Real root of an equation

Problem. Find the real root of the equation f(x)=x^{3}-2x^{2}+3x-5=0

with an accuracy upto 10^{-4}. Use the method of chords.


Let us first make sure that the given equation has only one real root. This follows from the fact that the derivative

f^{'}(x)=3x^{2}-4x+3>0 (use your knowledge of theory of equations here!!!)

Then, from f(1)=-3<0 and f(2)=1>0, it follows that the given polynomial has a single positive root, which lies in the interval (1,2).

Using the method of chords, we obtain the first approximation:


Since f(1.75)=-0.5156<0 and f(2)=1>0, then 1.75<\xi<2.

The second approximation:

x_{2}=1.75+(0.5156/1.5156).o.25=1.8350 and

since f(1.835)=-0.05059<0, then 1.835<\xi<2.

The sequence of approximations converges very slowly. Let us try to narrow down the interval, taking into account that the value of the function f(x) at the point x_{2}=1.835 is considerably less in absolute value than f(2). We have


Hence, 1.835<\xi<1.9.

Applying the method of chords to the interval (1.835,1.9), we will get a new approximation:


Further calculations by the method of chords yield

x_{4}=1.8437 and x_{5}=1.8438

and since f(1.8437)<0 and f(1.8438)>0, then \xi \approx 1.8438 with required accuracy of 10^{-4}.

You can also solve this problem by the method of tangents. Try it!

More later,

Nalin Pithwa

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