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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-




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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