Question:
If the quadratic equation 2x2 – (a3 + 8a – 1) x + a2 – 4a = 0 possesses roots of opposite signs, then a lies in the interval ____________.
Solution:
For $2 x^{2}-\left(a^{3}+8 a-1\right) x+a^{2}-4 a=0$ has roots of opposite sign.
Let $\alpha$ and $\beta$ be 2 roots of above equation.
Then, according to give condition,
$\alpha \beta<0$
also, since sum of roots $=\frac{a^{3}+8 a-1}{2}$
$\Rightarrow \alpha+\beta=\frac{a^{3}+8 a-1}{2}$
and product of roots is $a^{2}-4 a$
i. e. $\alpha \beta=a^{2}-4 a$
since $\alpha \beta<0$
$\Rightarrow a^{2}-4 a<0$
i.e. $a(a-4)<0$
Since a(a – 4) < 0 is true for 0 < a < 4
Hence a ∈ (0, 4)