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Question Number 4847 by sanusihammed last updated on 17/Mar/16

(√(6+(√(6+(√(6+(√(6+(√6))))))))) SOLUTION let x = (√(6+(√(6+(√(6+(√(6+(√6))))))))) therefore.. x^(2 ) = 6+(√(6+(√(6+(√(6+(√(6 )))))))) the equation is a continuos funtion Thus x^2 = 6+(√(6+(√(6+(√(6+(√(6+(√6) ))))))))...... since x = (√(6+(√(6+(√(6+(√(6+(√6))))))))) Therdfore x^2 = 6 + x x^2 − x − 6 = 0 x^2 − 3x + 2x − 6 = 0 (x^2 − 3x) + (2x − 6) = 0 x(x − 3) + 2(x − 3) = 0 (x − 3)(x + 2) = 0 x − 3 = 0 or x − 2 = 0 x = 3 or x = −2 since negative is not allowed Thus x = 6 Meaning that (√(6+(√(6+(√(6+(√(6+(√(6 )))))))))) = 3 DONE THANK YOU SO MUCH. I UNDERSTAND THE SOLUTION.

$$\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}} \\ $$$$ \\ $$$${SOLUTION} \\ $$$$ \\ $$$${let}\:{x}\:=\:\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}} \\ $$$$ \\ $$$${therefore}.. \\ $$$$ \\ $$$${x}^{\mathrm{2}\:} =\:\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}\:\:}}}} \\ $$$$ \\ $$$${the}\:{equation}\:{is}\:{a}\:{continuos}\:{funtion} \\ $$$${Thus} \\ $$$$ \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}\:}}}}...... \\ $$$$ \\ $$$${since}\:\:\:{x}\:=\:\:\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}}\: \\ $$$$ \\ $$$${Therdfore} \\ $$$$ \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{6}\:+\:{x} \\ $$$$ \\ $$$${x}^{\mathrm{2}} \:−\:{x}\:−\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$ \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{3}{x}\:+\:\mathrm{2}{x}\:−\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\left({x}^{\mathrm{2}} \:−\:\mathrm{3}{x}\right)\:+\:\left(\mathrm{2}{x}\:−\:\mathrm{6}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$${x}\left({x}\:−\:\mathrm{3}\right)\:+\:\mathrm{2}\left({x}\:−\:\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$$\left({x}\:−\:\mathrm{3}\right)\left({x}\:+\:\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$${x}\:−\:\mathrm{3}\:=\:\mathrm{0}\:{or}\:{x}\:−\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$ \\ $$$${x}\:=\:\mathrm{3}\:{or}\:{x}\:=\:−\mathrm{2} \\ $$$$ \\ $$$${since}\:{negative}\:{is}\:{not}\:{allowed} \\ $$$${Thus} \\ $$$$ \\ $$$${x}\:=\:\mathrm{6} \\ $$$$ \\ $$$${Meaning}\:{that} \\ $$$$ \\ $$$$\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}\:}}}}}\:\:\:=\:\:\mathrm{3} \\ $$$$ \\ $$$${DONE} \\ $$$$ \\ $$$${THANK}\:{YOU}\:{SO}\:{MUCH}.\:{I}\:{UNDERSTAND}\:{THE}\:{SOLUTION}. \\ $$

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