Question

If the roots of the equation $z^2+(a+ib)z+c+id=0$ are real, then

• A. d² - b²c = abd
• B. d² + b²c = acd
• C. d² + b²c = abd
• D. d² + b²a = abd

Answer 1

The correct option is C d² + b²c = abd

Let $xϵR$ be a root of the given equation.Putting x for z,the equation
reduces to

$x^2+(a+ib)x+c+id=0\Rightarrow (x^2+ax+c)+i(bx+d)=0$

Equating the real and imaginary parts,we have

$x=-\frac{d}{b}$ and $x^2+ax+c=0.$

Eliminating x from the above equations,we have $\frac{d^2}{b^2}-\frac{ad}{b}+c=0$

$\Rightarrow d^2+b^{2}c=abd.$