Evaluation of \( \dfrac{1}{\alpha} – \dfrac{1}{\beta} \)

Video Explanation

Question

If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial

\[ f(x) = ax^2 + bx + c, \]

evaluate

\[ \frac{1}{\alpha} – \frac{1}{\beta}. \]

Solution

Step 1: Use Relations Between Zeros and Coefficients

For the quadratic polynomial \( ax^2 + bx + c \),

\[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \]

Step 2: Simplify the Given Expression

\[ \frac{1}{\alpha} – \frac{1}{\beta} = \frac{\beta – \alpha}{\alpha\beta} \]

Step 3: Find \( \beta – \alpha \)

\[ (\alpha – \beta)^2 = (\alpha + \beta)^2 – 4\alpha\beta \]

\[ = \left(-\frac{b}{a}\right)^2 – 4\left(\frac{c}{a}\right) = \frac{b^2 – 4ac}{a^2} \]

\[ \Rightarrow \alpha – \beta = \frac{\sqrt{b^2 – 4ac}}{a} \]

Hence,

\[ \beta – \alpha = -\frac{\sqrt{b^2 – 4ac}}{a} \]

Step 4: Substitute the Values

\[ \frac{1}{\alpha} – \frac{1}{\beta} = \frac{-\dfrac{\sqrt{b^2 – 4ac}}{a}} {\dfrac{c}{a}} \]

\[ = -\frac{\sqrt{b^2 – 4ac}}{c} \]

Conclusion

The required value is:

\[ \boxed{\frac{1}{\alpha} – \frac{1}{\beta} = -\frac{\sqrt{b^2 – 4ac}}{c}} \]

\[ \therefore \quad \frac{1}{\alpha} – \frac{1}{\beta} = -\dfrac{\sqrt{b^2 – 4ac}}{c}. \]