Question

Use bisection method to find root of the equation x3 – 2x – 5 = 0

Answer 1

Answer:

The root of the equation upto 4 places of decimal = 2.0625

Step-by-step explanation:

Here the equation is  x³ - 2·x - 5 = 0

Let f(x) = x³ - 2·x - 5

Now,

x       0              1          2          3

f(x)     -5           -6          -1         16

1st iteration :

Here f(2) = -1 < 0 and f(3) = 16 > 0

∴ Root lies between 2 and 3

$x_0=\frac{2+3}{2}=2.5\\\\f(x_0)=f(2.5)=(2.5)\cdot 3-2\cdot (2.5)-5=5.625>0$

2nd iteration :

Here f(2) = -1 < 0 and f(2.5) = 5.625 > 0

∴ Now, Root lies between 2 and 2.5

$x_1=\frac{2+2.5}{2}=2.25\\\\f(x_1)=f(2.25)=(2.25)\cdot 3-2\cdot (2.25)-5=1.89062>0$

3rd iteration :

Here f(2) = -1 < 0 and f(2.25) = 1.89062 > 0

∴ Now, Root lies between 2 and 2.25

$x_2=\frac{2+2.25}{2}=2.125\\\\f(x_2)=f(2.125)=(2.125)\cdot 3-2\cdot (2.125)-5=0.3457>0$

4th iteration :

Here f(2) = -1 < 0 and f(2.125) = 0.3457 > 0

∴ Now, Root lies between 2 and 2.125

$x_3=\frac{2+2.125}{2}=2.0625\\\\f(x_3)=f(2.0625)=(2.0625)\cdot 3-2\cdot (2.0625)-5=-0.35132>0$

So, the root of the given equation upto 4 places of decimal is 2.0625

Answer 2

Given :    x³ -  2x  - 5  = 0  

To find :  root of the equation  by Bisection method correct up to three decimal places

Solution:

x³ -  2x  - 5  = 0

x = 1  => 1  - 2 - 5  = - 6

x = 2 => 8 - 4 -  5 = - 1

x = 3 =>  27 - 6 - 5 = 16

0 lies between -1  & 16

Hence x lies between 2 & 3

(2 + 3)/2 = 2.5

x = 2.5  =>     x³ -  2x  - 5  = 5.625

0 lies between -1  & 5.625

=> x lies between 2  &  2.5

(2 + 2.5)/2 = 2.25

x = 2.25  =>  x³ -  2x  - 5  = 1.89

0 lies between -1  & 1.89

=>  x lies between 2  &  2.25

=> x = 2.125  => x³ -  2x  - 5  = 0.34

0 lies between -1  & 0.34

=> x lies between 2  * 2.125

=> x = 2.0625   => x³ -  2x  - 5  = -0.35

0 lies between  -0.35 & 0.34

=> x lies between  2.0625   & 2.125

x = 2.09375    => x³ -  2x  - 5  =    -0.0089 ≈  0

x =  2.094    up to three decimal places

x =  2.094  

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