Question

If the roots of $a{x^2} + bx + c = 0$ are equal in magnitude but opposite in sign, then a possible condition is
$
  {\text{A}}{\text{. a < 0,c < 0}} \\
  {\text{B}}{\text{. a < 0,c > 0,b > 0}} \\
  {\text{C}}{\text{. a > 0,b = 0,c < 0}} \\
  {\text{D}}{\text{. a > 0,b = 0,c > 0}} \\
$

Answer 1

Hint:Here it is given that both the roots are equal in magnitude and opposite in sign so let us assume roots are p and –p and if these are roots then on putting these in equation it will satisfy the equation and we will find out the condition for a,b, c.

Complete step-by-step answer:
We have given quadratic equation
$a{x^2} + bx + c = 0$
Now using relation between coefficients and roots we will find condition for a,b,c
We know
Sum of roots = $\dfrac{{ - b}}{a}$
$
  \therefore p + \left( { - p} \right) = \dfrac{{ - b}}{a} \\
   \Rightarrow 0 = \dfrac{{ - b}}{a} \\
   \Rightarrow b = 0 \\
$
Product of roots = $\dfrac{c}{a}$
$\left( p \right) \times \left( { - p} \right) = \dfrac{c}{a}$
$ \Rightarrow - {\left( p \right)^2} = \dfrac{c}{a}$
Here we can see one side is always negative because the square of a number is also positive and multiplying with negative it becomes always negative so $\dfrac{c}{a}$ have to be negative and for sign to be negative c and a have to have opposite signs.
So with b = 0, and opposite signs of a and c we have only option.
So option C is the correct option.

Note:Whenever we get this type of question the key concept of solving is we have many options as one is described in this question or we may proceed through the graph but the method discussed here is very simple and easy. Relation between roots and coefficients of x are very helpful in solving questions like this.