1.

Find the value of `x` in `sqrt(x+2sqrt(x+2sqrt(x+2sqrt(3x))))=x.`

Answer» Rewrite the given equation
`sqrt(x+2sqrt(x+2sqrt(x+………….+2sqrt(x+2sqrt(x+2x)))))=x`……i
On replacing the last letter x on the LHS of Eq i by the value of `x` expressed by Eq. i w get
`x=ubrace(sqrt(x+2sqrt(x+2sqrt(x+……….+2sqrt(x+2sqrt(3x))))))_(2n"radical signs")`
Further let us replace the last letter x by the same expression again and again yields.
`:. `x=ubrace(sqrt(x+2sqrt(x+2sqrt(x+……….+2sqrt(x+2sqrt(3x))))))_(3n"radical signs")`
`x=ubrace(sqrt(x+2sqrt(x+2sqrt(x+……….+2sqrt(x+2sqrt(3x))))))_(4n"radical signs")=......`
We can write
`x=sqrt(x+2sqrt(x+2sqrt(x+...)))`
`=ubrace(lim_(Ntooo)sqrt(x+2sqrt(x+2sqrt(x+……….+2sqrt(x+2sqrt(3x))))))_(N"radical signs")`
it follows that
`x=sqrt(x+2sqrt(x+2sqrt(x+....)))`
`=sqrt(x+2(sqrt(x+2sqrt(x+...))))=sqrt((x+2x))`
Hence `x^(2)=x+2x`
`impliesx^(2)-3x=0`
`:.x=0,3`


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