Solve 16x^4-44x^2+204=0: Math Help Needed

  • Thread starter aisha
  • Start date
In Summary, Aisha is a name that is derived from Persian origin and is commonly used by others cultures.
  • #1
aisha
584
0
solve (4x^2-9)^2-11(4x^2-9)+24=0
I tried simplifying this and got 16x^4-44x^2+204=0 I am not sure if this is right, and I don't know what to do to solve. Please help! :confused:
 
Physics news on Phys.org
  • #2
Is there not a simple substitution that will make things easier ?
 
  • #3
Gokul43201 said:
Is there not a simple substitution that will make things easier ?

There is nothing but the question, what do I do?
 
  • #4
Actually, I was making a suggestion there. Try a substitution...look for a repeating quantity and call it something.
 
  • #5
let W = (4x^2-9), solve for W using some method, and then replace W with (4x^2-9), and now solve for x

:)
 
  • #6
Done needs to be checked please!

:smile: Thanks I get how to do it but want to make sure I did it right can you please check? Here is what I did:
(4x^2-9)-11(4x^2-9)+24=0
Let w=(4x^2-9)
w^2-11w+24=0
factored to become (w-8)(w-3)=0
sub in w value (4x^2-9-8)(4x^2-9-3)=0
(4x^2-17) (4x^2-12)=0

so solving for x
4x^2-17=0
4x^2=17
x=the square root of +or- 17/4

or 4x^2-12=0
4x^2=12
x=the square root of +or-3
 
  • #7
off the subject aisha, did you get your name from the ebaumsworld video aisha, I am sure it is on otherwebsites as well, but it is the video fo the skinny dude singing about some girl in want to be backstreet boys way?
 
  • #8
aisha said:
:smile: Thanks I get how to do it but want to make sure I did it right can you please check? Here is what I did:
(4x^2-9)-11(4x^2-9)+24=0
Let w=(4x^2-9)
w^2-11w+24=0
factored to become (w-8)(w-3)=0
sub in w value (4x^2-9-8)(4x^2-9-3)=0
(4x^2-17) (4x^2-12)=0

so solving for x
4x^2-17=0
4x^2=17
x=the square root of +or- 17/4

or 4x^2-12=0
4x^2=12
x=the square root of +or-3

Correct...except that you mean x=+or- the square root of 17/4 NOT x=the square root of +or- 17/4

And Tom : 'Aisha' is a reasonably common name; originally of Persian origin, it has spread into other cultures as well.
 
  • #9
Thanks EVERYONE! :smile:
 

1. What is the degree of the polynomial?

The degree of a polynomial is the highest exponent of the variable in the expression. In this case, the degree of the polynomial is 4.

2. How do you solve a polynomial equation?

To solve a polynomial equation, you need to factor the expression and set each factor equal to zero. Then, solve for the variable in each factor. The solutions for the variable will be the roots of the polynomial equation.

3. Can this polynomial equation be factored?

Yes, this polynomial equation can be factored. It is a fourth-degree polynomial, so it can be factored into two quadratic equations.

4. What is the difference between a root and a solution of a polynomial equation?

A root of a polynomial equation is a value of the variable that makes the equation true when substituted. A solution of a polynomial equation is a value of the variable that satisfies all the conditions of the equation, including any restrictions such as the domain of the equation.

5. How many solutions does this polynomial equation have?

Since this polynomial is a fourth-degree equation, it has up to four solutions or roots.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
821
  • Introductory Physics Homework Help
Replies
10
Views
863
  • Introductory Physics Homework Help
Replies
17
Views
751
  • Introductory Physics Homework Help
3
Replies
88
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
646
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
32
Views
959
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
1K
Back
Top