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The sum of non-real roots of the polynomial equation x3 + 3x2 + 3x + 3 = 0. 1. lies between 0 and 1 2. equals 0 3. lies between -1 and 0 4. has absolute value bigger than 1
The sum of non-real roots of the polynomial equation x3 + 3x2 + 3x + 3 = 0.
Let the the roots be a,b,c....in which b and c be the 2 non real roots...we need to find b+c...On the other hand we have:::a+b+c = -3....ab+bc+ca = 3...abc = -3...Now we use these 3 equations...bc = -3/a...putting this in 2nd eq...we get....-> a(b+c) - 3/a = 3-> a² (b+c) - 3a - 3 = 0...which is a quadratic in `a`..Discriminant of this should be 0 because only one possible value of a can be root of the given cubic equation...hence on doing so...-> 9 + 12(b+c) = 0-> (b+c) = -3/4....ans....option 3...
Write the equation as (x+1)3 = -2. Then the roots are (-2)1/3-1, (-2)1/3w-1, and (-2)1/3w2 – 1, where w is the primitive cube root of unity. Then sum of non-real roots = [ (-2)1/3w-1]+ [ (-2)1/3w2 – 1] = -[ (2)1/3w +(2)1/3w2] -2 = 21/3 - 2 [Since 1+w+w2= 0] = -[2-21/3]. 2>21/3 >1. Hence 1>2-21/3>0 and hence -11/3 - 2 Hence the option C is correct
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