# Using bisection method, find the root of the equation f(x)=x2 -6logx-3=0(2<=x<=3)'correct to 4 decimal places with an accuracy of 10 -2 .

Using bisection method, find the root of the equation f(x)=x2 -6logx-3=0(2<=x<=3)'correct to 4 decimal places with an accuracy of 10 -2 .

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Here,  f(x)=x2 -6 logx-3=0

f(2)=4-6 log2-3=-0.806

f(3)=9-6 log3-3=3.1373

f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.

Hence, the root lies between 2 and 3

c0 =(2+3)/2=2.5

f(2.5)=6.25-6 log 2.5-3=0.8623

Now

 n a(-ve) b(+ve) cn f(cn) 0 2 3 2.5 0.8623 1 2 2.5 2.25 -0.050595 2 2.25 2.5 2.375 0.38664 3 2.25 2.375 2.3125 0.1631658 4 2.25 2.3125 2.28125 0.05506 5 2.25 2.28125 2.265625 0.001925

From the table,

f(2.265625)=0.001928<10-2

Therefore, the root of the given equation correct upto four paces of decimal=2.2656

Here,  f(x)=x2 -6 logx-3=0

f(2)=4-6 log2-3=-0.806

f(3)=9-6 log3-3=3.1373

f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.

Hence, the root lies between 2 and 3

c0 =(2+3)/2=2.5

f(2.5)=6.25-6 log 2.5-3=0.8623

Now

 n a(-ve) b(+ve) cn f(cn) 0 2 3 2.5 0.8623 1 2 2.5 2.25 -0.050595 2 2.25 2.5 2.375 0.38664 3 2.25 2.375 2.3125 0.1631658 4 2.25 2.3125 2.28125 0.05506 5 2.25 2.28125 2.265625 0.001925

From the table,

f(2.265625)=0.001928<10-2

Therefore, the root of the given equation correct upto four paces of decimal=2.2656