# Thread: Finding root of equation

1. ## Finding root of equation

Show that the equation x^3 + x = 3 has exactly one root, ∝. Find an integer n such that ∝ lies between n and n + 1. Starting with x0 = n as a first approximation, use Newton's Method to find 2 further approximations to ∝.

f(x) = x^3 + x - 3, [n, n + 1]

f'(x) = 3x^2 + 1

x2 = ( x1 + x1^3 + x1 - 3 ) / ( 3x1^2 + 1 )
x2 = ( n + n^3 + n - 3 ) / ( 3n^2 + 1 )
x2 = ( n^3 + 2n - 3 ) / ( 3n^2 + 1 )
x2 = ( n( n^2 + 2 ) - 3 ) / ( 3n^2 + 1 )

I dont you what to do from here.

2. Originally Posted by SyNtHeSiS
Show that the equation x^3 + x = 3 has exactly one root, ∝. Find an integer n such that ∝ lies between n and n + 1. Starting with x0 = n as a first approximation, use Newton's Method to find 2 further approximations to ∝.

f(x) = x^3 + x - 3, [n, n + 1]

f'(x) = 3x^2 + 1

x2 = ( x1 + x1^3 + x1 - 3 ) / ( 3x1^2 + 1 )
x2 = ( n + n^3 + n - 3 ) / ( 3n^2 + 1 )
x2 = ( n^3 + 2n - 3 ) / ( 3n^2 + 1 )
x2 = ( n( n^2 + 2 ) - 3 ) / ( 3n^2 + 1 )

I dont you what to do from here.
If there is only one root then the function does not turn anywhere. So you need to show that you can't solve the derivative = 0

So if $\displaystyle f(x) = x^3 + x - 3$

$\displaystyle f'(x) = 3x^2 + 1$

$\displaystyle 3x^2 + 1 = 0$

$\displaystyle 3x^2 = -1$

$\displaystyle x^2 = -\frac{1}{3}$.

There does not exist any $\displaystyle x$ that solves this equation, so the function does not turn anywhere. Therefore there's only one root.

3. Originally Posted by Prove It
If there is only one root then the function does not turn anywhere. So you need to show that you can't solve the derivative = 0

So if $\displaystyle f(x) = x^3 + x - 3$

$\displaystyle f'(x) = 3x^2 + 1$

$\displaystyle 3x^2 + 1 = 0$

$\displaystyle 3x^2 = -1$

$\displaystyle x^2 = -\frac{1}{3}$.

There does not exist any $\displaystyle x$ that solves this equation, so the function does not turn anywhere. Therefore there's only one root.
But there is a root. The answer in my memo is x2 = 17/14 = 1.2142857

4. Originally Posted by TsAmE
But there is a root. The answer in my memo is x2 = 17/14 = 1.2142857

It is close , the solution is $\displaystyle x = 1.213411663...$

5. Originally Posted by TsAmE
But there is a root. The answer in my memo is x2 = 17/14 = 1.2142857
The point is, if there's more than one root, then the function has to turn somewhere. But we have shown that it doesn't turn, so there is at most one root.

To prove that there is a root, substitute some values in and see if you get both positive and negative values. If so then it must have crossed the $\displaystyle x$ axis somewhere.

6. Originally Posted by Prove It
The point is, if there's more than one root, then the function has to turn somewhere. But we have shown that it doesn't turn, so there is at most one root.

To prove that there is a root, substitute some values in and see if you get both positive and negative values. If so then it must have crossed the $\displaystyle x$ axis somewhere.
Yeah I recognize what you saying: use the intermediate value theorem. I subbed in n = 1 into x2 = ( n( n^2 + 2 ) - 3 ) / ( 3n^2 + 1 ), and got x2 = 0, which means this is the root right? How in the answer did they get x2 = 17/14?