how to Solve equations in the complex number system

[georgebaseball]

Hello forum how are you doing.

Could you please correct this exercise
- Solve the given equation in the complex number system. x^4+5x^3+2x^2-x+6=0 hint, find ) p(-2)

so I just use polynomial division,I would show you what I did but I don't know how to type those lines used for polynomial division

and I got -8 is that the answer that I'm supposed to get?

thank you.

 

[galactus]

You're being asked to find the roots. This particular polynomial has 2 real and 2 complex roots.

Maybe I am missing something, but I fail to see what the -2 has to do with it. When entered into the equation it does equal -8, but that is not a root. Unless they are hinting at the I.V.T.

You can try the Intermediate Value Theorem to find the real roots, provided you have a calculator. If you try -1 and -2, you see p(-1)=5 and p(-2)=-8. There is a sign change, so one of the real roots lies between -1 and -2. Keep narrowing in on it.

Also, there is a formula for a quartic equation, just as there is a quadratic formula.

It's a monstrous thing, but will give you the roots. What to check it out?. You'll be amazed that someone actually derived this leviathan. Go here:

http://planetmath.org/encyclopedia/QuarticFormula.html

 

[Count Iblis]

Solving the quatric for a given polynomial (coefficents are numbers not variables)is actually much simpler than using that general formula. This is for a similar reason as why calculating a determinant of, say, a given 30 by 30 matrix (matrix elements are numbers not variables) can be done pretty fast using row and column manipulations. But you should not write down the general formula of a general 30 by 30 matrix where all the matrix elements are independent variables and insert your numbers in there.

In this case we first put x = y - 5/4 to eliminate the x^3 term. You get an equation of the form:

y^4 + b y^2+ cy +d = 0

You write this as:

y^4 = -b y^2- cy -d

We want to take the square root of both sides, but the right hand side is not, in general, a perfect square. Now y^4 = (y^2)^2. If you add a parameter p to y^2 you get:

(y^2+p )^2 = y^4+2py^2 + p^2 =

-b y^2- cy -d + 2py^2 + p^2=

(2p-b)y^2 - cy +p^2-d

Choose p such that the discriminant of the quadratic is zero. This gives a third degree equation for p.

To solve a third degree equation of the form:

x^3 + bx^2 + cx+d =0

You eliminate the x^2 term by substituting x = y-b/3

You get an equation of the form

x^3 + ex + f = 0

This equation is very easy to solve. You just compare the equation with the formula:

(a+b)^3 = a^3 + 3a^2b+ 3ab^2 + b^3

Write this as:

(a+b)^3 - 3ab(a + b) - (a^3 + b^3) = 0

So, if we can find a and b such that

-3ab = e and

a^3 + b^3 = -f

then a + b will be a solution of the cubic equation x^3 + ex + f = 0. Now:

-3ab = e --->

27 a^3 b^3 = e^3

So, if we put A = a^3 and B = b^3 then:

A*B = (e/3)^3

A + B = -f

Solving for A and B thus amounts to solving a quadratic equation.

Then you extract cube roots to find a and b. Add them up to find the solution to the cubic equation. There are three complex cube roots. But note that you if you choose one of the three for, say, a, then b is fixed because you need to satisfy the equation:

-3ab = e

So, you see that solving the quatric or cubic equation involves only a few simple steps. The general formula is complicated, just like the general solution for 5 linear equation with 5 unknowns is also extremely complicated. The reason is that iof you work out a solution of an equation in which the copefficients are variables the formula will grow bigger and bigger after each step. If the coefficients are numbers then after each step the formulas won't get much bigger.