Solve Polynomial Problem: Find k (3x^2+kx-8) w/ 2 as Zero

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In summary, Problem 16 is about finding a cubic polynomial with integral coefficients having 3i and -6 as zeroes. You provided the solution of x^2-6x-3ix+(-18i), which is correct. However, -18i, being an imaginary number, is not what we would normally consider an integer, and in order to have integer (in particular "real") solutions, you would have to have -3i as a root too.
  • #1
mustang
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Problem 16.
Find a cubic polynomial with integral coefficients having 3i and -6 as zeros.
This is what I have done:
(x-r1)(x-r2)
(x-3i)(x+6)
x^2-6x-3ix+(-18i)
x^2-(6-3i)x+(-18i)
Is this right?

Problem 18.
If 2 is a zero of the polynomial P(x)=3x^2+kx-8, find k.
 
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  • #2
Is this right?

Is it cubic? Is it a polynomial? Does it have integral coefficients? Does it have 3i and -6 as zeroes? If so, then it's right!


Problem 18.
If 2 is a zero of the polynomial P(x)=3x^2+kx-8, find k.

What are your thoughts on this?
 
  • #3
My thoughts on problem 18.

ON problem 18 my thoughts are that since 2 is a zero of the polynomial
P(x)=3x^2+kx-8, find k. Then one of the x values is 2...so 3(2)^2+k(2)-8=28+k^2=sqrt(-28)=2isqrt(7).

I also have a question on problem 15.
A rectangular enclosure is to be made using a barn as one side and 80m of fencing to form the other three sides. What is the maximum area of such an enclosure?
 
  • #4
3(2)^2+k(2)-8=28+k^2=sqrt(-28)=2isqrt(7).

I find this confusing (and you probably do too); why don't you do it one step at a time?


A rectangular enclosure is to be made using a barn as one side and 80m of fencing to form the other three sides. What is the maximum area of such an enclosure?

Again, what are your thoughts?


(P.S. what kind of homework is this?)
 
  • #5
Problem 18.
3(2)^2+k(2)-8 ----I inserted the value of x equaling 2.
36+k(2)-8
28+k^2
k^2=-28
k=sqrt(-28)
k=2isqrt(7)

For the rectangle problem I believe that the problem is in the ax^2+bx+c=0.

This is Algebra 2 homework
 
  • #6
The attempted solution to the second problem is pretty poorly presented. First you have some sort of expression that you simplify, then you magically move the 28 to the previously non-existant right hand side of the "equation"! Yeah, we can figure out what you mean, but still. Also, remember that k(2) does not equal k^2...
 
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  • #7
mustang said:
Your step 2: 36+k(2)-8
Your step 3: 28+k^2
I think we found your problem. And don't forget to thank Hurkyl for the extremely helpful suggestion in presentation.




mustang said:
A rectangular enclosure is to be made using a barn as one side and 80m of fencing to form the other three sides. What is the maximum area of such an enclosure?
It depends on the size of the barn (whether it is greater than some critical value, hint, hint).
 
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  • #8
Regarding problem 22 & 14.

Problem 14.
This is what i have done:
(sqrt[x-4]+10)^2=(x+4)^2
x-4+20sqrt(x-4)+100=x+4
20sqrt(x-4)=92
400(x-4) = 8,464 --I squared both sides.
400x-1600=8464
400x=10,064
x=25.16

Is this right??

Problem 22.
This is what I have done:
x^3+bx^2+cx-10= (x-a)(x-b)(x-c)
Would i input the values -2+i and -2-i for a and b. TO find the value for c?
 
  • #9
Sorry!

I made a mistake, SORRY!
 
  • #10
Let's go back to problem 16:
Problem 16.
"Find a cubic polynomial with integral coefficients having 3i and -6 as zeros."

You gave, as your solution "x^2-(6-3i)x+(-18i)" and then Hurkyl asked several
questions, in particular, "is this a cubic equation?" which you never answered.
(In case you were wondering: no, x^2 is not cubic!)

Also, "-18i", being an imaginary number, is not what we would normally consider an integer (it is a "complex integer" or "Gaussian integer"). In order to have integer (in particular "real") solutions, you would have to have -3i as a root too. What does
(x-3i)(x+3i)(x+6) give you?
 
  • #11
THanks!

Thanks a lot for everything.!
 

1. What is the value of k in the given polynomial equation?

The value of k can be found by substituting 2 as the zero of the polynomial and solving for k.

2. How do I solve for k in a polynomial equation?

To solve for k, you need to use the fact that 2 is a zero of the polynomial. This means that when x=2, the polynomial will equal to 0. You can then set up an equation using the given polynomial and the value of 2 as the zero, and solve for k.

3. Can I use any other value as the zero to solve for k?

No, in this specific problem, the given value of 2 must be used as the zero to solve for k. This is because the problem requires finding the value of k that will make the given polynomial have 2 as a zero.

4. What is the significance of having a zero in a polynomial equation?

A zero of a polynomial equation is the value of x that makes the polynomial equal to 0. In other words, it is the x-intercept of the graph of the polynomial. In this problem, the significance of having 2 as a zero is that it helps us solve for the value of k, which is a coefficient in the polynomial.

5. Is there a specific method for solving for k in a polynomial equation?

Yes, there is a specific method for solving for k in this type of polynomial equation. You need to set up an equation using the given polynomial and the value of 2 as the zero, and then use algebraic methods to solve for k. The solution may involve factoring, substitution, or other algebraic techniques.

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