I need to use the given root to find the solution set of the polynomial equation??

Question

The question is:

x^4-21x^2-100=0;-2i

Does anyone understand this? Because I certainly don't...

Thanks for your help!

Answer 1

Use the given root:

2i is a root.
The equation has real coefficient so its complex conjugate must be a root.
Therefore -2i will be a root

(x+2i) and (x-2i) will be factors.
(x+2i)(x-2i) = x^2 - (2i)^2 = x^2 + 4

Divide x^4-21x^2-100 by x^2 + 4 to get x^2 - 25
Factor x^2 - 25 to get (x+5)(x-5)


x^4-21x^2-100=0
(x+5)(x-5)(x+2i)(x-2i) = 0
x = {-5, 5, 2i, -2i}

Answer 2

If you really don't understand that, then you have no business in the World being in the class. You really haven't met the required prerequisites and putting you into the class is not doing you any favors. But..... On to the question.

x^4 + 21x^2 -100 = 0 is a 'quartic' equation. Solve this one by making the substitution g = x² so that the equation becomes
g² - 21g - 100=0 which is a simple quadratic with roots at
g=25 and g=-4.
Now, since g=x², make the reverse substitution and get
x² - 25 = 0 and
x² + 4 = 0
The first equation factors easily into
(x+5)(x - 5) = 0 so two of the roots are
x = 5 and
x = -5
x² + 4 factors into
(x + 2i)(x - 2i) = 0 so the last 2 roots are
x = 2i andx = -2i

If that totally lost you, you need to get out of that class and get back to one that will teach you the proper prequisites for this class.

Doug

Answer 3

Hi Kitty,

You need to factor this guy:

x^4 -21x^2 -100 = 0
(x^2 - 25)(x^2 + 4) = 0

now let's factor some more :-)

(x+5)(x-5)(x^2 + 4) = 0

now the (x^2 + 4) is where the -2i comes from.

The solutions are:
x = 5
x = -5
x = 2i
x = -2i

("i" is the square root of -1 in case you didn't know that part).

hth.

REgards,
Chas.

Answer 4

If you let v = x^2, then the eqn
v^2 - 21v -100 = 0
(v+4)(v-25) = 0
Now you have (x^2+4)(x^2-25) = 0
Factoring this
x^2 = -4 or x^2 = 25
x = +- 5 from the latter and
x = +-2i from the former...the answer

Answer 5

Since one root is 2i, another root has to be -2i; The other two roots are 2 and -2.