Question

find the real root of he following x3-9x+1=0 correct to three significant figure by regula falsi method

Answer 1

Step-by-step explanation:

Here, x3−9x+1=0

Let f(x)=x3−9x+1

First Iteration:

Here, f(2)=−9<0 and f(4)=29>0

Now, Root lies between x0=2 and x1=4

x2=x0−f(x0)×f(x1)−f(x0)x1−x0=2−(−9)×29−(−9)4−2=2.47368

Second Iteration:

Here, f(2.47368)=−6.1264 and f(2)=29>0

Now, Root lies between x0=2.47368 and x1=4

x3=x0−f(x0)×f(x1)−f(x0)

Answer 2

Step-by-step explanation:

Given:x³-9x+1=0

To find: find the real root correct to three significant figure by regula falsi method.

Solution:

Formula:

$\bold{x = \frac{bf(a) - af(b)}{f(a) - f(b)} }$

Step 1:Find a and b,put x=1

$f(1) = {(1)}^{3} - 9(1) + 1 \\ \\ f(1) = 1 - 9 + 1 \\ \\\bold{\red{ f(1) = - 7}}$

Thus, a=1 and f(a)= -7

put x=3

$f(3) = {(3)}^{3} - 9(3) + 1 \\ \\ f(3) = 27 - 27 + 1 \\ \\ \bold{\green{f(3) = 1}}$

Thus, b=3 and f(b)=1

There is a real root between x=1 and 3.

Step 2: Find next a,by putting value in formula

$x_1 = \frac{3( - 7) - 1(1)}{( - 7) - (1)} \\ \\ x_1 = \frac{ - 21 - 1}{ - 7 - 1} \\ \\ x_1 = \frac{ - 22}{ - 8} \\ \\ x_1 = 2.75$

find the value of f(2.75)

$f(2.75) = {(2.75)}^{3} - 9(2.75) + 1 \\ \\ f(2.75) = 20.7969 - 24.75 + 1 \\ \\\bold{\pink{ f(2.75) = - 2.9531}}$

Step 3:For next iteration a=2.75 and f(a)=-2.9531

$x_2 = \frac{3( - 2.9531) - 2.75(1)}{ - 2.9531- (1)} \\ \\ x_2 = \frac{ - 8.593 - 2.75}{ - 2.9531 - 1} \\ \\ x_2 = \frac{ - 11.609375}{ - 3.9531} \\ \\ x_2 = 2.9367$

$f(2.9367) = ( {2.9367)}^{3} - 9(2.9367) + 1 \\ \\ f(2.9367) = 25.3287 - 26.303 + 1 \\ \\ f(2.9367) = - 0.1016$

Step 4: For next iteration

a=2.9367

f(a)=-0.1016

$x_3= \frac{3( - 0.1016) - 2.9367(1)}{ - 0.1016 - 1} \\ \\ x_3 = \frac{ - 3.2415}{ - 1.1016} \\ \\ x_3 = 2.9425$

Final answer:

One real root of x³-9x+1 is 2.942 upto 3 significant digits.

Hope it helps you.

Remark: This method can find one real root.Although the equation has three real roots.Other two can be find using other method like Newton-Raphson method.