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Find regula falsi method 2x^3-2x-5

Solution:
Your problem `->` regula falsi method 2x^3-2x-5


Here `2x^3-2x-5=0`

Let `f(x) = 2x^3-2x-5`

Here
`x`012
`f(x)`-5-57



`1^(st)` iteration :

Here `f(1) = -5 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1` and `x_1 = 2`

`x_2 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_2 = 1 - (-5) * (2 - 1)/(7 - (-5))`

`x_2 = 1.4167`

`f(x_2)=f(1.4167)=2(1.4167)^(3)-2(1.4167)-5=-2.147 < 0`


`2^(nd)` iteration :

Here `f(1.4167) = -2.147 < 0` and `f(2) = 7 > 0`






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