Question

Determine the number and type of complex solutions and possible real solutions for each of the following equations.
2x^4 + x^2 – x + 6 = 0

Answer 1

@CGGURUMANJUNATH

Answer 2

@Harkirat

Answer 3

@eliassaab

Answer 4

@tkhunny

Answer 5

Descartes Rules of Signs will get you there quickly. Do you know it?

Answer 6

2 sign changes

Answer 7

@tkhunny

Answer 8

@mathmale

Answer 9

This means what as far as Positive Real roots are concerned?

Answer 10

I am not sure @tkhunny

Answer 11

2 sign changes. There are AT MOST 2 Positive Real Roots. If there are not 2, it must be less by a multiple of 2. Thus, positive real roots are either 2 or 0. No other options.

Now, consider f(-x) = 2x^4 + x^2 + x + 6

NO sign changes. There are NO Negative Real roots.

Recap.

Positive Real: ??
Negative Real: ??
Complex: ??

Answer 12

no real roots but idk how many complex @tkhunny

Answer 13

Why would it be no real roots? We just said there might be two positive real roots.

It's either this (2 + real, 0 - real, 2 complex) or this (0 + real, 0 - real, 4 complex)

The existence of the word "possible", tells me that we are done.

Answer 14

There are 0 real roots and 4 complex roots how do i find those @tkhunny

Answer 15

I graphed it and ynever=0 so that means no real roots

Answer 16

Why? The problem statement makes no such request.

Answer 17

So would i just put a graph into the problem and show that there is no real solutions then say there must be 4 complex @tkhunny

Answer 18

There IS a way to find all the roots of a Quartic Polynomial, but I'm pretty sure you are not expected to know it. Numerical methods are the way to go.

Answer 19

ok so then the answer to my question is there are no real roots but there are 4 complex roots

Answer 20

That's too much. The question asks for POSSIBLE. The CORRECT answer is:

(2 + real, 0 - real, 2 complex) or (0 + real, 0 - real, 4 complex)

If you would also like to say that you used a graphical method and determined that it was

(0 + real, 0 - real, 4 complex)

That would be extra.

If you used a numerical algorithm to actually find the roots

(0.863-0.921i, -0.863-1.067i, 0.863+0.921i, -0.863+1.067i)

That would be still extra. If you JUST want to answer the question, you should quit at the first answer where we talk about "POSSIBLE" real roots.

Answer 21

4x3 – 12x + 9 = 0
Would this be 1 real solution and two complex? @tkhunny

Answer 22

@pooja195

Answer 23

Sign changes of f(x)?
Sign changes of f(-x)?
Just enumerate.