and should i jump in too


let x=square root of 25
so that x2-25=0 ....(1)
taking f(x)=x2-25 , newton raphson method gives
xn+1 = xn - [ f(xn) /f '(xn) ]
=xn - [ x2n - 25 /2xn ]
= ½ [ xn + (25 / xn ) ] .....(2)
now since f(4) = -9 , f(6) = 11 ,
and also since f(-4) = -9 and f(-6) = 11
a root of (1) lies between 4 and 6 and -4 and -6
Case 1
therefore , taking x0 = 4.5 , equ (2) gives
x1= ½ [ x0 + (25 /x0 ) ]
= ½ [ 4.5 +(25/4.5)]
= 5.0277778
x2= ½ [ x1 + (25 /x1 ) ]
= ½ [5.0277778 +(25/5.0277778 )]
= 5.0000767
x3= ½ [ x2 + (25 /x2 ) ]
= ½ [5.0000767 +(25/5.0000767 )]
= 5
x4=½ [ x3 + (25 /x3 ) ]
= ½ [5 +(25/5 )]
= 5
since x3=x4
we take square root of 25 = 5
Case 2
therefore , taking x0 = -4.5 , equ (2) gives
x1= ½ [ x0 + (25 /x0 ) ]
= ½ [ -4.5 +(25/-4.5)]
= -5.0277778
x2= ½ [ x1 + (25 /x1 ) ]
= ½ [-5.0277778 +(25/-5.0277778 )]
= -5.0000767
x3= ½ [ x2 + (25 /x2 ) ]
= ½ [5.0000767 +(25/5.0000767 )]
= -5
x4=½ [ x3 + (25 /x3 ) ]
= ½ [-5 +(25/-5 )]
= -5
since x3=x4
we take square root of 25 = -5
Hence there are 2 cases -5 and 5 . Thus, every positive number has two square roots, one positive and the other negative.Hope Helps !! Silly,ab,eric lets stop the love story here, now , in GMAT its important to refer to both the roots. So this is very clear .





