- #1

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To Solve:-

p^7 + p^3 - p^2 + 1 =0

here p = dy/dx

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- Thread starter heman
- Start date

- #1

- 349

- 0

To Solve:-

p^7 + p^3 - p^2 + 1 =0

here p = dy/dx

- #2

HallsofIvy

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My suggestion- go insane!

- #3

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I asked my tutor and he said that there will be atleast 1 real root

=> p=k

or y=kx + c

From here i am thinking ahead!

- #4

saltydog

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So we look for solutions of the form y=mx+b, plug it in, get a polynomial in m and solve for the roots. So there's one real root. So we find it numerically (I ain't proud), and then we get the solution:

[tex]y(x)=m_rx+b[/tex]

So, how do we know if there are other solutions?

- #5

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saltydog said:So, how do we know if there are other solutions?

The same thing i asked him!

He said our aim is to get just 1 family of curves of solution!

The other six can be complex,real who knows!

- #6

HallsofIvy

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(Yes, the polynomial p^7 + p^3 - p^2 + 1 =0, by DesCartes' "rule of signs", has one

- #7

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Your Predictions!!

(1) 1 negative real root and 3 pair of conjugate complex roots

I understand that it can have 3 pair of conjugate complex roots-but how do you say that real root will be negative?

(2) 1 negative real root, 2 positive real roots, and 2 pair of conjugate comples roots

How are you able to tell the signs??

I have taken the Complex Analysis Course last semester but i haven't come across any such theorem which tells about the sign!

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