How to get the equation from no root (imaginary roots)?

Question

How can to get the equation from no root?
For example: like 4i and -6i (I don't know if it is possible, just to give a general idea)

Need steps and explanations

Does the root have to be an even number?

Answer 1

If 4i and -6i are your roots, the binomials they came from are
(x - 4i)(x + 6i) = 0 Multiply using FOIL
x^2 + 2ix - 24i^2 = 0 Since i^2 = -1 you'll have
x^2 + 2ix + 24 = 0

Answer 2

You mean an equation with real coefficients, right? If not you just use

(x - 4i)(x + 6i) = x^2 + 2i + 24

as if they were real.

To get real coefficients you need to get rid of the imaginary part, and you can do this with the difference of square technique. That is, for 4i use

(x - 4i)(x + 4i) = x^2 + 16

For -6i use (x - 6i)(x + 6i) = x^2 + 36.

So for this example (x^2 + 16)(x^2 + 36) = x^4 + 53x^2 + 576 = 0

For mixed complex numbers it is the same idea, but a little messier math. For instance, if you have 1 + 2i

(x + 1 + 2i)(x + 1 - 2i) = x^2 + 2x + 1 +4 = x^2 + 2x + 5 is the factor you need.

Over the reals, every polynomial factors uniquely into a product of linear and quadratic factors with real coefficients, where the quadratics each represent a pair of conjugate complex numbers.

Answer 3

I'm guessing you mean that you are given some roots and asked to find an equation that has those roots. If so, and all the roots are imaginary, then you plug in the roots into the formlula:
(x-r1)(x-r2)...
When dealing with imatinaries, you must remember that each imaginary root has a conjugate. In math lingo:
a+bi is a root iff a-bi is a root.

In this case you would take each root and its conjugate and subtract them each from x and multiply them together:
(x-4i)(x+4i)(x-6i)(x+6i)
Each pair of terms is the sum of squares so you can simplify to get:
(x^2+16)(x^2+36)
Then multiply out to get:
x^4+52x^2+576

Answer 4

You really need to work on your English. If I understand you correctly, you want to know how to formulate a homogeneous polynomial equation of the form f(x) = 0 which has only imaginary roots? If ythe roots you want are (let's say) r1 and r2, then f(x) = (x-r1)*(x-r2). Example: You wanted roots of 4i and -6i, so form
(x-4i)*(x+6i) = x²+2ix+24. Then the quadratic
x²+2ix+24 = 0 will have roots of 4i and -6i. This also works for real or complex valued roots.


Doug

Answer 5

Given a quadratic equation,

ax^2 + bx + c = 0

if b^2 < 4ac, then the roots are imaginary
_____

If you are are given the roots,

x = 4i and x = -6i

then the equation is,

(x - 4i)(x + 6i)

= x^2 -4xi + 6xi -24i^2

= x^2 + 2xi -24(-1)

= x^2 + 2xi + 24

We can use the general equation to find the roots of the above equation and the roots will be the same.

Solve for,

x^2 + 2xi + 24 (taken from above)

a=1
b=2i
c=24

Use the general quadratic equation,

x = [-b +/- sqrt(b^2 - 4ac)]/(2a)

x = [-(2i) +/- sqrt((2i)^2 - 4(1)(24)]/(2(1))

= [-2i +/- sqrt(-100)]/2

= [-2i +/- 10i]/2

= -i +/- 5i

x = 4i or x = -6i

Answer 6

You put them into a factored setting (i.e.)
4i ; -6i
(x-4i)(x+6i)=0
then foil them:
x^2 +2i -24i^2 = x^2 + 2i + 24 = 0

Answer 7

you can get the equation from imaginery roots. In order for the equation to have real number coefficients, the roots would have to be in conjugate pairs. ex. (5+3i) and (5-3i)

if those are the roots the equation can be found like this

(x-(5+3i))(x-(5-3i))=0
x^2-(5+3i)x-(5-3i)x+(25-15i+15i-9i^2)=0
x^2-5x-3ix-5x+3ix+(25+9)=0
x^2-10x+36=0