Question
If $$\alpha \,\,{\text{and}}\,\,\beta \left( {\alpha < \beta } \right)$$ 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 $$c < 0 < b$$ and $$\alpha + \beta = - b\,\,\,\,\,\,\,\,\,.....\left( 1 \right)$$
$$\alpha \beta = c\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right)$$
From (2), $$c < 0$$
$$ \Rightarrow \,\,\alpha \beta < 0$$
⇒ either $$\alpha $$ is $$- ve$$ or $$\beta $$ is $$- ve$$ and second ;
quantity is positive.
from (1), $$b > 0$$
⇒ $$- b < 0$$
$$ \Rightarrow \,\,\alpha + \beta < 0$$
⇒ the sum is negative
⇒ modules of negative quantity is > modulus of positive quantity but $$\alpha $$ < $$\beta $$ is given. Therefore, it is clear that $$\alpha $$ is negative and $$\beta $$ is positive and modulus of $$\alpha $$ is greater than modulus of $$\beta $$
$$ \Rightarrow \,\,\alpha < 0 < \beta < \left| \alpha \right|$$