Question

If $a,b$ and $c$ are non-zero real numbers and $a{z}^2+bz+c+i=0$ has purely imaginary roots, then $a$ is equal to

• A. $bc$
• B. $b^{2}c$
• C. $-b^{2}c$
• D. $\frac{1}{2}{b}^{2}c$

Answer 1

Answer: (B)

$b^{2}c$

$a{z}^2+bz+c+i=0$

Let one of its root be $z=i\alpha$

$a{\left(i\alpha \right)}^2+b\left(i\alpha \right)+c+i=0$

$-a{\alpha }^2+i\alpha b+c+i=0$

$(c-a{\alpha }^2)+i(b\alpha +1)=0$

$(c-a{\alpha }^2)+i(b\alpha +1)=0+0i$

Comparing imaginary coefficients, we get

$b\alpha +1=0,$

$\Rightarrow \alpha =-\frac{1}{b}$

comparing real coefficients, we get

$c-a{\alpha }^2=0$

$c=a{\alpha }^2$

$c=a{(-\frac{1}{b})}^2$

$c=\frac{a}{b^2}$

$\Rightarrow a=b^{2}c$