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3-49-3-49-3-49-x-4-4-4-4-y-x-2-y-




Question Number 10206 by konen last updated on 30/Jan/17
^3 (√(49+^3 (√(49+^3 (√(49....)) )))) =x  (√(4(√(4(√(4(√(4....)))))))) =y  ⇒x^2 −y=?
$$\:^{\mathrm{3}} \sqrt{\mathrm{49}+^{\mathrm{3}} \sqrt{\mathrm{49}+^{\mathrm{3}} \sqrt{\mathrm{49}….}\:}}\:=\mathrm{x} \\ $$$$\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}….}}}}\:=\mathrm{y} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{y}=? \\ $$
Answered by FilupSmith last updated on 30/Jan/17
y=(√(4(√(4(√(4(√(...))))))))  y^2 =4y  y^2 −4y=0  y(y−4)=0  y=0, 4  (y=0 is illogical)  ∴ y=4  x=^3 (√(49+^3 (√(49+^3 (√(49....)) ))))  x^3 =49+x  x^3 −x−49=0  x ≈ 3.7504   (from WolframAlpha)     x^2 −y ≈ (3.7504)^2 −4  x^2 −y ≈ 14.0655−4  x^2 −y ≈ 10.0655
$${y}=\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{…}}}} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{y} \\ $$$${y}^{\mathrm{2}} −\mathrm{4}{y}=\mathrm{0} \\ $$$${y}\left({y}−\mathrm{4}\right)=\mathrm{0} \\ $$$${y}=\mathrm{0},\:\mathrm{4} \\ $$$$\left({y}=\mathrm{0}\:\mathrm{is}\:\mathrm{illogical}\right) \\ $$$$\therefore\:{y}=\mathrm{4} \\ $$$${x}=^{\mathrm{3}} \sqrt{\mathrm{49}+^{\mathrm{3}} \sqrt{\mathrm{49}+^{\mathrm{3}} \sqrt{\mathrm{49}….}\:}} \\ $$$${x}^{\mathrm{3}} =\mathrm{49}+{x} \\ $$$${x}^{\mathrm{3}} −{x}−\mathrm{49}=\mathrm{0} \\ $$$${x}\:\approx\:\mathrm{3}.\mathrm{7504}\:\:\:\left(\mathrm{from}\:\mathrm{WolframAlpha}\right) \\ $$$$\: \\ $$$${x}^{\mathrm{2}} −{y}\:\approx\:\left(\mathrm{3}.\mathrm{7504}\right)^{\mathrm{2}} −\mathrm{4} \\ $$$${x}^{\mathrm{2}} −{y}\:\approx\:\mathrm{14}.\mathrm{0655}−\mathrm{4} \\ $$$${x}^{\mathrm{2}} −{y}\:\approx\:\mathrm{10}.\mathrm{0655} \\ $$

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