Question

Find a positive real root of x^3-2x-5=0 by bisection method?

Answer 1

Answer:

$given = {x}^{3} - 2x - 5 = 0 \\ x = 1 = 1 - 2 - 5 = - 6 \\ x = 2 = 8 - 4 - 5 = - 1 \\ x = 3 = 27 - 6 - 5 = 16 \\ = 0 \: lies \: between \: - 1 \: and \: 16 \\ = hence \: x \: lies \: between \: 2 \: and \: 3 \\ = \frac{(2 + 3)}{2} = 2.5 \\ x = 2.5 = {x}^{3} - 2x - 5 = 5.625 \\ = 0 \: lies \: between \: - 1 \: and \: 5.625 \\ = x \: lies \: between \: 2 \: and \: 2.5 \\ = \frac{(2 + 2.5)}{2} = 2.25 \\ x = 2.25 = {x}^{3} - 2x - 5 = 1.89 \\ = 0 \: lies \: between \: - 1 \: and \: 1.89 \\ = x \: lies \: between \: 2 \: and \: 2.25 \\ x = 2.125 = \: {x}^{3} - 2x - 5 = 0.34 \\x \: lies \: between \: 2 \times 2.125 \\ 0 \: lies \: between \: - 1 \: and \: 0.34 \\ 0 \: lies \: between \: - 0.35 \: and \: 0.34 \\ x = 2.0625 = {x}^{3} - 2x - 5 = - 0.0089 = 0 \\ x = 2.094 \: up \: to \: three \: decimal \: places \: \\ = x = 2.094$

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