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- 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

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- anonymous

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- anonymous

Because I believe if you factor the equation it becomes x^5-4

- anonymous

qudratic form : x^2 + x + C
x^5 is def not quadratic

- anonymous

so in other words you can't solve this equation?

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- anonymous

This x^5-3x+7=0 is in standard form?

- anonymous

that is not in standard form for quadratic,

- toxicsugar22

hey can u help me

- 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

5^2+3x+7=0 yea this is quadratic

- anonymous

the answer would be x=-3+i \[\sqrt{131}/10\]

- anonymous

and -3-i sqrt 131/10

- anonymous

suzi20 is that correct what I just posted for the answer

- anonymous

sorry, i miss read, 5^2+3x+7=0 is not quadratic because 5^2 = 25 and this is constanta

- anonymous

3x + 32 = 0
x = -32/3

- anonymous

how would you create a quadratic equation that was not constata?

- anonymous

5x^+3x-7=0 would this be a quadratic equation since the last sign was changed to minus instead of addition

- anonymous

suzi20 or would it not make a difference would it still be constata?

- 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

then the answer would be x=-3+sqrt149/10, -3-sqrt149/10

- 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

Thank you suzi20 you was a big help!

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