anonymous
  • anonymous
How many Real solution does the equation x^7 + 14x^5 + 16x^3 + 30x = 0 have ?
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
x(x^6+14x^4+16x^2+30)=0 so x=0 or x^6+14x^4+16x^2+30=0 but if x is a real number x^6+14x^4+16x^2+30>0 so it has one solution: x=0
tkhunny
  • tkhunny
One at x = 0. Factor this out and move on. If we define the LHS (with x factored out) as f(x), all coefficient signs are positive. There cannot be a positive Real Solution. No sign change if f(-x), either. We are done. x = 0 is the only Real Solution. marsss identical conclusion is much simpler, but less general.
anonymous
  • anonymous
i explained the solution of this question so it doesn't need to be general, @tkhunny

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tkhunny
  • tkhunny
@marss Absolutely agree with your comment. I just wanted the rest along for the ride. Not a complaint or criticism at all. :-)

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