Question

O
find a root of the equation ox-2x-5-
in the internal [2,3] correct to 3
decimal
places by wing bidection
method.​

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

\begin{lgathered}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\end{lgathered}

x

0

=

2

2+3

=2.5

f(x

0

)=f(2.5)=(2.5)⋅3−2⋅(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

\begin{lgathered}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\end{lgathered}

x

1

=

2

2+2.5

=2.25

f(x

1

)=f(2.25)=(2.25)⋅3−2⋅(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

\begin{lgathered}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\end{lgathered}

x

2

=

2

2+2.25

=2.125

f(x

2

)=f(2.125)=(2.125)⋅3−2⋅(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

\begin{lgathered}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\end{lgathered}

x

3

=

2

2+2.125

=2.0625

f(x

3

)=f(2.0625)=(2.0625)⋅3−2⋅(2.0625)−5=−0.35132>0

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