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Question Number 192958 by Mingma last updated on 31/May/23
Answered by Frix last updated on 01/Jun/23
Question 191675 x=((1+(√5))/2) (2x−1)^4 =((√5))^4 =25
$$\mathrm{Question}\:\mathrm{191675} \\ $$$${x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{4}} =\left(\sqrt{\mathrm{5}}\right)^{\mathrm{4}} =\mathrm{25} \\ $$
Answered by BaliramKumar last updated on 01/Jun/23
(√(x−(1/x))) + (√(1−(1/x))) = x (√((x^2 −1)/x)) + (√((x−1)/x)) = x ((√(x^2 −1)) + (√(x−1)))^2 = (x(√x))^2 x^2 −1+x−1+2(√((x^2 −1)(x−1))) = x^3 2(√(x^3 −x^2 −x+1)) = x^3 −x^2 −x+1+1 put x^3 −x^2 −x+1 = t^2 2(√t^2 ) = t^2 +1 t^2 −2t+1 = 0 ⇒ (t−1)(t−1)= 0 t = 1 ⇒ t^2 = 1^2 ⇒ t^2 = 1 x^3 −x^2 −x+1 = 1 x^3 −x^2 −x=0 x^2 −x−1=0 x ≠ 0 [∵ x ≥1 ] x = ((1±(√5))/2) x ≠ ((1−(√5))/2) [∵ (√(x−1)) ≥ 0] x−1≥ 0 or x ≥ 1 ∴ x = ((1+(√5))/2) ⇒ 2x−1 = (√5) (2x−1)^4 = ((√5))^4 = 25
$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:+\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\:=\:{x} \\ $$$$\sqrt{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}}\:+\:\sqrt{\frac{{x}−\mathrm{1}}{{x}}}\:=\:{x} \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:+\:\sqrt{{x}−\mathrm{1}}\right)^{\mathrm{2}} \:=\:\left({x}\sqrt{{x}}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{1}+{x}−\mathrm{1}+\mathrm{2}\sqrt{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−\mathrm{1}\right)}\:=\:{x}^{\mathrm{3}} \\ $$$$\mathrm{2}\sqrt{{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:=\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −{x}+\mathrm{1}+\mathrm{1} \\ $$$${put}\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −{x}+\mathrm{1}\:=\:{t}^{\mathrm{2}} \\ $$$$\mathrm{2}\sqrt{{t}^{\mathrm{2}} }\:=\:{t}^{\mathrm{2}} +\mathrm{1} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\:=\:\mathrm{0}\:\:\:\Rightarrow\:\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{t}−\mathrm{1}\right)=\:\mathrm{0} \\ $$$${t}\:=\:\mathrm{1}\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\mathrm{t}^{\mathrm{2}} \:=\:\mathrm{1}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{t}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$${x}^{\mathrm{3}} −{x}^{\mathrm{2}} −{x}+\cancel{\mathrm{1}}\:=\:\cancel{\mathrm{1}} \\ $$$${x}^{\mathrm{3}} −{x}^{\mathrm{2}} −{x}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\neq\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\:{x}\:\geq\mathrm{1}\:\right] \\ $$$${x}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\neq\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\left[\because\:\sqrt{{x}−\mathrm{1}}\:\geq\:\mathrm{0}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}−\mathrm{1}\geq\:\mathrm{0}\:\mathrm{or}\:{x}\:\geq\:\mathrm{1} \\ $$$$\therefore\:{x}\:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\Rightarrow\:\:\:\mathrm{2}{x}−\mathrm{1}\:=\:\sqrt{\mathrm{5}} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{4}} \:=\:\left(\sqrt{\mathrm{5}}\right)^{\mathrm{4}} \:=\:\mathrm{25} \\ $$$$ \\ $$
Commented by Mingma last updated on 01/Jun/23
Perfect ��
Answered by ajfour last updated on 01/Jun/23
