# 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.

Solution:

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: $x_{1}=1-(-3/4).1=1.75$

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 $f(1.9)=0.339>0$.

Hence, $1.835<\xi<1.9$.

Applying the method of chords to the interval $(1.835,1.9)$, we will get a new approximation: $x_{3}=1.835-(\frac{-0.05059}{0.339+0.05059})(0.065)=1.8434$

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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