Question

If the roots of the quadratic equation $2x^2-(a^3+1)x+(a^2-2a)=0$ are real and opposite signs, then the set of possible values of a in the interval

• A. (0, 2)
• B. (0, 1)
• C. (1, 2)
• D. (-1, 2)

Answer 1

Answer: (A)

$2x^2-(a^3+1)x+(a^2-2a)=0$ has real and opposite signs roots.
Sum of roots i.e., $x_1+x_2=\frac{a_3+1}{2}$
Products of roots i.e., $x_1\cdot x_2=\frac{a^2-2a}{2}$
Since $x_1$ and $x_2$ are of opposite signs
$\therefore x_1\cdot x_2<0$
$\Rightarrow \frac{a^2-2a}{2}<0\Rightarrow a^2-2a<0$
$\Rightarrow a(a-2)<0\Rightarrow a\in (0,2).$