p+q=x p−q=t p^2 −q^2 =x−1 p^2 +q^2 =x+1−(2/x) tx=x−1 x^2 +t^2 =2x+2−(4/x) x^2 +(1−(1/x))^2 =2x+2−(4/x) x^2 +(1/x^2 )=1+2(x−(1/x)) (x−(1/x))^2 −2(x−(1/x))+1=0 x−(1/x)−1=0 x^2 −x−1=0 x=((1+(√5))/2) 2x−1=(√5) (2x−1)^4 =25
$${p}+{q}={x} \\ $$$${p}−{q}={t} \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} ={x}−\mathrm{1} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={x}+\mathrm{1}−\frac{\mathrm{2}}{{x}} \\ $$$${tx}={x}−\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{t}^{\mathrm{2}} =\mathrm{2}{x}+\mathrm{2}−\frac{\mathrm{4}}{{x}} \\ $$$${x}^{\mathrm{2}} +\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{2}{x}+\mathrm{2}−\frac{\mathrm{4}}{{x}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{1}+\mathrm{2}\left({x}−\frac{\mathrm{1}}{{x}}\right) \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\left({x}−\frac{\mathrm{1}}{{x}}\right)+\mathrm{1}=\mathrm{0} \\ $$$${x}−\frac{\mathrm{1}}{{x}}−\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{2}{x}−\mathrm{1}=\sqrt{\mathrm{5}} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{4}} =\mathrm{25} \\ $$
Commented by York12 last updated on 01/Jun/23
can you make it more clear
$${can}\:{you}\:{make}\:{it}\:{more}\:{clear} \\ $$
Commented by Mingma last updated on 01/Jun/23
Perfect ��
Answered by York12 last updated on 01/Jun/23
(√(x−(1/x))) +(√(1−(1/x)))=x (√(x^3 −x))+(√(x^2 −x))=x^2 ....... (i) ⇒ (√(x^3 −x))−(√(x^2 −x))=(x−1) ......(ii) (i) − (ii) ⇒ 2(√(x^2 −x_(λ) ))=x^2 −x_(λ) +1 ⇒ 2(√λ)=λ + 1 ⇒ 4λ = λ^2 +2λ +1 λ^2 −2λ +1 =0 ⇒ (λ−1)^2 =0 , λ =1 x^2 −x−1=0 ⇒ x= ((1∓(√5))/2) but ((1−(√5))/2) doesn′t satisfy the given equation ⇒ x =((1+(√5))/2) ★ (2x−1)=(√(5 ))⇒(2x−1)^4 = 25 ★
$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x}\: \\ $$$$\sqrt{{x}^{\mathrm{3}} −{x}}+\sqrt{{x}^{\mathrm{2}} −{x}}={x}^{\mathrm{2}} \:\:…….\:\left({i}\right) \\ $$$$\Rightarrow\:\sqrt{{x}^{\mathrm{3}} −{x}}−\sqrt{{x}^{\mathrm{2}} −{x}}=\left({x}−\mathrm{1}\right)\:……\left({ii}\right) \\ $$$$\left({i}\right)\:−\:\left({ii}\right)\: \\ $$$$\Rightarrow\:\mathrm{2}\sqrt{\underset{\lambda} {\underbrace{{x}^{\mathrm{2}} −{x}}}}=\underset{\lambda} {\underbrace{{x}^{\mathrm{2}} −{x}}}+\mathrm{1}\: \\ $$$$\Rightarrow\:\mathrm{2}\sqrt{\lambda}=\lambda\:+\:\mathrm{1}\:\Rightarrow\:\mathrm{4}\lambda\:=\:\lambda^{\mathrm{2}} \:+\mathrm{2}\lambda\:+\mathrm{1}\: \\ $$$$\lambda^{\mathrm{2}} −\mathrm{2}\lambda\:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\:\left(\lambda−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:,\:\lambda\:=\mathrm{1}\: \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:{x}=\:\frac{\mathrm{1}\mp\sqrt{\mathrm{5}}}{\mathrm{2}}\: \\ $$$${but}\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:{doesn}'{t}\:{satisfy}\:{the}\:{given}\: \\ $$$${equation}\:\Rightarrow\:{x}\:=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\bigstar \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)=\sqrt{\mathrm{5}\:}\Rightarrow\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{4}} \:=\:\mathrm{25}\:\bigstar \\ $$$$ \\ $$
Commented by Mingma last updated on 01/Jun/23
Perfect ��

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