## AM-GM problem

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rakeen
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### AM-GM problem

Kulna08-hisec-Q6

the AM(arithmetic mean) of x and y is a. and GM(geometric mean) is g. if a+g = y-x then what is $\frac{x}{y}$
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AntiviruShahriar
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### Re: AM-GM problem

rakeen
Posts: 384
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### Re: AM-GM problem

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AntiviruShahriar
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### Re: AM-GM problem

$a+g=y-x$
$\frac{x+y+2\sqrt{xy}}{2}=y-x$
$(9x-y)(x-y)=0$
if $x=y$ then 1st line is wrong...........s_____
avror bepare onno vaiara valoo bolte parbe

rakeen
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### Re: AM-GM problem

how did u get (9x-y)(x-y) = 0
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AntiviruShahriar
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### Re: AM-GM problem

rakeen wrote:how did u get (9x-y)(x-y) = 0
after the 2nd line of my last post::::
$\sqrt{xy} =\frac{y-3x}{2}$
$4xy=y^2-6xy+9x^2$
$(9x-y)(x-y)=0$
if $x=y$ then $a+g =y-x$ line is wrong...........
so $\frac{x}{y}=\frac{1}{9}$

Dipan
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### Re: AM-GM problem

Thanks to rakeen and anti...I have got a huge benefit from this two mathematicians(pam)

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### Re: AM-GM problem

Bismillahir Rahmanir Raheem.

If x=y then is a+g=y−x wrong?

(0+0)/2 + sqrt(0*0) = 0-0.

ibrahim
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### Re: AM-GM problem

AM bujhi, GM ta keo bujhaya diba?

ibrahim
Posts: 25
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Location: Dhaka

### Re: AM-GM problem

AM bujhi, GM ta keo bujhaya diba?

akashe shurjota uthe nirobe