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Find bisection method 2x^3-2x-5

Solution:
Your problem `->` bisection 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 `1` and `2`

`x_0=(1+2)/2=1.5`

`f(x_0)=f(1.5)=2*1.5^(3)-2*1.5-5=-1.25 < 0`


`2^(nd)` iteration :

Here `f(1.5) = -1.25 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `1.5` and `2`

`x_1=(1.5+2)/2=1.75`

`f(x_1)=f(1.75)=2*1.75^(3)-2*1.75-5=2.2188 > 0`






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