My Math Forum prove prove prove. currently dont know where to post

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 July 29th, 2008, 07:09 PM #1 Newbie   Joined: Jul 2008 Posts: 1 Thanks: 0 prove prove prove. currently dont know where to post 3<(13)^1/2< 4, deduce that 0<(13)^1/2 -3<1 b) [(13^1/2)+3]^4 + [(13^1/2)-3)]^4 =1904 c)1903<[(13^1/2)+3]^4 <1904 here is another question expresss [(p^1/2 +q(r^1/2)]^2 in the form =a+b(c^1/2). without evaluating the square root or using table or calculator,show that 10^1/2 +2(2^1/2) <6
July 31st, 2008, 07:27 AM   #2
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Re: prove prove prove. currently dont know where to post

Moving this to the geometry section.

***
Quote:
 Originally Posted by qweiop90 3<(13)^1/2< 4, deduce that 0<(13)^1/2 -3<1
In the first inequality (3<...<4), subtract 3 from each expression. What happens?

Quote:
 b) [(13^1/2)+3]^4 + [(13^1/2)-3)]^4 =1904
Notice that if we expand out both exponents, everything with 3^1 and 3^3 will cancel. Do you know the binomial theorem?

Quote:
 c)1903<[(13^1/2)+3]^4 <1904
Use the binomial theorem for this one also.

here is another question

expresss [(p^1/2 +q(r^1/2)]^2 in the form =a+b(c^1/2). without evaluating the square root or using table or calculator,show that 10^1/2 +2(2^1/2) <6[/quote]

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