anonymous
  • anonymous
I have a question we are doing quadratic equation and I have an example which I do not think it can be an quadratic equation. x^5+3x-7=0
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Because I believe if you factor the equation it becomes x^5-4
anonymous
  • anonymous
qudratic form : x^2 + x + C x^5 is def not quadratic
anonymous
  • anonymous
so in other words you can't solve this equation?

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anonymous
  • anonymous
This x^5-3x+7=0 is in standard form?
anonymous
  • anonymous
that is not in standard form for quadratic,
toxicsugar22
  • toxicsugar22
hey can u help me
anonymous
  • anonymous
suzi20 I have a question for you but if you would take 5^2+3x+7=0 this is a quadratic equation correct?
anonymous
  • anonymous
5^2+3x+7=0 yea this is quadratic
anonymous
  • anonymous
the answer would be x=-3+i \[\sqrt{131}/10\]
anonymous
  • anonymous
and -3-i sqrt 131/10
anonymous
  • anonymous
suzi20 is that correct what I just posted for the answer
anonymous
  • anonymous
sorry, i miss read, 5^2+3x+7=0 is not quadratic because 5^2 = 25 and this is constanta
anonymous
  • anonymous
3x + 32 = 0 x = -32/3
anonymous
  • anonymous
how would you create a quadratic equation that was not constata?
anonymous
  • anonymous
5x^+3x-7=0 would this be a quadratic equation since the last sign was changed to minus instead of addition
anonymous
  • anonymous
suzi20 or would it not make a difference would it still be constata?
anonymous
  • anonymous
5x^2 + 3x - 7 is quadratic since u have variables x^2 and x 5^2 + 3x - 7 is linear because u just have 1 variable which is x
anonymous
  • anonymous
then the answer would be x=-3+sqrt149/10, -3-sqrt149/10
anonymous
  • anonymous
5x^2 + 3x - 7 a = 5 b= 3 c=-7 D = b^2 - 4ac = 9 + 140 = 149 x = -b +/- sqrt (D) over 2a = -3 +/- sqrt (149) over 10 x[1] = -3 + sqrt(149) over 10 x[2] = -3 - sqrt(149) over 10
anonymous
  • anonymous
Thank you suzi20 you was a big help!

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