Question

Let $x^2-ax+b=0$, where $a,b\in R$ be a quadratic equation such that the roots are opposite in sign and the magnitude of one root is twice the other. Then which of the following options is/are always true ?

• A. $2b^2+a=0$
• B. $2a^2+b=0$
• C. $b<0$
• D. $b>0$

Answer 1

Answer: (B), (C)

$b<0$

$x^2-ax+b=0...\left(1\right)$
Roots are opposite in sign. This means the roots are real.
Let $\alpha ,\beta$ be the roots.
$\alpha +\beta =a,\alpha \beta =b;a,b\ne 0$
Let $\alpha =-2\beta$
$\Rightarrow \beta =-a$
$\therefore \alpha =2a$

$\Rightarrow \alpha \beta =b=-2a^2<0$

Now, substituting either $\alpha =2a$ or $\beta =-a$ in equation $\left(1\right),$ we get
$2a^2+b=0$