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Solve equation (sq root(5+2square root 6))^x+(sq root(5-2square root 6))^x=98?

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minlux | Honors

Posted July 27, 2013 at 1:21 PM via web

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Solve equation (sq root(5+2square root 6))^x+(sq root(5-2square root 6))^x=98?

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aruv | High School Teacher | Valedictorian

Posted July 27, 2013 at 3:08 PM (Answer #1)

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`(sqrt(5+2sqrt(6)))^x+(sqrt(5-2sqrt(6)))^x=98`

`(x^m)^n=x^(mn)`

`therefore`

`((5+2sqrt(6))^(1/2))^x+((5-2sqrt(6))^(1/2))^x=98`

`(5+2sqrt(6))^(x/2)+(5-2sqrt(6))^(x/2)=98`

`5+2sqrt(6)=3+2sqrt(6)+2=sqrt(3)^2+2sqrt(2)sqrt(3)+sqrt(2)^2`

`=(sqrt(3)+sqrt(2))^2`

`and`

`5-2sqrt(6)=(sqrt(3)-sqrt(2))^2`

`Thus`

`(5+2sqrt(6))^(x/2)+(5-2sqrt(6))^(x/2)=((sqrt(3)+sqrt(2))^2)^(x/2)+((sqrt(3)-sqrt(2))^2)^(x/2)`

`=(sqrt(3)+sqrt(2))^x+(sqrt(3)-sqrt(2))^x`

Define a function:

find the zero of f(x) , which will be root of equation f(x)=0.So draw the graph of f(x)

Thus the zero of f(x) is 4.

Thus solution of equation f(x)=0 is x=4

It is easy to verify that

`(sqrt(3)+sqrt(2))^4+(sqrt(3)-sqrt(2))^2=2(9+36+4)=98`

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aruv | High School Teacher | Valedictorian

Posted September 26, 2013 at 7:37 PM (Reply #1)

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Please read `(sqrt(3)-sqrt(2))^4`  instead of `(sqrt(3)-sqrt(2))^2` . It was typographical error.

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