Zero's of a function.

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hannerkinz

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Aug 16, 2010
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So I know how to find a zero of a regular polynomial. (binomial, trinomial, etc.)

Here's the problem.
Find the zero's of this function.

g(x)=x^2+8x-20
g(x)=x^5+4

What do I do? :eek:
Do I set them equal to eachother and factor??
HELP!!
 
hannerkinz said:
I know how to find a zero of a regular polynomial. (binomial, trinomial, etc.)

Notes about terminology: Polynomials do not have zeros; they have roots. Functions have zeros. Equations have solutions.

Here's the problem.
Find the zero's of this function. Which function? I see two definitions for function g.

g(x)=x^2+8x-20 ? This version of function g is defined by a "regular" trinomial.

Use the Quadratic Formula.

g(x)=x^5+4 ? Is this version of function g a separate exercise ?

Is function g a piecewise function defined by both of the posted polynomials ?

Is something else going on ?

Please clarify. 8-)
Click to expand...
 
I'm not to sure.
My teacher didn't really clarify.
I'm not sure if she wants me to set them equal to eachother since they're both functions of g or not.

ex: x^2+8x-20=-x^5+4
giving me: x^5+x^2-24=0


but it doesn't make sense that way!
I'm so confused. D:

What do you suggest??
 
Hi:

More notes about terminology.

The two functions that you posted are not "functions of g". They are both functions of x, the independent variable. The symbol g is just the name of each function. The symbol g(x) is the dependent variable; it stands for y.

Well, I'm not sure what to suggest.

Did you copy the assignment off of the blackboard? Did it come from your textbook? Is it a handout? Is it on your instructor's web page? What exactly are you looking at?

Did you type the information exactly as you see it?

Let me know, and we'll go from there.

Cheers,

~ Mark
 


Okay. Assuming that you copied the information from the overhead projection verbatim, I would say that these are two different exercises, and that your instructor mistakenly wrote "this function" versus "each of these functions".

Whenever a single exercise contains more than one function, each of the functions must have different names. (I mean, if everybody in the world was named Jack, how would anybody know who Jack is?)

So, if you have no way to contact your instructor for clarification, before this assignment is due, I would treat it as two separate exercises.

It's easy to find the roots of x^2 + 8x - 20; use the Quadratic Formula.

Gosh, I just realized that the five roots of x^5 - 4 are not so straightforward. The single Real root is easy enough, but finding the four Complex roots (with imaginary parts) is quite a bit more involved.

Hmmm. Let me ask you another question. What particular algebra topics are you currently studying in class? What did your instructor lecture about, when this assignment was given?

I'm beginning to think that you might have missed some given information on that overhead projection, or, perhaps, you missed some additional verbal instruction(s).

 
I think s/he was supposed to find intersection points of those two functions.
 


Subhotosh, the x-coordinate of the intersection point (between these two functions with the same name) is an Irrational root of a fifth-degree polynomial.

The instruction to "Find the zero's of this function" seems like a strange way to request the intersection point.

I suppose anything is possible, though. :roll:

< It's almost time, today, to increase b(T) by one … yet again >

 
hannerkinz said:
My teacher didn't really clarify.
I'm not sure if she wants me to set them equal to eachother since they're both functions of g or not.
Click to expand...
Is your teacher a History teacher on loan?
 
\(\displaystyle If \ we \ are \ given \ g(x) \ = \ x^2+8x-20 \ and \ we \ want \ to \ find \ its \ two \ roots, \ then\)

\(\displaystyle x^2+8x-20, \ we \ set \ to \ zero \ and \ solve. \ Now, \ any \ quadratic \ can \ be \ solve \ by \ the \ quadratic\)

\(\displaystyle formula, \ however \ since \ we \ are \ dealing \ with \ integers, \ perhaps \ grouping \ will \ alleviate \ the\)

\(\displaystyle amount \ of \ grunt \ work \ necessary. \ Now \ what \ multiplies \ to \ -20 \ and \ adds \ to \ 8? \ Why\)

\(\displaystyle 10 \ and \ -2 \ does. \ Hence, \ x^2+8x-20 \ = \ x^2+10x-2x-20 \ = \ x(x+10)-2(x+10) \ = \ (x+10)(x-2).\)

\(\displaystyle Hence, \ the \ roots \ of \ g(x) \ are \ x \ = \ -10, \ 2.\)
 
What's your point, BigG? Even the original poster knows that...
 
BigGlenntheHeavy said:
Denis, I see by your profile that you are retired.
Hence, my question,how come you never grew up?
Click to expand...
Ah gee, me ain't too good at seplling: I meant "retarded".

"The adult with nary a smile
Is but the skeleton of his inner child"

By the way, Oh Big One, I've seen quite a few "ungrown" posts of yours,
so all is not lost: keep it up :idea:
 


I'm waiting for hannerkinz to tell me what the instructor was talking about, at the time this assignment was given.

Maybe the class is learning how to zoom in on calculator graphs, to obtain numerical approximations for function zeros.

(I also noticed that the second function g is defined by x^5 + 4 in one post and -x^5 + 4 in another. More confusion.)

 
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