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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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