Answer to Question #180118 in Quantitative Methods for Oliver

Question

solve the following question by using bisection method.

correct to the four decimal places.

x² ⁻ 3 = 0

Answer 1

x²-3=0

Our function becomes f(x) = x²-3 ,to get interval substitute various values as follow

When x = 0 then f(0) =0²-3 = -3

When x =1 then f(1) = 1²-3 = -2

When x = 2 then f(2) = 2²-3 =1

The result changes from Negative value (-2) to positive value (1) hence our interval is (1,2) where a =1 and b=2

Using bisection method,x value is obtained as

x=(a+b)÷2 where a is the lower limit of our interval and b is the upper limit of our interval

We can Perform iterations in the following ways:

x₁=(1+2)÷2 =1.5

We know that f(x)=x²-3 ,when x=1.5,then we have

f(1.5)=(1.5)²-3=-0.75

The interval changes based on value of f(x) such the the lower limit of interval (a,b) is replaced by (a+b)÷2 when the value of f(x) is negative and upper limit b of our interval is replaced by (a+b)÷2 when the value of f(x) is positive

Since (-0.75) is a negative value,we replace lower limit of interval (1,2) with x₁ value 1.5  to obtain new interval as (1.5,2) to get x₂ we use new interval (1.5,2)

x₂=(1.5+2)÷2 =1.75

We know that f(x)=x²-3

When x=1.75 then we have

f(1.75)=(1.75)²-3=0.0625 since 0.0625 is a positive values we replace upper limit of interval (1.5,2) with x₂ value 1.75 to obtain new interval as (1.5,1.75) and to get x₃ we use our new interval (1.5,1.75)

x₃=(1.5+1.75)÷2=1.625

We know that f(x)=x²-3

When x=1.625 then we have

f(1.625)=(1.625)²-3=-0.359375 since -0.359375 is a negative value,then we replace the lower limit of interval (1.5,1.75) when our new x value 1.625 to obtain new interval as (1.625,1.75)

To get x₄ we use our new interval (1.625,1.75)

x₄=(1.625+1.75)÷2=1.6875

We know that f(x)=x²-3

When x=1.6875

f(1.6875)=(1.6875)²-3=-0.15234375 since -0.15234375 is a negative value,then we replace lower limit of our interval (1.625,1.75) with our new x value 1.6875 to obtain new interval as (1.6875,1.75) to get x₅ we use our new interval (1.6875,1.75)

x₅=(1.6875+1.75)÷2=1.71875

We know that f(x)=x²-3

When x=1.71875

f(1.71875)=(1.71875)²-3=-0.0458984375 since -0.0458984375 is a negative value,then we replace lower limit of interval (1.6875,1.75) with our new x value 1.71875 to obtain new interval as (1.71875,1.75) to get x₆ we use our new interval (1.71875,1.75)

x₆=(1.71875+1.75)÷2=1.734375

We know that f(x)=x²-3

When x=1.734375

f(1.734375)=(1.734375)²-3=0.008056640625 since 0.008056640625 is a positive value,then we replace upper limit of interval (1.71875,1.75) with our new x value 1.734375 to obtain new interval as (1.71875,1.734375) to get x₇ we use our new interval (1.71875,1.734375)

x₇=(1.71875+1.734375)÷2=1.7265625

We know that f(x)=x²-3

When x=1.7265625

f(1.7265625)=(1.7265625)²-3=-0.01898193359375 since -0.01898193359375 then we replace lower limit of interval (1.71875,1.734375) with new x value 1.7265625 to obtain new interval as (1.7265625,1.734375) to get x₈ we use new interval (1.7265625,1.734375)

x₈₌(1.7265625+1.734375)÷2=1.73046875

We know that f(x)=x²-3

When x=1.73046875

f(1.73046875)=(1.73046875)²-3=-0.00547790527344 since -0.00547790527344 is a negative value then we replace lower limit of interval (1.7265625,1.734375)with our new x value 1.73046875 to obtain new interval as (1.73046875,1.734375) to get x₉ we use new interval (1.73046875,1.734375)

x₉=(1.73046875+1.734375)÷2=1.732421875 we know that f(x)=x²-3

When x=1.732421875

f(1.732421875)=(1.732421875)²-3=0.00128555297852 since 0.00128555297852 is a positive value,then we replace upper limit of interval (1.73046875,1.734375) with our new x value 1.732421875 to obtain new interval as (1.73046875,1.732421875) to get x₁₀ we use new interval (1.73046875,1.732421875)

x₁₀=(1.73046875+1.732421875)÷2=1.7314453125 we know that f(x)=x²-3

When x=1.7314453125

f(1.7314453125)=(1.7314453125)²-3=-0.00209712982178 since -0.00209712982178 is a negative value,then we replace lower limit of interval (1.73046875,1.732421875)with new x value 1.7314453125 to obtain new interval as (1.7314453125,1.732421875) to get x₁₁ we use new interval (1.73144553125,1.732421875)

x₁₁=(1.7314453125+1.732421875)÷2=1.73193359375 we know that f(x)=x²-3

When x=1.73193359375 then

f(1.73193359375)=(1.73193359375)²-3=-0.00040602684021 since -0.00040602684021 is a negative value,then we replace lower limit of interval (1.7314453125,1.732421875) with our new x value 1.73193359375 to obtain new interval as (1.73193359375,1.732421875) to get x₁₂ we use new interval (1.73193359375,1.732421875)

x₁₂=(1.73193359375,1.732421875)÷2=1.732177734375 we know that f(x)=x²-3

When x=1.732177734375

f(1.732177734375)=(1.732177734375)²-3=0.00043970346451 since 0.00043970346451 is a positive value,then we replace upper limit of interval (1.73193359375,1.732421875) with our new x value 1.732177734375 to obtain new interval as (1.73193359375,1.732177734375)

To get x₁₃ we use new interval (1.73193359375,1.732177734375)

x₁₃=(1.73193359375,1.732177734375)÷2=1.7320556640625 we know that f(x)=x²-3 ,when x=1.7320556640625,then

f(1.7320556640625)=(1.7320556640625)²-3=0.00006823410987854 since 0.00006823410987854 is a positive value,then we replace upper limit of interval (1.73193359375,1.732177734375) with our new x value 1.7320556640625 to obtain new interval as (1.73193359375,1.7320556640625) to get x₁₄ we use new interval (1.73193359375,1.7320556640625)

x₁₄=(1.73193359375,1.7320556640625)÷2=1.73199462871875

therefore x = 1.7320 correct to four decimal places