3-5-2-3-5-2-

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

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

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

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

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