# 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

This site uses Akismet to reduce spam. Learn how your comment data is processed.