Question

Find by the method of Regula Falsi a root of the equation
x3 + x2 – 3x -3 =0 lying between 1 and 2.​

Answer 1

isa ka answer hai ya = ???????

Answer 2

Answer:

We have to calculate the root of equation by using Regula Falsi method.

Step-by-step explanation:

Regula Falsi Method-It is a numerical method to estimate the roots of the given polynomial.

In this method we use approximation as;

$x_{i} =\frac{af(b)-bf(a)}{f(b)-f(a)}$        .....(1)

Given:

$x^{3}+x^{2} -3x-3=0\\ a=1 , b=2$

First approximation is written as;

$f(1)=1^{3}+1^{2}-3(1)-3\\ =1+1-3-3\\ =-4\\ f(2)= 2^{3}+2^{2}-3(2)-3\\ =8+4-6-3\\ =12-9\\ =3$

Therefore we have,

$x_{1}=\frac{3+8}{3+4} \\x_{1}=\frac{11}{7} \\x_{1}=1.57$

Again we have,

a= 1.57 and f(a)=-1.3753

$x_{2}=\frac{4.71+2.7506}{3+1.3753} \\x_{2}=\frac{7.4606}{4.3753} \\x_{2}=1.7051\\ f(1.7051)=4.957+2.9073-5.1153-3\\f(1.7051)=-0.251$

a= 1.7051 and f(a)=-0.251

$x_{3}= \frac{5.1153+0.502}{3.251} \\x_{3}=1.7278\\ f(1.7278)=5.1579+2.98529-5.1834-3\\ =-0.04021$

a=1.7278 and f(a)=-0.04021

$x_{4}=\frac{5.1834+0.08042}{3.040} \\x_{4}=1.7315\\ f(1.73)=5.1911+2.9980-5.19-3\\ =-0.0009$

$a= 1.7315 ,f(a)=-0.0009\\x_{5}=\frac{5.1945+0.0018}{3.0009} \\x_{5}=1.7315$

Here we can see that approximation 4th and 5th are having same values up to 4 decimal places therefore we can take a root of equation is equal to x= 1.7315

Conclusion:

The root of equation, x = 1.7315