Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 4825 by sanusihammed last updated on 16/Mar/16

Find the value of (√(6+(√(6+(√(6+(√(6+(√6)))))))))

$${Find}\:{the}\:{value}\:{of}\: \\ $$$$ \\ $$$$\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}} \\ $$

Commented by prakash jain last updated on 16/Mar/16

Assuming the series goes infinitely x=(√(6+(√(6+(√(6+(√(6+...)))))))) x=(√(6+x))⇒x=3 Now if the value is only for 6 nested roots. (√(6+(√(6+(√(6+(√(6+(√6)))))))))=2.997≈3

$$\mathrm{Assuming}\:\mathrm{the}\:\mathrm{series}\:\mathrm{goes}\:\mathrm{infinitely} \\ $$$${x}=\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+...}}}} \\ $$$${x}=\sqrt{\mathrm{6}+{x}}\Rightarrow{x}=\mathrm{3} \\ $$$$\mathrm{Now}\:\mathrm{if}\:\mathrm{the}\:\mathrm{value}\:\mathrm{is}\:\mathrm{only}\:\mathrm{for}\:\mathrm{6}\:\mathrm{nested}\:\mathrm{roots}. \\ $$$$\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}}=\mathrm{2}.\mathrm{997}\approx\mathrm{3} \\ $$$$ \\ $$

Commented by 123456 last updated on 16/Mar/16

y=−(√(6−y)) y^2 =6−y y^2 +y−6=0 Δ=(1)^2 −4(1)(−6)=1+24=25 y=((−1±5)/2) y_1 =((−1−5)/2)=−(6/2)=−3 y_2 =((−1+5)/2)=(4/2)=2

$${y}=−\sqrt{\mathrm{6}−{y}} \\ $$$${y}^{\mathrm{2}} =\mathrm{6}−{y} \\ $$$${y}^{\mathrm{2}} +{y}−\mathrm{6}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{6}\right)=\mathrm{1}+\mathrm{24}=\mathrm{25} \\ $$$${y}=\frac{−\mathrm{1}\pm\mathrm{5}}{\mathrm{2}} \\ $$$${y}_{\mathrm{1}} =\frac{−\mathrm{1}−\mathrm{5}}{\mathrm{2}}=−\frac{\mathrm{6}}{\mathrm{2}}=−\mathrm{3} \\ $$$${y}_{\mathrm{2}} =\frac{−\mathrm{1}+\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{4}}{\mathrm{2}}=\mathrm{2} \\ $$

Answered by Rasheed Soomro last updated on 16/Mar/16

Let x=(√(6+(√(6+(√(6+(√(6+(√6))))))))).... x^2 =6+(√(6+(√(6+(√(6+(√(6+(√6)))))))))..... x^2 =6+x⇒x^2 −x−6=0 x=((−(−1)±(√((−1)^2 −4(1)(−6))))/(2(1)))=((1±(√(25)))/2) x=3,−2 −2 is extraneous root. ∴ x=3

$${Let}\:{x}=\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}}.... \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} =\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}}..... \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} =\mathrm{6}+{x}\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{6}=\mathrm{0} \\ $$$${x}=\frac{−\left(−\mathrm{1}\right)\pm\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{6}\right)}}{\mathrm{2}\left(\mathrm{1}\right)}=\frac{\mathrm{1}\pm\sqrt{\mathrm{25}}}{\mathrm{2}} \\ $$$${x}=\mathrm{3},−\mathrm{2} \\ $$$$−\mathrm{2}\:{is}\:{extraneous}\:{root}. \\ $$$$\therefore\:{x}=\mathrm{3} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com