# Question 12.8: An approximate root of x³ − 2x² − 5 = 0 is x = 3. By using t......

An approximate root of

x³ − 2x² − 5 = 0

is x = 3. By using the Newton–Raphson technique repeatedly, determine the value of the root correct to two decimal places.

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## Script File

We have

$\begin{array}{ll}x_1=3 & \\f(x)=x^3-2 x^2-5 & f\left(x_1\right)=4 \\f^{\prime}(x)=3 x^2-4 x & f^{\prime}\left(x_1\right)=15\end{array}$

Hence

$x_2=3-\frac{4}{15}=2.733$

An improved estimate of the value of the root is 2.73 (2 d.p.). The method is used again, taking  $x_1$  = 2.73 as the initial approximation:

\begin{aligned}& x_1=2.73 \quad f\left(x_1\right)=0.441 \quad f^{\prime}\left(x_1\right)=11.439 \\& x_2=2.73-\frac{0.441}{11.439}=2.691\end{aligned}

An improved estimate is x = 2.69 (2 d.p.). The method is used again:

$x_1=2.69 \quad f\left(x_1\right)=-0.007 \quad f^{\prime}\left(x_1\right)=10.948$

So

$x_2=2.69-\frac{(-0.007)}{10.948}=2.691$

There is no change in the value of the approximate root and so to two decimal places the root of f (x) = 0 is x = 2.69.

The calculation can be performed in  $MATLAB^{\circledR}$  using the roots function:

roots([1 -2 0 -5])

which will produce the real root of 2.69, agreeing with our numerical  calculation using the Newton–Raphson method. The two complex roots of the equation will also be re-turned. Technical computing languages make use of a variety of numerical methods and to some extent the user has to take it on trust that they are correctly implemented and tested so that they always produce the correct result.

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