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