Solving \[x^{4}-4x^{3}+10=0\] with a "Binary Search

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In summary: Wikipedia, the free encyclopediaIn summary, the conversation is about solving the equation x^4-4x^3+10=0 by using numerical approximation methods. The suggested methods are binary search, goal seek in excel, and using a numerical approximation tool like Wolfram. The best method suggested is Newton's Method.
  • #1
Yankel
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Hello all

I am trying to draw a graph of a function. On the way, I wanted to see where the function meet the x axis, so I put y=0. It gave me this:

\[x^{4}-4x^{3}+10=0\]

How do I solve this equation ?

Thanks !

I tried a "binary search" and got really close to the answer, but I guess there must be a better way...
 
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  • #2
Yankel said:
Hello all

I am trying to draw a graph of a function. On the way, I wanted to see where the function meet the x axis, so I put y=0. It gave me this:

\[x^{4}-4x^{3}+10=0\]

How do I solve this equation ?

Thanks !

I tried a "binary search" and got really close to the answer, but I guess there must be a better way...

It's not easy...
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A binary search or goal seek in excel are probably your best bets with a numerical approximation being next. Honestly this is one of those questions I'd just put into wolfram and use their answer(s)
 
  • #3
Yankel said:
Hello all

I am trying to draw a graph of a function. On the way, I wanted to see where the function meet the x axis, so I put y=0. It gave me this:

\[x^{4}-4x^{3}+10=0\]

How do I solve this equation ?

Thanks !

I tried a "binary search" and got really close to the answer, but I guess there must be a better way...
You can occasionally use the rational root theorem, but it doesn't work for this equation.

-Dan
 
  • #4
I understand, thank you !

What numerical approximation methods do we have to solve such equations ?
 
  • #5
Yankel said:
I understand, thank you !

What numerical approximation methods do we have to solve such equations ?

Perhaps the best known is:

Newton's Method
 

FAQ: Solving \[x^{4}-4x^{3}+10=0\] with a "Binary Search

What is a "Binary Search" and how is it used to solve equations?

A "Binary Search" is a search algorithm commonly used in computer science. It involves dividing a sorted list in half, and then checking which half the desired item is in. This process is repeated until the item is found. In the context of solving equations, a "Binary Search" involves dividing the search space of possible solutions in half and checking which half contains the solution. This process is then repeated until the solution is found.

Why is a "Binary Search" useful for solving equations?

A "Binary Search" is useful for solving equations because it is a systematic and efficient method that can quickly narrow down the search space of possible solutions. This can save time and effort compared to other methods, especially for complex equations with multiple solutions.

What are the steps involved in using a "Binary Search" to solve the equation x^{4}-4x^{3}+10=0?

The steps involved in using a "Binary Search" to solve the equation x^{4}-4x^{3}+10=0 are as follows:

  1. Arrange the equation in the form f(x)=0, in this case x^{4}-4x^{3}+10=0.
  2. Choose an initial range of values for x that contains the solution.
  3. Divide the range in half and evaluate the function at the midpoint.
  4. If the midpoint is the solution, stop. Otherwise, determine which half of the range contains the solution.
  5. Repeat steps 3 and 4 until the solution is found within a desired level of accuracy.

Can a "Binary Search" be used to solve any equation?

No, a "Binary Search" is most commonly used for solving equations that have a single real solution. It may not be effective for equations with multiple solutions, complex solutions, or equations that are not easily rearranged in the form f(x)=0.

Are there any limitations or drawbacks to using a "Binary Search" to solve equations?

Yes, there are a few limitations and drawbacks to using a "Binary Search" to solve equations. It may not be effective for equations with multiple solutions, complex solutions, or equations that are not easily rearranged in the form f(x)=0. It also requires the equation to be sorted in a specific way, which may not always be possible. Additionally, it may not be as accurate as other methods for finding solutions, as the search space is divided in half each time and may miss smaller or more complex solutions.

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