# Newton-Raphson Method

We have find the solution of *f* (*x*, *y*) = 0 and *g* (*x*, *y*) = 0

First find an approximate solution of equation (*x*_{0}, *y*_{0}) by graphical or trial and error method.

From this we find more approximate values (*x*_{1}, *y*_{1}) such that *x*_{1} = *x*_{0 }+ *h* and *y*_{1 }= *y*_{0} + *k*

âˆ´ *f* (*x*_{0} + *h*, *y*_{0} + *k*) = 0 and *g* (*x*_{0} + *h*, *y*_{0} + *k*) = 0

From Taylorâ€™s series we get,

= 0 and = 0

Solving these equations we get *h* and *k*.

Thus we find *x*_{1} and *y*_{1}

From this value of *x*_{1} and *y*_{1} we find *x*_{2} and *y*_{2} in similar way.

# Bisection Method

For a non linear equation *f* (*x*) and for any two number *a* and *b* if *f* (*a*), *f* (*b*) > 0, then there is atleast one root for the equation *f* (*x*) in between *a* and *b*.

Now let *x*_{1} = *a* and *x*_{2} = *b*

We find the mid point of *x*_{1} and *x*_{2} i.e. *x*_{0}

*x _{0}* =

If *f* (*x*_{0}) = 0, *x*_{0} is the root

If *f* (*x*_{0}) *f* (*x*_{1}) < 0, root is in between *x*_{0} and *x*_{1}

If *f* (*x*_{0}) *f* (*x*_{2}) < 0, root is in between *x*_{0} and *x*_{2 }and this process in continued with either (*x*_{0}, *x*_{1}) or with (*x*_{0}, *x*_{2}).