Question

find all zeros In the equation
x^4-8x^2-9=0

Answer 1

Temporarily replace x^2 with n. You now have a quadratic in "n"

n^2 - 8n - 9 = 0

Can you start by solving the quadratic here? @chelseahearts159

Answer 2

is it
x(x-8x-9)

Answer 3

This quadratic in "n" can be solved by factoring. Look for solutions of the form:

(n + )(n - )

You are looking for 2 numbers that when multiplied give you -9 but when added give you -8.

Start by listing all factors of -9, both the positive and negative factors.

Answer 4

positive 1 and negative nine added give you negative 8

Answer 5

That's good! Can you fill those 2 numbers in the format I gave you?

(n + )(n - )

Answer 6

(n+1)(n-9)

Answer 7

You're doing really well here. Now, the next step is to substitute the x^2 back for the n. The second factor will be the difference of 2 perfect squares, so you can factor that. The first factor will be solvable with 2 complex numbers. You will have a total of 2 real and 2 imaginary solutions.

Answer 8

so (x+1)(x-9)

Answer 9

Not "x", but "x^2" as follows

(x^2 + 1)(x^2 - 9)

(x^2 + 1)(x - 3)(x + 3)

The first factor has "i" and "-i" as solutions.

Answer 10

okay

Answer 11

So you have 4 solutions. 2 real and 2 imaginary. 3, -3, i, and -i

Answer 12

what do the "i"s stand for

Answer 13

These solutions are what is called your "zeros". They are the values of "x" such that they will give you "0" when you multiply out the factors by substituting for "x" any one of them.

"i" is the sqrt(-1)

Answer 14

So,

(x + 3)(x - 3)(x - i)(x + i)

Answer 15

okay, are the zeros 3, -3, 1, and -1? or do I actually put "I"

Answer 16

-1 is not the same as sqrt(-1). You use those 4 values as your "zeros". When it comes to the imaginary solutions, you can write them either as:

i and -i or as sqrt(-1) and -sqrt(-1)

Answer 17

okay thank you! would you be able to help me a couple other problems?

Answer 18

Just post your problems to a whole new thread (not here) and either I or another helper will come by.

And you're welcome!

Answer 19

Well, it's just me and you left. Any questions on this problem? @Farheen28

Answer 20

Good luck to you in all of your studies and thx for the recognition! @chelseahearts159