Question-183711

Question Number 183711 by Michaelfaraday last updated on 29/Dec/22

Commented by Matica last updated on 29/Dec/22

$${to}\:{mathematize} \\ $$

Answered by HeferH last updated on 29/Dec/22

((x^2 ((x^2 ((x^2 ...))^(1/3) ))^(1/3) ))^(1/3) = H ((x^2 H))^(1/3) = H Hx^2 =H^3 x = H (√(x^2 + x(√(x^2 + x(√(x^2 + x(√(...)))))))) = P (√(x^2 + x∙P)) = P x^2 + Px = P^2 x^2 + Px − P^2 = 0 (x/P) + 1 − (P/x) = 0 let (x/P) = t t + 1 − (1/t) = 0 t^2 + t − 1 = 0 t = ((−1 ±(√(1 − 4(1)(−1))))/(2(1))) = ((−1±(√5))/2) (x/P) = ((−1 + (√5))/2) ((2x)/( (√5) −1)) = P ⇒ (x/((2x)/( (√5) − 1))) = (x/5) ⇒ ((2x)/( (√5) − 1)) = 5 x = ((5(√5) − 5)/2)

$$\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} …}}}\:=\:{H} \\ $$$$\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} {H}}\:=\:{H} \\ $$$$\:{Hx}^{\mathrm{2}} ={H}^{\mathrm{3}} \\ $$$$\:{x}\:=\:{H} \\ $$$$\sqrt{{x}^{\mathrm{2}} \:+\:{x}\sqrt{{x}^{\mathrm{2}} \:+\:{x}\sqrt{{x}^{\mathrm{2}} \:+\:{x}\sqrt{…}}}}\:=\:{P} \\ $$$$\:\sqrt{{x}^{\mathrm{2}} \:+\:{x}\centerdot{P}}\:=\:{P} \\ $$$$\:{x}^{\mathrm{2}} \:+\:{Px}\:=\:{P}^{\mathrm{2}} \\ $$$$\:{x}^{\mathrm{2}} \:+\:{Px}\:−\:{P}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\frac{{x}}{{P}}\:+\:\mathrm{1}\:−\:\frac{{P}}{{x}}\:=\:\mathrm{0} \\ $$$$\:{let}\:\frac{{x}}{{P}}\:=\:{t} \\ $$$$\:{t}\:+\:\mathrm{1}\:−\:\frac{\mathrm{1}}{{t}}\:=\:\mathrm{0} \\ $$$$\:{t}^{\mathrm{2}} \:+\:{t}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:{t}\:=\:\frac{−\mathrm{1}\:\pm\sqrt{\mathrm{1}\:−\:\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{1}\right)}}{\mathrm{2}\left(\mathrm{1}\right)}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\: \\ $$$$\:\frac{{x}}{{P}}\:=\:\frac{−\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{5}}\:−\mathrm{1}}\:=\:{P}\:\:\Rightarrow\: \\ $$$$\:\frac{{x}}{\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{5}}\:−\:\mathrm{1}}}\:=\:\frac{{x}}{\mathrm{5}}\:\Rightarrow\:\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{5}}\:−\:\mathrm{1}}\:=\:\mathrm{5} \\ $$$$\:{x}\:=\:\frac{\mathrm{5}\sqrt{\mathrm{5}}\:−\:\mathrm{5}}{\mathrm{2}} \\ $$

Commented by JDamian last updated on 29/Dec/22

In this step x^2 +Px=P^2 why don′t you solve quickly as a quadratic equation? You have changed into another quadratic equation in t and solved to get P, which is unnecessary.

$${In}\:{this}\:{step} \\ $$$$\:{x}^{\mathrm{2}} +{Px}={P}^{\mathrm{2}} \\ $$$${why}\:{don}'{t}\:{you}\:{solve}\:{quickly}\:{as}\:{a}\:{quadratic} \\ $$$${equation}?\:{You}\:{have}\:{changed}\:{into}\:{another} \\ $$$${quadratic}\:{equation}\:{in}\:{t}\:{and}\:{solved}\:{to}\:{get}\:{P}, \\ $$$${which}\:{is}\:{unnecessary}. \\ $$

Commented by Michaelfaraday last updated on 29/Dec/22

thanks sir but sir i think you should have solve it inform of quadratic eqn from x^2 +px−p^2 =0 while your solving from eqn which is not necessary.

$${thanks}\:{sir} \\ $$$${but}\:{sir}\:{i}\:{think}\:{you}\:{should}\:{have}\:{solve}\:{it} \\ $$$${inform}\:{of}\:{quadratic}\:{eqn}\:{from} \\ $$$${x}^{\mathrm{2}} +\boldsymbol{{px}}−\boldsymbol{{p}}^{\mathrm{2}} =\mathrm{0}\:\boldsymbol{{while}}\:\boldsymbol{{your}}\:\boldsymbol{{solving}}\:\boldsymbol{{from}}\:\boldsymbol{{eqn}} \\ $$$$\boldsymbol{{which}}\:\boldsymbol{{is}}\:\boldsymbol{{not}}\:\boldsymbol{{necessary}}. \\ $$

Commented by HeferH last updated on 29/Dec/22

Yeah you are right , in my defense it was very late (00:48) when i did it

$$\:{Yeah}\:{you}\:{are}\:{right}\:,\:{in}\:{my}\:{defense}\: \\ $$$$\:{it}\:{was}\:{very}\:{late}\:\left(\mathrm{00}:\mathrm{48}\right)\:{when}\:{i}\:{did}\:{it} \\ $$

Commented by Michaelfaraday last updated on 29/Dec/22

$${okay}\:{sir} \\ $$

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