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( a ) 3 + i

( b ) 3 + 2i

( c ) -1 + i

( d ) -1 - i

Answer
Verified

Now, firstly we will find the coefficients of $ {{x}^{2}} $, $ x $ and constant c from the given polynomial $ i{{x}^{2}}-2(1+i)x+(2-i)=0 $ by comparing it with the general form of quadratic polynomial which is expressed as $ a{{x}^{2}}+bx+c=0 $ .

On comparing given polynomial with general form of quadratic polynomial, we get coefficients of quadratic polynomial $ i{{x}^{2}}-2(1+i)x+(2-i)=0 $ equals to,

a = i, b = -2( i + 1 ), c = 2 - i

let, \[\alpha ,\beta \] be two roots of quadratic polynomials.

Now,

Here, we know that $ \alpha +\beta =-\dfrac{b}{a} $ ……( i ),

So, we can obtain the value of $ \beta $ which represents the another root of quadratic equation, easily by substituting the values of b and a in an equation ( i )

Substituting values of a = i and b = -2( i + 1 ) in $ \alpha +\beta =-\dfrac{b}{a} $ , we get

$ \alpha +\beta =-\dfrac{(-2-2i)}{i} $

On simplifying signs, we get

$ \alpha +\beta =\dfrac{(2+2i)}{i} $

We know, $ \alpha =3-i $ as it is one of the root of quadratic equation, then substituting value of $ \alpha =3-i $ in $ \alpha +\beta =\dfrac{(2+2i)}{i} $, we get

$ (3-i)+\beta =\dfrac{(2+2i)}{i} $

Multiplying numerator and denominator on right hand side by i, we get

$ (3-i)+\beta =\dfrac{(2+2i)}{i}\times \dfrac{i}{i} $

On solving, we get

$ (3-i)+\beta =\dfrac{(2i+2{{i}^{2}})}{{{i}^{2}}} $

As $ {{i}^{2}}=-1 $ , so

$ (3-i)+\beta =\dfrac{(2i-2)}{-1} $

On simplifying, we get

$ (3-i)+\beta =-2i+2 $

$ \beta =-i-1 $