Question

if $0<a<b<c$, and the roots $\alpha ,\beta$ of the equation $a{x}^2+bx+c=0$ are non real complex numbers, then

• A. $\left|\alpha \right|=\left|\beta \right|$
• B. $\left|\alpha \right|>1$
• C. $\left|\beta \right|<1$
• D. None of these

Answer 1

Answer: (A), (B)

The correct options are
A $\left|\alpha \right|=\left|\beta \right|$
B $\left|\alpha \right|>1$

R.E.F image

$a{x}^2+bx+c=0$

$D<0$

$b^2-4ac<0$ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$|\alpha |=\sqrt{{\alpha }^2+{\beta }^2}$

$|\beta |=\sqrt{{\alpha }^2+{\beta }^2}$

$|\alpha |=|\beta |$

$\alpha =\frac{-b+!\sqrt{4ac-b^2}}{2a}$

$|\alpha {|}^2=\frac{b^2}{4a^2}+\frac{4ac-b^2}{4a}=\frac{4ac}{4a^2}=\frac{c}{a}>1$

$|\alpha |>1$

$|\beta |>1$

Hence option A & B are correct