Find the root of the equation using Bisection Method, Engineering Mathematics

    • March 17, 2020
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    Latest Questions Update › Category: Mathematics › Find the root of the equation using Bisection Method, Engineering Mathematics
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    learning4u Staff asked 5 years ago
    Using the method of bisection, find the root of the equation x3 – 9x + 1 = 0 between x = 2 and x = 4. Solve it up to four approximation four decimal places.
    Numerical Analysis, Engineering Mathematics 
    1 Answers
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    Education Desk Staff answered 5 years ago
    Given equation is y = f(x) = x3 – 9x + 1 and it is continuous in the interval 2 ≤ x ≤ 4. Now, first determine the value of a function at the points 2 and 4 i.e.
    f(2) = 23 – 9(2) + 1 = 8 – 18 + 1 = – 9
    f(4) = 43 – 9(4) + 1 = 64 – 36 + 1 = +29
    f(2) f(4) < 0
    Since f(2) is negative and f(4) is a positive value, therefore the root f(x) lies in between 2 and 4. Now go to the First Approximation.

    First Approximation value of x
    x1 = (2+4)/2 = 3
    Again, check the value of f(x) at x = 3,
    f(3) = 33 – 9(3) + 1 = 27 – 27 + 1 = 1
    f(2)f(3) < 0
    Since f(3) is a positive value and f(2) is a negative value, therefore the root of f(x) lies in between 2 and 3. Now go to Second Approximation.

    Second Approximation value of x
    x2 = (2+3)/2 = 2.5
    Again, check the value of f(x) at x = 2.5,
    f(2.5) = 2.53 – 9(2.5) + 1 = – 5.87
    f(2.5)f(3) < 0
    Since f (3) is a positive value and f (2.5) is a negative value, therefore the root of f(x) lies in between 2.5 and 3. Now go to Third Approximation.

    Third Approximation value of x
    x3 = (2.5+3)/2 = 2.75
    Again, check the value of f(x) at x = 2.75,
    f(2.75) = 2.753 – 9(2.75) + 1 = – 2.96
    f(2.75)f(3) < 0
    Since f(3) is a positive value and f(2.75) is a negative value, therefore the root of f(x) lies in between 2.75 and 3 and go to Fourth Approximation.

    Fourth Approximation value of x
    x4 = (2.75+3)/2 = 2.88
    Again, check the value of f(x) at x = 2.88,
    f(2.88) = 2.883 – 9(2.88) + 1 = – 1.03
    f(2.88)f(3) < 0
    Since f(3) is a positive value and f(2.88) is a negative value, therefore the root of f(x) lies in between 2.88 and 3. Now go to Fifth Approximation.

    Fifth Approximation value of x
    x5 = (2.88+3)/2 = 2.94
    Again, check the value of f(x) at x = 2.94,
    f(2.94) = 2.943 – 9(2.94) + 1 = – 0.05
    f(2.94)f(3) < 0
    Proceeding similarly we obtain x6 = 2.9375. Therefore, x = 2.9375 is our approximate root of f(x).

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