Menu Close

If-x-1-5-5-1-find-5x-2-5x-1-




Question Number 188073 by Shrinava last updated on 25/Feb/23
If   x = (√((1 + (√5))/( (√5) − 1)))     find    5x^2 −5x−1=?
$$\mathrm{If}\:\:\:\mathrm{x}\:=\:\sqrt{\frac{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}\:−\:\mathrm{1}}}\:\:\:\:\:\mathrm{find}\:\:\:\:\mathrm{5x}^{\mathrm{2}} −\mathrm{5x}−\mathrm{1}=? \\ $$
Answered by som(math1967) last updated on 25/Feb/23
x=(√((((√5)+1)^2 )/4))=(((√5)+1)/2)  (1/x)=(2/( (√5)+1))=((2((√5)−1))/4)=(((√5)−1)/2)  x−(1/x)=1   ((x^2 −1)/x)=1⇒x^2 −1=x   5x^2 −5−5x+4=5(x^2 −1)−5x+4  =5x−5x+4=4
$${x}=\sqrt{\frac{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{4}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$$\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}=\mathrm{1}\Rightarrow{x}^{\mathrm{2}} −\mathrm{1}={x} \\ $$$$\:\mathrm{5}{x}^{\mathrm{2}} −\mathrm{5}−\mathrm{5}{x}+\mathrm{4}=\mathrm{5}\left({x}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{5}{x}+\mathrm{4} \\ $$$$=\mathrm{5}{x}−\mathrm{5}{x}+\mathrm{4}=\mathrm{4} \\ $$
Commented by Shrinava last updated on 26/Feb/23
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Frix last updated on 25/Feb/23
((1+(√5))/( (√5)−1))=(((1+(√5))^2 )/4) ⇒ x=((1+(√5))/2)=ϕ  ϕ^2 =ϕ+1 ⇒  5(ϕ+1)−5ϕ−1=4
$$\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}=\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow\:{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\varphi \\ $$$$\varphi^{\mathrm{2}} =\varphi+\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{5}\left(\varphi+\mathrm{1}\right)−\mathrm{5}\varphi−\mathrm{1}=\mathrm{4} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *