limit of sqrs

[math0]

What is the limit of [tex]\sqrt6,\sqrt{6+\sqrt6},\sqrt{6+\sqrt{6+\sqrt6}},\cdots[/tex]

I would appreciate an answer that includes both a way and a final answer.

Thanks

[Baltuilhe]

Good afternoon!

[tex]x=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}\\
x^2=6+\underbrace{\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}}_{x}\\\\
x^2=6+x\\
x^2-x-6=0\\
\Delta=(-1)^2-4(1)(-6)=1+24\\
\Delta=25\\
x=\frac{-(-1)\pm\sqrt{25}}{2(1)}\\
x=\frac{1\pm 5}{2}\\
x'=\frac{1+5}{2}=\frac{6}{2}=3\\
x''=\frac{1-5}{2}=\frac{-4}{2}=-2[/tex]

Solution:
x=3

Hope to have helped!

[HallsofIvy]

Nicely done, Baltuilhe!

Just a couple of comments.

Rather than use the "quadratic formula", it is easy to factor [tex]x^2- x- 6= (x- 3)(x+ 2)= 0[/tex] so x= 3 or -2 and, since this sum of square roots is clearly positive, x= 3 is the solution.