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[問題] pure 2005IQ10ai (solved)

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pure 2005IQ10ai (solved)

Let f(x)=x^4+2ax^2+4bx+c, where a,b,c are real numbers with b is not equal to 0. It known that f(x)=(x^2+2tx+r)(x^2-2tx+s), where r,s,t are real numbers

Prove that t is not equal to 0.

Marking shemce said:put x^4+2ax^2+4bx+c=(x^2+2tx+r)(x^2-2tx+s)
we have t(s-r)=2b
b is not equal to 0, so t is not equal to 0.

but how about (s-r)? I think there's probability that s-r=0.
So, I would like to ask how to prove (s-r) is also not equal to 0.

[ 本帖最後由 lamppast 於 2013-2-15 08:47 PM 編輯 ]
   

TOP

if s-r=0  => b = 0

since b <> 0 , s-r <> 0

actually it's obvious that both t and s-r <> 0

TOP

oh, i see.
thanks very much!!

TOP

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