Question

If the roots of \[a{x^2} + bx + c = 0\] are both negative and \[b < 0\] then
A) \[a < 0,c < 0\]
B) \[a < 0,c > 0\]
C) \[a > 0,c < 0\]
D) \[a > 0,c > 0\]

Answer 1

Hint: The sum and product of the root of a quadratic equation \[a{x^2} + bx + c = 0\] is given by \[\dfrac{{ - b}}{a}\& \dfrac{c}{a}\] respectively. Use them to to get the relations of a,b and c.
Complete Step by Step Solution:
Let \[\alpha \& \beta \] be the roots of a quadratic equation \[a{x^2} + bx + c = 0\]
Now we know that
\[\begin{array}{l}
\alpha + \beta = \dfrac{{ - b}}{a}\\
\alpha \beta = \dfrac{c}{a}
\end{array}\]
It is given that \[b < 0\]
Now if b itself is negative then -b will naturally be positive
So it should be noted that we are given \[\alpha ,\beta < 0\]
Which is both of them are negative
Therefore \[\alpha + \beta \] which is also \[( - ) + ( - )\] will naturally be a negative number
Which means \[\dfrac{{ - b}}{a}\] individually must be a negative number as we have seen that b is negative and -b is positive.
Therefore a has to be negative for \[\alpha + \beta < 0\]
Now as we have that a is less than 0 then we know that \[\alpha \beta \] will basically be a negative number times negative number i.e., \[( - ) \times ( - )\] which will basically be a positive number now as we know that the whole \[\dfrac{c}{a}\] must be positive then as a is less than 0 which means c must be less than 0 to then only a negative divided by negative will be a positive number
So from here we can say that \[a < 0,c < 0\] which means option A is the correct option here.

Note: Observing the values is a key factor here. One can also approach this problem by taking discriminant to be greater than equal to zero but the problem will be at one point we will be struck at \[b \ge 2\sqrt {ac} \] Now according to this both a and c must be either negative or positive. So it cannot give a definite answer.