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If the roots of the equation $a{{x}^{2}}+bx+c=0$ are real and distinct, then find the characteristics of roots.(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 the above.

Last updated date: 11th Jun 2024
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Hint: We start solving the problem by using the fact that discriminant should be greater than zero for roots to become real and distinct. We now assume a positive number for discriminant and substitute this in the roots of the quadratic equation. We get two cases related to the denominator of the roots to find the characteristics of the roots of the equation.

Given that we have a quadratic equation $a{{x}^{2}}+bx+c=0$ and it has real and distinct roots. We need to find characteristics of both roots.
We know that discriminant should be greater than zero for a quadratic equation to have real and distinct roots.
We know that discriminant $D=\sqrt{{{b}^{2}}-4ac}$.
We have got $\sqrt{{{b}^{2}}-4ac}>0$---(1).
We know that roots of quadratic equation $a{{x}^{2}}+bx+c=0$ are $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
We have the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ as $\dfrac{-b}{2a}\pm \dfrac{\sqrt{{{b}^{2}}-4ac}}{2a}$.
Let us assume that the value of the discriminant be ‘p’$\left( p>0 \right)$. So, we get the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ as $\dfrac{-b}{2a}\pm \dfrac{p}{2a}$.
We need to solve for two cases of ‘a’. (i) ‘a’ is positive (ii) ‘a’ is negative.
Let us assume the value of ‘a’ is positive. This makes $\dfrac{p}{2a}$ also a positive number.
We have the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ as $\dfrac{-b}{2a}\pm \dfrac{p}{2a}$. If $\dfrac{p}{2a}$ is added to $\dfrac{-b}{2a}$, then the value of the root exceeds $\dfrac{-b}{2a}$. If $\dfrac{p}{2a}$ is subtracted to $\dfrac{-b}{2a}$, then the value of the root will be less than $\dfrac{-b}{2a}$.
Let us assume the value of ‘a’ is negative. This makes $\dfrac{p}{2a}$ also a negative number
We have the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ as $\dfrac{-b}{2a}\pm \dfrac{p}{2a}$. If $\dfrac{p}{2a}$ is subtracted to $\dfrac{-b}{2a}$, then the value of the root exceeds $\dfrac{-b}{2a}$. If $\dfrac{p}{2a}$ is added to $\dfrac{-b}{2a}$, then the value of the root will be less than $\dfrac{-b}{2a}$.
We can see that from both cases of ‘a’, we got the value of one of the roots exceeds $\dfrac{-b}{2a}$ and other root is less than $\dfrac{-b}{2a}$.
∴ We get one of the roots exceed $\dfrac{-b}{2a}$ and another root is not.

So, the correct answer is “Option C”.

Note: We should not take the value of discriminant greater than or equal to zero here $\left( D\ge 0 \right)$ for this problem as the problem clearly stated that the roots are real and distinct. From this problem we have seen that both roots will not be greater than or less than the value of $\dfrac{-b}{2a}$. Similarly we can expect problems to find about properties of roots by giving about the properties of the ‘b’.