Question

The real values of $a$ for which the quadratic equation $2x^2-(a^3+8a-1)x+a^2-4a=0$ possesses roots of opposite signs are given by :

• A. $a>6$
• B. $a>9$
• C. $0<a<4$
• D. $a<0$

Answer 1

Answer: (C)

$0<a<4$

Let $\alpha$ and $\beta$ be the roots of the equation.

$\therefore \alpha \beta =\frac{a^2-4a}{2}$

If the roots are of opposite sign then the product of roots is negative

$\therefore \alpha \beta <0$

$\therefore a^2-4a<0$

$\therefore a(a-4)<0$

$\therefore a\in (0,4)$

Hence, $0<a<4$.