Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 38332 by ajfour last updated on 24/Jun/18

((3+(√5))/2) −(√((3+(√5))/2)) = ?!

$$\:\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:−\sqrt{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\:=\:?! \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18

x=((3+(√(5 )))/2) =((6+2(√5) )/4) =((((√5) +1)/2))^2 =±(((√5) +1)/2) ((3+(√5) )/2)−(((√5) +1)/2) =((3+(√5) −(√5) −1)/2)=1 and ((3+(√5) )/2)−((−((√5) +1))/2) ((3+(√5) +(√5) +1)/2)=2+(√5)

$${x}=\frac{\mathrm{3}+\sqrt{\mathrm{5}\:}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:}{\mathrm{4}} \\ $$$$=\left(\frac{\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\pm\frac{\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}+\sqrt{\mathrm{5}}\:}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}+\sqrt{\mathrm{5}}\:−\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}}=\mathrm{1} \\ $$$${and} \\ $$$$\frac{\mathrm{3}+\sqrt{\mathrm{5}}\:}{\mathrm{2}}−\frac{−\left(\sqrt{\mathrm{5}}\:+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}+\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{2}}=\mathrm{2}+\sqrt{\mathrm{5}}\: \\ $$

Commented by MJS last updated on 24/Jun/18

we don′t have 2 solutions in cases like this one 3−(√4)=3−2=1 it′s not an equation like x^2 =4, where we get 2 solutions. otherwise (√2)−(√2)={−2(√2); 0; 2(√2)} it′s tbe golden ratio φ=(1/2)+((√5)/2) and its square φ^2 =(3/2)+((√5)/2) ⇒ φ^2 −φ=1

$$\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{in}\:\mathrm{cases}\:\mathrm{like}\:\mathrm{this}\:\mathrm{one} \\ $$$$\mathrm{3}−\sqrt{\mathrm{4}}=\mathrm{3}−\mathrm{2}=\mathrm{1} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{like}\:{x}^{\mathrm{2}} =\mathrm{4},\:\mathrm{where}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{2}\:\mathrm{solutions}. \\ $$$$\mathrm{otherwise}\:\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}=\left\{−\mathrm{2}\sqrt{\mathrm{2}};\:\mathrm{0};\:\mathrm{2}\sqrt{\mathrm{2}}\right\} \\ $$$$ \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{tbe}\:\mathrm{golden}\:\mathrm{ratio}\:\phi=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{its}\:\mathrm{square} \\ $$$$\phi^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:\phi^{\mathrm{2}} −\phi=\mathrm{1} \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com