Question
If $$\alpha $$ and $$\beta \left( {\alpha < \beta } \right)$$ are the roots of the equation $$x^2 + bx + c = 0 ,$$ where, $$c < 0 < b,$$ then
A.
$$0 < \alpha < \beta $$
B.
$$\alpha < 0 < \beta < \left| \alpha \right|$$
C.
$$\alpha < \beta < 0$$
D.
$$\alpha < 0 < \left| \alpha \right| < \beta$$
Answer :
$$\alpha < 0 < \beta < \left| \alpha \right|$$
Solution :
Given, $$\alpha < \beta ,c < 0,b < 0,$$
$$\therefore \alpha + \beta = - b < 0\,\,{\text{and }}\alpha \beta = c < 0$$
Clearly, $$\alpha$$ and $$\beta$$ have opposite signs and $$\alpha < \beta$$
$$\eqalign{
& \therefore \alpha < 0\,\,{\text{and }}\beta > 0 \cr
& \Rightarrow \alpha < 0 < \beta \cr} $$
Further, $$\alpha + \beta < 0$$
$$\eqalign{
& \Rightarrow \beta < - \alpha \cr
& \Rightarrow \left| \beta \right| < \left| { - \alpha } \right| \cr
& \Rightarrow \beta < \left| \alpha \right|\left( {\beta > 0 \Rightarrow \left| \beta \right| = \beta } \right) \cr} $$
Hence, $$\alpha < 0 < \beta < \left| \alpha \right|$$