can someone help me with Nature of roots and zeros? Find all the zeros of each polynomial function. f(x) = x4 – 5x2 – 36 1) Select one: a. ±3, ±2i b. ±3 only c. ±2i only d. none of the above To find the roots of a polynomial equation you should: Select one: a. only use Descartes Rule of Signs. b. only use the Quadratic Formula. c. only factor it. d. use any method that gives the roots of the function
Nikki: One option available to you would be substituting x^2=y. If you do this, you end up with y^2 - 5y -36. What are the roots of this quadratic equation? Write out all that apply.
For second I think d is correct
Depends on question which method suits it.
yeah I say D too
@mathmale the root would be 2 right?
Plug it into the equation then if you get its value =0 then its correct
i dont know how to?
the function you have can be factored \[x^4 -5x^2 - 36 = (x^2 - 9)(x^2 + 4)\] now you can solve each factor \[x^2 - 9 = 0~~~~~~~~~~~~~~ and~~~~~~~~~~~x^2 + 4 = 0\]
the equation is called an equation that is reducible to a quadratic.... you can make a substitution as some people have suggested...and then factor , or simply factor... hope it helps
so if there both 0 its correct but then how do you know?
well to find roots or zeros... then y or f(x) or P(x) needs to be equal to zero... what you have written in the question is simply an algebraic expression that can be factored...
huh?
wait the roots for both of those is 2?
well can you solve \[x^2 - 9 = 0\] that is one factor... so if that factor equals zero, the whole equation equals zero... so solve for x. the other factor is \[x^2 + 4 = 0\] same thing, find the values of x that make the factor zero... remember the question asks you to find the values of x, often called roots or zeros that make the equation equal zero.
-2i?
well remember you are solving a degree 2 equation so 2 solutions...
-3 is the other one
The original equation is of the fourth order (the highest power is 4). Thus, there should be FOUR solutions.
how do you get 4 solutions?
If the highest x power in the given equation is n, then the order of the polynomial is "n th." Thus, because the highest power of x that you're working with here is 4, it's a fourth-order polynomial and has 4 solutions.
x^2 - 9 = 0 is a 2nd order poly and thus has 2 solutions.
how do i find 4 solutions if theres only 2 mini equations to work with?
Please review campbell_st's first comment. He was right on target there. He factored the original function into two factors, each of which has 2 factors. 2 * 2 = 4 factors total.
Each "mini-equation" has 2 factors and thus two solutions.
You've already tried to factor and then solve x^2+4=0. You got -2i as one root. The other is +2i. We call these roots "complex pairs."
and the other one is -3 and +3?
Yes, provided that you began with x^2-9=0.
I'm assuming you're satsified and finished with this problem. If not, please say so. Have to get off the computer!
so its +3 and +2 ?
- you've forgotten the negatives!
But that reply contains only 2 solutions. Just earliler, we got 4! You have to list all four solutiions.
List them as a set: {3i, 2, [and two more]}
complex roots such as 3i stay complex; do not drop the "i".
2i right?
First of all, Nikki, how many solutions does your original problem have?
4
Yes. So you MUST list all four. If the first complex root is 3i, what is the accompanying complex root?
2i
There is no complex root containing a '2' both complex roots contain a '3'. If the first complex root is 3i, what is the accompanying (negative) complex root?
complex roots just about always come in PAIRS.
-3i? and the other one is 2i and -2i
2i and -2i are the roots of x^2+4=0. The roots of x^2-9=0 are real. Again, please go back and review Campbell_st's first post. Looks like you and I have both made a few minor mistakes. Campbell is correct. What are the two (complex) roots of x^2+4=0? What are the two (real) roots of x^2-9=0? The four solutions of the original equation are { you fill in }
huh? so 3 isnt correct?
Nikki, please recognize that this equation has FOUR solutions. Please don't type out just one. 3 is one of the two real roots of x^2-9=0. What is the other real root? What are the two (complex) roots of x^2+4=0?
Nikki: Please fill this out: The four roots of the original equation are { }.
I dont know
ive told you what i know
What are the two roots of x^2+4=0? We've gone over this before.
-2 and +2
Actually, 2i and -2i. So the set of your complex roots is {2i,-2i} (a pair of roots).
what are the roots of x^2-9=0? Write them as a set, please.
as a set?
{2i,-2i} is a SET of roots, your complex roots. { ? , ? } is a SET of your real roots. Please fill in.
If I ask for a set, list the numeric values and enclose all of the values within { } to show that they are a SET.
Nikki, what are the roots of x^2+4=0? Hint: there are two of them, and both are complex. Write them as a set, please. Answer: {2i, -2i}.
what are the roots of x^2-9=0-? There are 2 of them, and both are real. Write them as a set, please.
Your answer:
By now you are probably tired and frustrated. Do you want to finish this problem now or finish it later? Your choice.
Sorry, Nikki, but I'm logging off. I apologize for the extreme length of this discussion. Both of us made some mistakes, which helped to extend the discussion. I'd suggest that you either return to this discussion when you're ready, or repost it, so all discussion will be fresh. You might want to review this discussion, summarize what you've learned from it, and make up a list of questions to ask for clarification. Take care. mathmale
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