We have $f(x)=x^2-6$. I have to find $p_3$ if $p_0 = 3$ and $p_1 = 2$ by using
a) Secant method
b) False position method
So for the first one I have
$p_2=p_0- \dfrac {f(p_0)(p_1-p_0)}{f(p_1)-f(p_0)}$= $3- \dfrac {3(2-3)}{(-2-3)}$=3-0.6=2.4
$p_3=p_1- \dfrac {f(p_1)(p_2-p_1)}{f(p_2)-f(p_1)}$=2.45454
Now how do I find an approximate root using false position method?