Question

If one root of the equation $x^2+(1-3i)x-2(1+i)=0$ is $-1+i$, then the other root is

• A. $-1-i$
• B. $\frac{\left(-1-i\right)}{2}$
• C. $i$
• D. $2i$

Answer 1

The correct option is D

$2i$

Explanation for the correct solution

From the question, we know that one root of the given equation is $-1+i$. Suppose that the other root is $b$.

We also know that the quadratic equation of the form $x^2+cx+d=0$ has sum of its roots equal to the negative of the coefficient of $x$ that is $-c$.

In this case, the sum of the roots will be equal to the $-\left(1-3i\right)$.

$\begin{array}{rcl}\therefore -1+i+b& =& -1+3i\\ & \Rightarrow & b=-1+1+3i-i\\ & \Rightarrow & b=2i\end{array}$

Hence, the correct option is (D).