Question

find a real root of x3-2-11=0 between 2 and 3 correct upto four decimal places using bisection method

Answer 1

Answer:

(p+2)(p-3)+(p+3)(p-4)=p(2p-5)

p²-3p+2p-6+p²-4p+3p-12=2p²-5p

2p²-2p-18=2p²-5p

-2p+5p=18

3p=18

p=6