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 157220 by MathSh last updated on 21/Oct/21

x = (1/2) find (√(x + 1 + (√(x^2 + 1)))) = ?

$$\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{find}\:\:\sqrt{\mathrm{x}\:+\:\mathrm{1}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}}}\:=\:? \\ $$

Answered by Rasheed.Sindhi last updated on 21/Oct/21

x=(1/2); (√(x + 1 + (√(x^2 + 1)))) = ? (√(x + 1 + (√(x^2 + 1)))) (√((1/2)+1+(√(((1/2))^2 +1)))) (√((3/2)+(√((1+4)/4)))) (√((3/2)+((√5)/2))) =(√((3+(√5))/2)) =((√(3+(√5)))/( (√2)))×((√2)/( (√2))) =((√(6+2(√5)))/2) ▶ (√(6+2(√5))) =p+q(√5) (Say) ((√(6+2(√5))) =p+q(√5) )^2 ; p,q∈Q =6+2(√5) =p^2 +2pq(√5) +5q^2 p^2 +5q^2 =6 ∧ 2pq=2 ((1/q))^2 +5q^2 =6 (1/q^2 )+5q^2 =6⇒5q^4 −6q^2 +1=0 (q−1)(q+1)(5q^2 −1_(q∉Q) )=0 q=±1⇒p=(1/q)=(1/(±1))=±1 (√(6+2(√5))) =p+q(√5) =±1±(√5) =1+(√5) ,−1−(√5) x=((√(6+2(√5)))/2)=((1+(√5) )/2),((−1−(√5))/2)

$$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}};\:\:\sqrt{\mathrm{x}\:+\:\mathrm{1}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}}}\:=\:? \\ $$$$\:\sqrt{\mathrm{x}\:+\:\mathrm{1}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}}}\: \\ $$$$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}+\sqrt{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\sqrt{\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}+\mathrm{4}}{\mathrm{4}}}} \\ $$$$\sqrt{\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}\:=\sqrt{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\:=\frac{\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}}}{\:\sqrt{\mathrm{2}}}×\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}} \\ $$$$=\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}} \\ $$$$\blacktriangleright\:\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}\:={p}+{q}\sqrt{\mathrm{5}}\:\:\:\left({Say}\right) \\ $$$$\:\:\:\:\:\:\left(\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}\:={p}+{q}\sqrt{\mathrm{5}}\:\right)^{\mathrm{2}} \:;\:{p},{q}\in\mathbb{Q} \\ $$$$\:\:\:=\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:={p}^{\mathrm{2}} +\mathrm{2}{pq}\sqrt{\mathrm{5}}\:+\mathrm{5}{q}^{\mathrm{2}} \\ $$$$\:\:\:{p}^{\mathrm{2}} +\mathrm{5}{q}^{\mathrm{2}} =\mathrm{6}\:\wedge\:\mathrm{2}{pq}=\mathrm{2} \\ $$$$\:\:\:\:\left(\frac{\mathrm{1}}{{q}}\right)^{\mathrm{2}} +\mathrm{5}{q}^{\mathrm{2}} =\mathrm{6} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\mathrm{5}{q}^{\mathrm{2}} =\mathrm{6}\Rightarrow\mathrm{5}{q}^{\mathrm{4}} −\mathrm{6}{q}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left({q}−\mathrm{1}\right)\left({q}+\mathrm{1}\right)\left(\underset{{q}\notin\mathbb{Q}} {\underbrace{\mathrm{5}{q}^{\mathrm{2}} −\mathrm{1}}}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{q}=\pm\mathrm{1}\Rightarrow{p}=\frac{\mathrm{1}}{{q}}=\frac{\mathrm{1}}{\pm\mathrm{1}}=\pm\mathrm{1} \\ $$$$\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}\:={p}+{q}\sqrt{\mathrm{5}}\:=\pm\mathrm{1}\pm\sqrt{\mathrm{5}} \\ $$$$\:\:\:\:\:\:=\mathrm{1}+\sqrt{\mathrm{5}}\:,−\mathrm{1}−\sqrt{\mathrm{5}} \\ $$$$\mathrm{x}=\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}\:}{\mathrm{2}},\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Commented by MathSh last updated on 21/Oct/21

Thank you so much dear Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{Ser} \\ $$

Commented by Tawa11 last updated on 21/Oct/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 22/Oct/21

Thanks miss for encouraging us!

$${Thanks}\:{miss}\:{for}\:{encouraging}\:{us}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com