x=6

d) x=9

Please tell the answer with complete explaination :) - Mar 30th 2009, 07:59 PMmollymcf2009
- Mar 30th 2009, 08:26 PMo_O
Let: $\displaystyle x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$

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

But the RHS is $\displaystyle x$ again!

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

which is simply 3. - Mar 30th 2009, 08:39 PMGaurav