Math, asked by piyushTiwari2831, 2 months ago

x^3-4x-9=0 find the bisection method

Answers

Answered by lokeshnandigam69
0

Answer:

Hence the required root is x_4=2.6875.

Step-by-step explanation:

Given:

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

3

−4x−9

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

3

−4(1)−9 = -12

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

3

−4(2)−9 = 8-8-9= -9

f(3)=(3)^3- 4(3) - 9(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} =2.5x

1

=

2

2+3

=2.5

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

1

=2.5

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

1

)=2.5

3

−4×2.5−9

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

1

)=15.625−10−9

= -3.375

Root lies between 2.5 and 3

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

2

=

2

2.5+3

=2.75

x_2=2.75x

2

=2.75

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

3

−4×2.75−9

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

2

)=20.79−11−9

f( x_2) =0.79f(x

2

)=0.79

Root lies between 2.5 and 2.75.

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

3

=

2

2.5+2.75

=2.625

x_3=2.625x

3

=2.625

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

3

)=2.625

3

−4×2.625−9

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

3

)=18.09−10.5−9

f(x_3)=-1.41f(x

3

)=−1.41

Root lies between 2.625 and 2.75.

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

4

=

2

2.625+2.75

=2.6875

x_4=2.6875x

4

=2.6875

Hence the required root is x_4=2.6875x

4

=2.6875 .

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