Question & Answer
QUESTION

If $\alpha $ and $\beta $ , are the roots of the quadratic 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 Verified Verified
Hint: We have to use the given condition along with the relationship between sum and product of roots with the coefficients and the constant term.

Complete step-by-step answer:
Given equation is ${x^2} + bx + c = 0$
It is given that $\alpha $ and $\beta $ are the roots of the equation.
$ \Rightarrow $ Sum of roots = $\alpha + \beta = - \dfrac{\text{coefficient of x}}{\text{coefficient of }x^2} = \dfrac{{ - b}}{1} = - b$
$ \Rightarrow $ Product of roots = $\alpha \beta = \dfrac{\text{constant}}{\text{coefficient of }x^2} = \dfrac{c}{1} = c $
It is given that c < 0 < b …… (1)
$ \Rightarrow $ c < 0
$ \Rightarrow $ $\alpha $$\beta $ < 0
$ \Rightarrow $ One root is negative
Let us suppose $\alpha $ is negative
$ \Rightarrow $ $\beta $ is positive
$ \Rightarrow $ $\alpha $ < 0 < $\beta $ …… (2)
From equation (1)
b > 0
$ \Rightarrow $ -b < 0
Now, $\alpha $ + $\beta $ = $ - b$
$ \Rightarrow $ $\alpha $ + $\beta $ < 0
$ \Rightarrow $ -$\alpha $ > $\beta $
$ \Rightarrow $ \[\left| \alpha \right| > \beta \] …… (3)
So, from equation (2) and (3)
$ \Rightarrow $$\alpha $ < 0 < $\beta $ < \[\left| \alpha \right|\]
So, option B is your correct answer.

Note: To solve these types of problems, we have to use the relationship between sum and product of roots with the coefficients and the constant term. Mistakes can be made in dealing with signs in presence of inequality, so try to avoid that.