Question

find the approximate root x3-4x-9=0 by using bisection method​

Answer 1

Hence the required root is x_4=2.6875.

Step-by-step explanation:

Given:

Let $f(x) =x^3- 4x - 9$

Then f(1) =$(1)^3- 4(1) - 9$= -12

f(2)=$(2)^3- 4(2) - 9$ = 8-8-9= -9

f(3)=$(3)^3- 4(3) - 9$=27-12-9= 6

Here f(2) is -ve and f(3) is positive. Therefore root lies in (2,3)

$x_1=\frac{2+3}{2} =2.5$

Hence the first approximation to the root is $x_1 =2.5$

$f( x_1) =2.5^3- 4\times2.5 - 9$

$f( x_1) =15.625-10 - 9$

                = -3.375

Root lies between 2.5 and 3

$x_2=\frac{2.5+3}{2} =2.75$

$x_2=2.75$

$f( x2) =2.75^3- 4\times2.75 - 9$

$f( x_2) =20.79- 11 - 9$

$f( x_2) =0.79$

Root lies between 2.5 and 2.75.

$x_3=\frac{2.5+2.75}{2} =2.625$

$x_3=2.625$

$f( x_3) =2.625^3- 4\times2.625 - 9$

$f(x_3)=18.09-10.5-9$

$f(x_3)=-1.41$

Root lies between 2.625 and 2.75.

$x_4=\frac{2.625+2.75}{2}=2.6875$

$x_4=2.6875$

Hence the required root is $x_4=2.6875$.

Answer 2

Answer:

The root of the given equation is 2.625.

Step-by-step explanation:

Given: $x^3-4x-9=0$

We have to find the approximate root by using the bisection method​.

• As we know, the bisection method is an approximation method to find the roots of the given equation by repeatedly dividing the interval.
• This method will divide the interval until the resulting interval is found, which is extremely small.

We are solving in the following way:

We have,

$x^3-4x-9=0$

First, we have to assume an interval [a, b], such that $f(a) \times f(b)<0$

So, we assume b=3 and x=1,

Then

$\begin{array}{l}f(b)=3^{3}-4 \times 3-9=6 \\f(a)=1-4-9=-12 \\f(a) \times f(b)=6 \times-12=-48<0,\end{array}$

So our assumption for the interval was correct.

Then,

$\begin{array}{l}C_{1}=\frac{a+b}{2}=\frac{3+1}{2}=2 \cdots\left(1^{\text {st }} \text { iteration }\right) \\\\f\left(C_{1}\right)=2^{3}-8-9=-9\end{array}\\\begin{array}{l}\\C_{2}=\frac{a+b}{2}=\frac{2+3}{2}=2.5 \cdots\left(2^{\text {nd }} \text { iteration }\right) \\\\f\left(C_{2}\right)=2.5^{3}-4 \times 2.5-9=-3.375\end{array}$

$\begin{array}{l}\mathrm{C}_{3}=\frac{a+b}{2}=\frac{2.5+3}{2}=2.75 \cdots\left(3^{\text {rd }} \text { iteration }\right) \\\\\mathrm{f}\left(\mathrm{C}_{3}\right)=0.79\end{array}\\\\\begin{array}{l}\mathbf{C}_{4}=\frac{a+b}{2}=\frac{2.5+2.75}{2}=2.625 \cdots\left(4^{\text {th }} \text { iteration }\right) \\\\\mathbf{f}\left(\mathbf{C}_{4}\right)=\mathbf{2 . 6 7}^{3}-4 \times 2.67-9 \approx 0\end{array}$

Hence, the root of the given equation is  2.625.