anonymous
  • anonymous
I need some help with understanding a basic mathematical proof.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Disprove the statement: There is a real number x such that \[x ^{6}+x ^{4}+1 = 2x ^{2}\]
anonymous
  • anonymous
Proof: For all real numbers, x, \[x ^{6}+x ^{4}+1\neq 2x ^{2}\].
anonymous
  • anonymous
Case 1: suppose x = 0, then x^6 + x^4 + 1 = 1 and 2x^2 = 0, so x^6 + x^4 +1 =/= 2x^2.

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anonymous
  • anonymous
Case 2 (this is the one I'm having a little trouble understanding) : Suppose \[x \neq0\], then x^2 >0. Since \[x ^{6}=x ^{2^{3}}\], x^6 >0. Therefore x^6+(x^2-1)^2 >x^6, so x^6 +x^4 +1 - 2x >0
anonymous
  • anonymous
(x^2-1)^2 is factored from x^6+x^4+1-2x^2...can I end the proof there?
jim_thompson5910
  • jim_thompson5910
x is nonzero ---> x^2 is positive x^4, x^6, etc are all positive as well
jim_thompson5910
  • jim_thompson5910
and as you pointed out, x^4-2x^2+1 turns into (x^2-1)^2 so (x^2-1)^2 is always positive when x is nonzero
anonymous
  • anonymous
Yes, so x^6+x^4+1 - 2x^2 >0 ...can I say that x^6+x^4+1 >2x^2? Thus x^6+x^4+1 =/= 2x^2
jim_thompson5910
  • jim_thompson5910
I would do it like this x^6+x^4+1 - 2x^2 >0 x^6+x^4- 2x^2+1 >0 x^6+(x^2- 1)^2 >0 when x is nonzero, x^6 is positive and (x^2- 1)^2 is positive so that's why x^6+(x^2- 1)^2 >0 is true and x^6+(x^2- 1)^2 = 0 will never be true
anonymous
  • anonymous
I understand what you did, but how (exactly) does that show x^6+x^4+1 =/= 2x^2?
anonymous
  • anonymous
Or is that supposed to be obvious so I don't have to write anything more?
jim_thompson5910
  • jim_thompson5910
it shows that x^6+(x^2- 1)^2 =/= 0
jim_thompson5910
  • jim_thompson5910
which consequently means x^6+x^4+1 - 2x^2 =/= 0
anonymous
  • anonymous
ahh, yes so bringing 2x^2 to the left side gives x^6+x^4+1 =/= 2x^2
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
thank you so much~
jim_thompson5910
  • jim_thompson5910
no problem

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