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Question Number 107484 by bemath last updated on 11/Aug/20

     ∦BeMath∦  (√(x+(√(x+(√(x+(√(x+...)))))))) = (√(4(√(4(√(4(√(4...))))))))  x=?

$$\:\:\:\:\:\nparallel\mathcal{B}{e}\mathcal{M}{ath}\nparallel \\ $$$$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...}}}}\:=\:\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}...}}}} \\ $$$${x}=?\: \\ $$

Answered by john santu last updated on 11/Aug/20

           ⋇JS⋇  ⇒x + (√(x+(√(x+(√(x+...)))))) = 4(√(4(√(4(√(4(√(4...))))))))  ⇒x+(√(4(√(4(√(4(√(4(√(4...)))))))))) = 4(√(4(√(4(√(4(√(4(√(...))))))))))  (1) let t = (√(4(√(4(√(4(√(4(√(4(√(4...))))))))))))                t^2  = 4t ⇒ t=4   (2) x+4 = 4.4 ⇒x=12

$$\:\:\:\:\:\:\:\:\:\:\:\divideontimes\mathcal{JS}\divideontimes \\ $$$$\Rightarrow{x}\:+\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...}}}\:=\:\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}...}}}} \\ $$$$\Rightarrow{x}+\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}...}}}}}\:=\:\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{...}}}}} \\ $$$$\left(\mathrm{1}\right)\:{let}\:{t}\:=\:\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}...}}}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}^{\mathrm{2}} \:=\:\mathrm{4}{t}\:\Rightarrow\:{t}=\mathrm{4}\: \\ $$$$\left(\mathrm{2}\right)\:{x}+\mathrm{4}\:=\:\mathrm{4}.\mathrm{4}\:\Rightarrow{x}=\mathrm{12}\: \\ $$

Answered by Dwaipayan Shikari last updated on 11/Aug/20

(√(x+(√(x+(√(x+(√(x.... ))))))))  =p  x+p=p^2   p^2 −p−x=0  p=((1+(√(1+4x)))/2)  (√(4(√(4(√(4(√(4(√(4(√4)))))))))))=a  4a=a^2   a=4  1+(√(1+4x)) =8  1+4x=49  x=12

$$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}....\:}}}}\:\:={p} \\ $$$${x}+{p}={p}^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} −{p}−{x}=\mathrm{0} \\ $$$${p}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}}}}}}}={a} \\ $$$$\mathrm{4}{a}={a}^{\mathrm{2}} \\ $$$${a}=\mathrm{4} \\ $$$$\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}}\:=\mathrm{8} \\ $$$$\mathrm{1}+\mathrm{4}{x}=\mathrm{49} \\ $$$${x}=\mathrm{12} \\ $$

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