Question

Find a root of the equation x3-5x-3=0 correct to three decimal place using bisection method.

Answer 1

We have to find the root of the equation x³ - 5x - 3 = 0 to correct to three decimal place using bisection method.

let find the value of x³ - 5x + 3 = 0 at 1, 2, 3

f(1) = 1 - 5 - 3 = -1

f(2) = 8 - 10 - 3 = -5

f(3) = 27 - 15 - 3 = 9

you see, f(2) . f(3) < 0 it means, there must be one root between 2 and 3.

now, let's take value of f(x) in the average of 2 and 3 i.e., 2.5

f(2.5) = (2.5)³ - 5(2.5) - 3 = 0.125.

∵ f(2) f(2.5) < 0

so the root must be between 2 and 2.5

now again take an average of 3 and 2.5. that is 2.75

now f(2.25) = (2.25)³ - 5(2.25) - 3 = -2.85

here, f(2.5) . f(2.25) < 0

so, the root must be lie between 2.25 and 2.5.

so, x = (2.25 + 2.5)/2 = 2.375

now, f(2.375) = (2.375)³ - 5(2.375) - 3 = -1.47

here, f(2.375).f(2.5) < 0

hence the root must be lie between 2.375 and 2.5.

now, x = (2.375 + 2.5)/2 = 2.4375

f(2.4375) = (2.4375)³ - 5(2.4375) - 3 = -0.7

so the root must be lie between 2.4375 and 2.5

now, x = (2.4375 + 2.5)/2 = 2.46875

so f(2.46875) = (2.46875)³ - 5(2.46875) - 3 =-0.297

so the root must be lie between 2.46875 and 2.5

now, x = (2.46875 + 2.5)/2 = 2.484375

now, f(2.484375) = (2.484375)³ - 5(2.484375) - 3 = -0.088 ≈ 0 [ we can take it now ]

hence the value of a root of given equation is 2.484375 approximately.

Therefore a root of equation x³ - 5x - 3 is 2.484 (approximately) which is to three decimal places.