1.

Using Graeffe's root-squaring method, find real roots of the equation \( x^{3}-6 x^{2}+11 x-6=0 \)Turn Over

Answer»

x3 −6x+11x−6=0 

Put X=1 

1−6+11−6=0 

1 is root of equation, so (x−1) is factor of x3 −6x2 +11x−6 

x2 −5x+6=0 

(x−3)(x−2)=0 

x=2,3 

root of cubic equation =1,2,3



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