Question

Find a real root of the equation f (x) = x3 x 1 = 0, using bisection method.

Answer 1

Find root using Bisection Method :

Algorithm :

Let , f(x)=0 and two real numbers a and b such that f(a)*f(b)<0

Set c = (a+b)/2

if f(c)<0 then b=c, otherwise c=a

Repeat step 2 and 3 until (a-b)<DOA where DOA means Degree of accuracy.

The given equation is f(x)=x^3+x-1=0

Let a=0, b=1

First approximate root :

x1 = (a + b)/2 = (0+1)/2 = 0.5

f(x1) = 0.5^3+0.5-1 = -0.375 (-ve)

Hence root exits between x1=0.5 and x2=1.0

Second approximate root :

x2 = (a + b)/2 = (0.5+1)/2 =0.75

f(x1) = 0.75^3+0.75-1 = +0.17 (+ve)

Hence root exits between x1=0.5 and x2=0.75

Third approximate root :

x3 = (a + b)/2 = (0.5+0.75)/2 =0.625

f(x1) = 0.625^3+0.625-1 = -0.130 (-ve)

Hence root exits between x1=0.625 and x2=0.75

Fourth approximate root :

x4 = (a + b)/2 = (0.625+0.75)/2 =0.6875

f(x1) =0.6875^3+0.6875-1 = +0.012 (+ve)

Hence root exits between x1=0.625 and x2=0.6875

Fourth approximate root :

x5 = (a + b)/2 = (0.625+0.6875)/2 =0.6562

f(x1) =0.6562^3+0.6562-1 = -0.0611 (-ve)

Hence root exits between x1=0.6562 and x2=0.6875

Sixth approximate root :

x6 = (a + b)/2 = (0.6562+0.6875)/2 =0.6718

f(x1) =0.6718^3+0.6718-1 = -0.0248(-ve)

Hence root exits between x1=0.6718 and x2=0.6875

Seventh approximate root :

x7 = (a + b)/2 = (0.6718+0.6875)/2 =0.6796

f(x1) =0.6796^3+0.6796-1 = -0.0063(-ve)

Hence root exits between x1=0.6796 and x2=0.6875

8th approximate root :

x8 = (a + b)/2 = (0.6796+0.6875)/2 =0.6835

f(x1) =0.6835^3+0.6835-1 = +0.0030(+ve)

Hence root exits between x1=0.6796 and x2=0.6835

9th approximate root :

x9 = (a + b)/2 = (0.6796+0.6835)/2 =0.6816

f(x1) =0.6816^3+0.0.6816-1 = -0.0016(-ve)

Hence root exits between x1=0.6816 and x2=0.6835

10th approximate root :

x10 = (a + b)/2 = (0.6816+0.6835)/2 =0.6826

Hence the approximate correct upto 3 decimal places. So the real root of the equation = 0.6826

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FROM
LALITH PRIYADARSHAN

Answer 2

Answer:

Did you mean " Find a real root of the equation $f (x) = x^3-x -1 = 0$, using bisection method?"

The bisection method is a straightforward method to find the roots of a polynomial. Identify two points x = a and x = b such that f (a) and f (b) are having opposite signs.

Let f (a) be negative and f (b) be positive. Then there will be a root of f (x) = 0 between a and b. Let the first approximation be the midpoint of the interval (a, b). i.e.

$x1 =\frac{(a+b)}{2}$.

If f (x1) = 0, then x1 is a root, otherwise root lies between a and x1 or x1 and b according as f (x1) is positive or negative. Then again we bisect the interval and continue the process until the root is found to the desired accuracy.

Step-by-step solution:

Let $f (x) = x^3-x -1 = 0$.

$f(0)= -1 ,\\f(1)= -1,\\f(2) =5.$

1st iteration:

Here f(1) = -1 < 0 and f(2) = 5 > 0.

Now, Root lies between 1 and 2

$x_0=\frac{1+2}{2}=1.5$,

$f(x_0)=f(1.5)=1.53-1.5-1=0.875 > 0$.

2nd iteration:

Here f(1) = -1 < 0 and f(1.5) = 0.875 > 0

Therefore, Now, the Root lies between 1 and 1.5

$x_1=\frac{1+1.5}{2}=1.25$,

$f(x_1)=f(1.25)=1.253-1.25-1=-0.2969 < 0$

3rd iteration:

Here f(1.25) = -0.2969 < 0 and f(1.5) = 0.875 > 0

Therefore, Root lies between 1.25 and 1.5

$x2=\frac{1.25+1.5}{2}=1.375$,

$f(x_2)=f(1.375)=1.3753-1.375-1=0.2246 > 0$

4th iteration:

Here f(1.25) = -0.2969 < 0 and f(1.375) = 0.2246 > 0

Therefore, Root lies between 1.25 and 1.375

$x_3=\frac{1.25+1.375}{2}=1.3125$

$f(x_3)=f(1.3125)=1.31253-1.3125-1=-0.0515 < 0$

5th iteration:

Here f(1.3125) = -0.0515 < 0 and f(1.375) = 0.2246 > 0

Therefore,  Root lies between 1.3125 and 1.375

$x_4=\frac{1.3125+1.375}{2}=1.3438$,

$f(x_4)=f(1.3438)=1.34383-1.3438-1=0.0826 > 0$

6th iteration :

Here f(1.3125) = -0.0515 < 0 and f(1.3438) = 0.0826 > 0

Therefore, Root lies between 1.3125 and 1.3438

$x_5=\frac{1.3125+1.3438}{2}=1.3281$,

$f(x_5)=f(1.3281)=1.32813-1.3281-1=0.0146 > 0$

7th iteration :

Here f(1.3125) = -0.0515 < 0 and f(1.3281) = 0.0146 > 0

Therefore, Root lies between 1.3125 and 1.3281

$x_6=\frac{1.3125+1.3281}{2}=1.3203$,

$f(x_6)=f(1.3203)=1.32033-1.3203-1=-0.0187 < 0$

8th iteration :

Here f(1.3203) = -0.0187 < 0 and f(1.3281) = 0.0146 > 0

Therefore, Root lies between 1.3203 and 1.3281

$x_7=\frac{1.3203+1.3281}{2}=1.3242$,

$f(x_7)=f(1.3242)=1.32423-1.3242-1=-0.0021 < 0$

9th iteration :

Here f(1.3242) = -0.0021 < 0 and f(1.3281) = 0.0146 > 0

Therefore, Root lies between 1.3242 and 1.3281

$x_8=\frac{1.3242+1.3281}{2}=1.3262$,

$f(x_8)=f(1.3262)=1.32623-1.3262-1=0.0062 > 0$

10th iteration :

Here f(1.3242)=-0.0021<0 and f(1.3262)=0.0062>0

Therefore, Root lies between 1.3242 and 1.3262

$x_9=\frac{1.3242+1.3262}{2}=1.3252$,

$f(x_9)=f(1.3252)=1.32523-1.3252-1=0.002 > 0$

11th iteration :

Here f(1.3242)=-0.0021<0 and f(1.3252)=0.002>0

Therefore, Root lies between 1.3242 and 1.3252

$x_{10}=\frac{1.3242+1.3252}{2}=1.3247$,

$f(x_{10})=f(1.3247)=1.32473-1.3247-1=0 < 0$

Hence, the approximate root of the equation $x^3-x-1=0$ using Bisection method is 1.3247.