Newton Raphson Method for Solving 6x3+7x2-9x-7=0

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Newton-Raphson method

The equation I am going to Solve is 6x3+7x2-9x-7=0

Newton-Raphson is a fixed point iteration method. The first approximation has to be near the root so the method will work. A tangent has to be drawn at the first point y=f(x) and when it touches the x-axis this gives another root. The tangent cuts the second point on the curve. This process is then repeated.  The Newton Raphson method uses an algebraically formula to calculate the value of roots.   .

The formula is first worked out by the gradient of the tangent there are steps of algebra to get to the final equation. Autograph will do these calculations automatically.

f `(xn) =              

(x1-x2)f `(x1) = f(x1)          

x1-x2 =

x2 =x1 -







                       f(x) is negative                                 The root lies between 1.1155 and 1.11555

1.11545                                                        1.1155                                                                1.11555

f(1.11545) = -0.002199645                       f(1.1155) = 0.00020309         f(1.11555)= 0.000701867.

X = 1.1155

Error Bounds = ±0.00005

Solution bounds = (1.11545≤ x ≤1.11555).

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Solving 6x3+7x2-9x-7=0 for second root





The root lies between    -0.63488 and   -0.634875         ...

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