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Question-172360




Question Number 172360 by Mikenice last updated on 25/Jun/22
Commented by mr W last updated on 25/Jun/22
(1)  4=(√(16))=(√(13+(√9)))=(√(13+(√(5+4))))=(√(13+(√(5+(√(13+(√(5+(√(13+...))))))))))
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{4}=\sqrt{\mathrm{16}}=\sqrt{\mathrm{13}+\sqrt{\mathrm{9}}}=\sqrt{\mathrm{13}+\sqrt{\mathrm{5}+\mathrm{4}}}=\sqrt{\mathrm{13}+\sqrt{\mathrm{5}+\sqrt{\mathrm{13}+\sqrt{\mathrm{5}+\sqrt{\mathrm{13}+…}}}}} \\ $$
Commented by mr W last updated on 25/Jun/22
(2)  see Q# 170948
$$\left(\mathrm{2}\right) \\ $$$${see}\:{Q}#\:\mathrm{170948} \\ $$
Answered by Rasheed.Sindhi last updated on 26/Jun/22
Let x=(√(13+(√(5+(√(13+(√(5+(√(13+...))))))))))  x^2 −13=(√(5+(√(13+(√(5+(√(13+...))))))))  (x^2 −13)^2 =5+(√(13+(√(5+(√(13+...))))))  (x^2 −13)^2 −5=x  See below the comments of mrW sir↓
$${Let}\:{x}=\sqrt{\mathrm{13}+\sqrt{\mathrm{5}+\sqrt{\mathrm{13}+\sqrt{\mathrm{5}+\sqrt{\mathrm{13}+…}}}}} \\ $$$${x}^{\mathrm{2}} −\mathrm{13}=\sqrt{\mathrm{5}+\sqrt{\mathrm{13}+\sqrt{\mathrm{5}+\sqrt{\mathrm{13}+…}}}} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{13}\right)^{\mathrm{2}} =\mathrm{5}+\sqrt{\mathrm{13}+\sqrt{\mathrm{5}+\sqrt{\mathrm{13}+…}}} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{13}\right)^{\mathrm{2}} −\mathrm{5}={x} \\ $$$${See}\:{below}\:{the}\:{comments}\:{of}\:{mrW}\:\boldsymbol{{sir}}\downarrow \\ $$
Commented by mr W last updated on 25/Jun/22
please recheck sir:  (x^2 −13)^2 −5=x  x^4 −26x^2 −x+164=0  (x−4)(x^3 +4x^2 −10x−41)=0  ...
$${please}\:{recheck}\:{sir}: \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{13}\right)^{\mathrm{2}} −\mathrm{5}={x} \\ $$$${x}^{\mathrm{4}} −\mathrm{26}{x}^{\mathrm{2}} −{x}+\mathrm{164}=\mathrm{0} \\ $$$$\left({x}−\mathrm{4}\right)\left({x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{10}{x}−\mathrm{41}\right)=\mathrm{0} \\ $$$$… \\ $$
Commented by Rasheed.Sindhi last updated on 25/Jun/22
You′re right sir!
$${You}'{re}\:{right}\:\boldsymbol{{sir}}! \\ $$
Commented by mr W last updated on 25/Jun/22
x=(√(13+...))>(√(13))>3.5  all roots of x^3 +4x^2 −10x−41=0   are <3.5, therefore x=4 is the only  suitable solution.
$${x}=\sqrt{\mathrm{13}+…}>\sqrt{\mathrm{13}}>\mathrm{3}.\mathrm{5} \\ $$$${all}\:{roots}\:{of}\:{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{10}{x}−\mathrm{41}=\mathrm{0}\: \\ $$$${are}\:<\mathrm{3}.\mathrm{5},\:{therefore}\:{x}=\mathrm{4}\:{is}\:{the}\:{only} \\ $$$${suitable}\:{solution}. \\ $$

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