
If $\alpha ,\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
161.7k+ views
Hint: Use the fact that we can calculate $\alpha + \beta $ and $\alpha \beta $ from the given equation ${x^2} + bx + c = 0$. Form a few inequalities with the given information in the question and from the sum and product of the roots. Combine all the inequalities to get a final answer.
Formula Used:
standard form $a{x^2} + bx + c = 0$ and its two roots are $m$ and $n$, then the product of the roots, $mn$ is given by $\dfrac{c}{a}$ and the sum of the roots, $m + n$ is given by $ - \dfrac{b}{a}$.
Complete step-by-step solution:
Given the quadratic equation ${x^2} + bx + c = 0$, we know that the sum of the roots is $ - b$ and the product of the roots is $c$.
$b$ is a positive number while $c$ is a negative number as it is given that $c < 0 < b$.
Since the product of the roots is negative, one of the roots must be negative and the other positive. Therefore, $\alpha $ is negative and $\beta $ is positive as it is given that $\alpha < \beta $.
Since the sum of the roots is also negative, the absolute value of the negative root must be greater than the positive root. Therefore, $\left| \alpha \right| > \beta $.
We know that $\alpha < \beta $, $\left| \alpha \right| > \beta $, $\alpha $ is negative and $\beta $ is positive.
Therefore, \[\alpha < 0 < \beta < \left| \alpha \right|\].
Note: Students should not get confused with the two quadratic equations $a{x^2} + bx + c = 0$ and ${x^2} + bx + c = 0$. The value of a can change our solution completely. So take care of it while simplifying.
Formula Used:
standard form $a{x^2} + bx + c = 0$ and its two roots are $m$ and $n$, then the product of the roots, $mn$ is given by $\dfrac{c}{a}$ and the sum of the roots, $m + n$ is given by $ - \dfrac{b}{a}$.
Complete step-by-step solution:
Given the quadratic equation ${x^2} + bx + c = 0$, we know that the sum of the roots is $ - b$ and the product of the roots is $c$.
$b$ is a positive number while $c$ is a negative number as it is given that $c < 0 < b$.
Since the product of the roots is negative, one of the roots must be negative and the other positive. Therefore, $\alpha $ is negative and $\beta $ is positive as it is given that $\alpha < \beta $.
Since the sum of the roots is also negative, the absolute value of the negative root must be greater than the positive root. Therefore, $\left| \alpha \right| > \beta $.
We know that $\alpha < \beta $, $\left| \alpha \right| > \beta $, $\alpha $ is negative and $\beta $ is positive.
Therefore, \[\alpha < 0 < \beta < \left| \alpha \right|\].
Note: Students should not get confused with the two quadratic equations $a{x^2} + bx + c = 0$ and ${x^2} + bx + c = 0$. The value of a can change our solution completely. So take care of it while simplifying.
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