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zmudz
 one year ago
Let \(f\) be a function such that \(\sqrt {x  \sqrt { x + f(x) } } = f(x) , \) for \(x > 1\). In that domain, \(f(x)\) has the form \(\frac{a+\sqrt{cx+d}}{b},\) where \(a,b,c,d\) are integers and \(a,b\) are relatively prime. Find \(a+b+c+d.\)
zmudz
 one year ago
Let \(f\) be a function such that \(\sqrt {x  \sqrt { x + f(x) } } = f(x) , \) for \(x > 1\). In that domain, \(f(x)\) has the form \(\frac{a+\sqrt{cx+d}}{b},\) where \(a,b,c,d\) are integers and \(a,b\) are relatively prime. Find \(a+b+c+d.\)

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freckles
 one year ago
Best ResponseYou've already chosen the best response.3I got a=1 and b=2 and 2c+d=5 ... I plugged into x=2 into both forms to get this I bet you can plug in another value for x to get another equation relating c and d and then solve that system for c and d

dan815
 one year ago
Best ResponseYou've already chosen the best response.3just 1 question.. 2,1 is relatively prime im assuming is 2,1 also said to be relatively prime

freckles
 one year ago
Best ResponseYou've already chosen the best response.3the gcd(2,1)=1 so 2 and 1 are relatively prime

dan815
 one year ago
Best ResponseYou've already chosen the best response.3oh okay that cool, i didnt realize gcd's protected u from negative factors till now lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.3plugging in x=3 will give you an easy solution to play with

freckles
 one year ago
Best ResponseYou've already chosen the best response.3well equation relating c and d I mean

freckles
 one year ago
Best ResponseYou've already chosen the best response.3gcd(x,y)=d where d>=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.3if x and y are negative you can just look at the absolute values of them

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\gcd(2,1)=\gcd(2,1)=\gcd(2,1)\]
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