The bisection method is one of the methods used to find out the solution of the linear equation.
So, without wasting any moment lets begin by discussing one simple example.
Note - Bisection formula : x= a+b/2
Q 1. Find the real root of equation x^3 -x-4 = 0 using bisection method correct up to 3 decimal places ?
Sol - Step 1 - suppose let f(x) = x^3 - x -4 =0 -------------- eq(1)
Step 2 - Find one negative value and one positive value by hit and trial method.
Put integer values in equation 1 to get one negative and positive value.
put f(1) = -4 <0
f(2) = 2 >0
Note - Put decimal values like 1.3, 1.4 to get positive and negative values
Put f(1.7)= -0.767 <0
f(1.8) = 0.032 >0
Step 3 - Choosing a= 1.7 and b =1.8
Step 4 - Now find approximate values using bisection formula
. 1st approximate value -
X1 = a+b/2 = (1.7 + 1.8)/2 = 1.75
f(X1) = (1.75)^3 - 1.75 -4 = -0.39 <0
Hence root lies between 1.75 and 1.8
. 2nd approximate value -
X2 = a+b/2 = (1.75+1.8)/2 = 1.775
f(X2) = (1.775)^3 - 1.775 - 4 = -0.182 <0
As 1 st approximate value and 2nd approximate values not match
Hence root lies between 1.775 and 1.8
. 3rd approximate value -
X3 = a+b/2 = (1.775+1.8)/2 = 1.7875
f(X3) = (1.7875)^3 - 1.7875 - 4 = -0.076 <0
As 2nd and 3rd approximate values not match
Hence root lies between 1.7875 and 1.8
. 4th approximate value -
X4 = a+b/2 = (1.7875+1.8)/2 = 1.79375
f(X4) = (1.79375)^3 - 1.79375 - 4 = -0.022 <0
As 3rd and 4th approximate values not match
Hence root lies between 1.79375 and 1.8
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This procedure will continue up to 7th approximate value
. 7th approximate value -
X7 = a+b/2 = (1.795312 + 1.796875)/2 = 1.796093
As X6 and X7 values are same hence
The approximate roots correct up to 3 decimal places is
ANSWER = X= 1.796
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