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The AM between two positive numbers A and B (A>B) is twice the GM .prove that A:B= 2+√ 3 : 2-√3

Divyansh , 8 Years ago
Grade 11
anser 1 Answers
Shailendra Kumar Sharma

Last Activity: 8 Years ago

given condition is that
(A+B)/2 = 2\sqrt{AB} 
(A+B)/\sqrt{AB}  =4
Applying dividendo componendo 
((A+B)+2\sqrt{AB})/(A+B-2\sqrt{AB})=(2+1)/(2-1)
It can be seen from now
(\sqrt{A } +\sqrt{B})^{2} /(\sqrt{A } -\sqrt{B})^{2} = 3/1
Square root both sides
(\sqrt{A } +\sqrt{B}) /(\sqrt{A } -\sqrt{B}) = \sqrt{3}
Components dividendo 
A/B =(1 +√ 3)/(-√ 3 +1)
 
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