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 Solution will be displayed step by step (In 4 parts) Solution Find bisection method 2x^3-2x-5, a=1, b=4Solution:Your problem -> bisection method 2x^3-2x-5, a=1, b=4Here 2x^3-2x-5=0Let f(x) = 2x^3-2x-51^(st) iteration :Here f(1) = -5 < 0 and f(4) = 115 > 0:. Now, Root lies between 1 and 4x_0=(1+4)/2=2.5f(x_0)=f(2.5)=2(2.5)^(3)-2(2.5)-5=21.25 > 02^(nd) iteration :Here f(1) = -5 < 0 and f(2.5) = 21.25 > 0:. Now, Root lies between 1 and 2.5x_1=(1+2.5)/2=1.75f(x_1)=f(1.75)=2(1.75)^(3)-2(1.75)-5=2.2188 > 03^(rd) iteration : Solution provided by AtoZmath.com Any wrong solution, solution improvement, feedback then Submit Here Want to know about AtoZmath.com and me