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At which approximation do we get an approximate root of equation x3 + 3x – 1 = 0 corrected to 2 decimal places?
    Correct answer is '6'. Can you explain this answer?
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    At which approximation do we get an approximate root of equation x3 + ...
    Solution to equation x3 + 3x – 1 = 0:
    Put x = 0 in f(x).
    f(0) = 0 + 0 – 1 = – 1
    x = 1
    f(1) = 13 + 3 – 1 = 3
    The root of equation lies between 0 and 1 as f(0) and f(1) has an opposite sign.
    First approximation:
    f(x0) = 0.53 + 3 x 0.5 – 1 = 0.625,
    As f(x0) is positive, next interval for root is 0 and 0.5.
    Second Approximation:

    f(x1) = 0.253 + 3 x 0.25 – 1 = - 0.234875
    As f(x1) is negative, the rest lies between 0.25 and 0.5.
    Third Approximation:

    f(x2) = 0.17773, which is positive.
    Hence, the root lies between 0.25 of 0.375.
    Fourth Approximation:

    f(x3) = – 0.03198 (–ve)
    Fifth Approximation:

    f(x4) = 0.07186

    f(x5) = 0.0197

    f(x6) = – 0.006197
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    At which approximation do we get an approximate root of equation x3 + 3x – 1 = 0 corrected to 2 decimal places?Correct answer is '6'. Can you explain this answer?
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