Answer to Question #306269 in Linear Algebra for parth

Question

Find the real root of the equation 𝑥𝑒𝑥=2 correct upto two decimal places using the method of false position.

Answer 1

$f(x)=xe^x-2$

For $x-1$ $f(1)=1x^1-2=0=0.718$

For $x=0.5$ $f(0.5)×e^{0.5}-2=-1.17$

As the value of $f(x)$ at $x=0.5$ is -ve and $x=1.0$ it is positive. The root lies between $x=0.5$ and $x=1$

1^st Step

$x_0=0.5$ and $x_1=1$

$x_2=x_0-\frac{(x_1-x_0)}{f(x_1)-(x_0)}$ $f(x_0)$

$=0.5-\frac{1-0.5}{0.718-(-1.17)}$ $(-1.17)$

$=0.8098$

$f(x_2)=f(0.8098)=0.8098e^{0.8098}-2$

$=0.8098$

2^nd Step

Root lies between $x_1=1$ and $x_2=0.8098$

$x_3=x_2-\frac{x_1-x_2}{f(x_1)-f(x_2)}$ $f(x_2)$

$=0 .8098-\frac{1-0.8098}{0.718+0.18}$ $(-0.18)$

$=0.8479$

$f(x_2)=f(0.8479)=0.8479-e^{0 8479}-2$

$=-0.02$

3^rd Step

Next root lies between $x_3=0 .8479$ and $x_1=1$

$x_4=x_3-\frac{x_1-x_3}{f(x_1)-f(x_3)}$ $f(x_3)$

$=0 .8479-\frac{1-0.8479}{0.718-(-0.02)}$ $(-0.02)$

$=0.8519$

$f(x_4)=0.8519e^{0.8519}-2=-0.0031$

4^th Step

Next value of $x$ lies in between $x_1=1$ and $x_4=0.8519$

$x_5=x_4-\frac{x_1-x_4}{f(x_1)-f(x_4)}$ $f(x_4)$

$=0.8519-\frac{1-0.8519}{0.718+0.0031}$ $(0.0031)=0.8525$

Approximate root to 4d.p $=0.8525$