Question

If the roots of the equation \[a{{x}^{2}}+bx+c=0\] are real and distinct, then
(a) both roots are greater than \[\dfrac{-b}{2a}\]
(b) both roots are less than \[\dfrac{-b}{2a}\]
(c) one of the roots exceeds \[\dfrac{-b}{2a}\]
(d) None of these

Answer 1

Hint: First find the condition for the roots being real and distinct. Then use this condition on the expression of the root in terms of a, b and c. Now, find the condition of the root on the\[\dfrac{b}{2a}\] term. This condition is our required result.

Complete step-by-step answer:
The given equation in the question can be written as:
\[a{{x}^{2}}+bx+c=0\]
The condition for roots being real and distinct is given by: D > 0
By substituting the value of D in the inequality, we get it as:
\[{{b}^{2}}-4ac>0\]
The above equation is proved by the following steps here. The roots of the given equation are said to be written as:
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
So, the term inside the square root has 3 cases. If it is negative, then the square root of the negative gives us the imaginary roots. If it is zero, then we get both the roots as equal. If it is positive, then the symbols \[\pm \] are given 2 roots, so we say real and distinct.
We are given in the question the real and distinct roots. So, we can write,
\[{{b}^{2}}-4ac>0\]
Its value is given to be positive for this solution. Let us assume the square root of this number is p. By assumption, we can write the value of p as
\[p=\sqrt{{{b}^{2}}-4ac}\]
By substituting this into roots expression, we get it as:
\[x=\dfrac{-b\pm p}{2a}\]
Let the two roots of the equation given be j, k. So, by this assumption and previous value, we get it as,
\[j=\dfrac{-b}{2a}+\dfrac{p}{2a};k=\dfrac{-b}{2a}-\dfrac{p}{2a}\]
p has 2 cases of being positive or negative as if it is root.
Case 1: p is positive. So,
p > 0
\[\sqrt{{{b}^{2}}-4ac}>0\]
As p is positive, we can say j’s value gets added up.
\[j>\dfrac{-b}{2a}\]
Case 2: The value of p is negative. So,
p < 0
\[\sqrt{{{b}^{2}}-4ac}<0\]
As p is negative, we can say k’s value gets added up
\[k>\dfrac{-b}{2a}\]
So, in both cases, one of them is greater than \[\dfrac{-b}{2a}\].
Therefore, option (c) is the correct option.

Note: Be careful while getting the conditions on discriminant as the whole answer depends on that value. While checking cases on p, be careful in which the value exceeds \[\dfrac{-b}{2a}\] because you have to see that the other value does not exceed. Only one must exceed or else the answer might change. So, be careful at this step of the categorization of p.