Question

If the equations $x^2+ix+a=0,x^2-2x+ia=0,a\ne 0$ have a common root then

• A. $a$ is real
• B. $a=\frac{1}{2}+i$
• C. $a=\frac{1}{2}-i$
• D. the other root is also common

Answer 1

The correct option is C $a=\frac{1}{2}-i$

Given,
$x^2+ix+a=0,$ $x^2-2x+ia=0$ have a common root.
Let $\alpha$ be the common root.
$\Rightarrow {\alpha }^2+i\alpha +a=0$
${\alpha }^2-2\alpha +ia=0$
Solving by Cramer's rule
$\frac{{\alpha }^2}{i^{2}a+2a}=\frac{\alpha }{a-ia}=\frac{1}{-2-i}$
$\Rightarrow \frac{{\alpha }^2}{a}=\frac{\alpha }{a(1-i)}=\frac{1}{-2-i}$
$\Rightarrow \alpha =\frac{1}{1-i},\alpha =\frac{a(1-i)}{-2-i}$
$\Rightarrow a=\frac{-2-i}{1+i^2-2i}=\frac{-2-i}{-2i}$
$a=\frac{1}{i}+\frac{1}{2}$
$a=-i+\frac{1}{2}$
Hence, option 'C' is correct.