Question

# If one root of the $i{{x}^{2}}-2(1+i)x+(2-i)=0$ is 3 – i; where $i=\sqrt{-1}$ , then the other root is( a ) 3 + i( b ) 3 + 2i( c ) -1 + i( d ) -1 - i

Hint: We will use the results based on of relationship between roots of quadratic equation to evaluate the values of $\alpha +\beta$, which is given by the relation of coefficients of quadratic equation such as $\alpha +\beta =-\dfrac{b}{a}$ for quadratic equation $a{{x}^{2}}+bx+c=0$ where $a\ne 0.$

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
$(3-i)+\beta =-2i+2$
$\beta =-i-1$
So, the correct answer is “Option D”.

Note: Remember these formulae as they are very helpful in solving questions. While calculating the coefficients of a quadratic equation, try to avoid signs of error as this makes the answer incorrect. Simplification of signs should be done carefully. We can also use the product of the root formula and simplify to get the answer.