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Q.
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
COMEDKCOMEDK 2005Complex Numbers and Quadratic Equations
Solution:
$2x^2 - (a^3 + 1)x + (a^2 - 2a) = 0$ 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 - 2 a < 0$
$ \Rightarrow \:\:\: a( a- 2) < 0 \: \Rightarrow a \in ( 0, 2).$