Solving x^3 +5x -4 to Get Root 0.3274

[RMHAZARD]

I am on my final part of my C3 coursework, doing the comparison of methods, i have found the root using the change of sign method, and the Newton raphson method but i am struggling with using the rearranging method. the equation i am using is x^3 +5x -4, i have tried rearranging it to get the root which is 0.3274, but i can not get it to converge to this figure, the coursework is due in tomorrow at 4pm and i just need to do this final bit and i am done but i can't i have been tring for 3 hours please help me.

Thanks

I am not asking for it to be done for me i am just asking to know whati need to do to my equation to get my root 0.3274.

 

[dirk_mec1]

Show me Newton-raphson with your equation.

 

[HallsofIvy]

I have no idea what you mean by the "rearranging method". Please explain.

 

[madmike159]

Hes talking about the g(x) method. You make your equation in the form x = ... then find points of interception on the graph y = x. These interceptions are the roots of the orignal equation.

You need to find a root between 2 points using a integer search. Take the x value from the search and put it into the x = equation. then take that value and put it into the equation again until it converges to 5/6 DP.

eg x=x^3-2x/3
x = (-1, 0)
x0 = -1
x1 = -1^3-2*-1/3 = n
x2 = n^3-2*n/3 etc.

 

[HallsofIvy]

Hes talking about the g(x) method. You make your equation in the form x = ... then find points of interception on the graph y = x. These interceptions are the roots of the orignal equation.

You need to find a root between 2 points using a integer search. Take the x value from the search and put it into the x = equation. then take that value and put it into the equation again until it converges to 5/6 DP.

eg x=x^3-2x/3
x = (-1, 0)
x0 = -1
x1 = -1^3-2*-1/3 = n
x2 = n^3-2*n/3 etc.

What function are you talking about? x= x³- 2x/3 is equivalent to x³- (2/3)x- x= 0 or x³- (5/3)x= 0 which has NO roots between -1 and 0. The method you give will eventually converge to 0 which is a root.

 

[madmike159]

No I was giving an exampe of the steps you take. I didn't want to give him somthing that worked because it is course work, he can get marked down if some on did it for him. It works in the same way s the Newton Raphson method (gets a closer value every time) but isn' as quick.