Evaluation of an Expression Using Zeros of a Quadratic Polynomial

Video Explanation

Question

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

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

evaluate

\[ \frac{1}{a\alpha + b} + \frac{1}{a\beta + b}. \]

Solution

Step 1: Write 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: Combine the Fractions

\[ \frac{1}{a\alpha + b} + \frac{1}{a\beta + b} = \frac{(a\beta + b) + (a\alpha + b)} {(a\alpha + b)(a\beta + b)} \]

Step 3: Simplify the Numerator

\[ (a\beta + b) + (a\alpha + b) = a(\alpha + \beta) + 2b \]

Substitute \( \alpha + \beta = -\frac{b}{a} \):

\[ a\left(-\frac{b}{a}\right) + 2b = -b + 2b = b \]

Step 4: Simplify the Denominator

\[ (a\alpha + b)(a\beta + b) = a^2\alpha\beta + ab(\alpha + \beta) + b^2 \]

Substitute \( \alpha\beta = \frac{c}{a} \) and \( \alpha + \beta = -\frac{b}{a} \):

\[ = a^2\left(\frac{c}{a}\right) + ab\left(-\frac{b}{a}\right) + b^2 \]

\[ = ac – b^2 + b^2 = ac \]

Step 5: Substitute Back

\[ \frac{1}{a\alpha + b} + \frac{1}{a\beta + b} = \frac{b}{ac} \]

Conclusion

The required value is:

\[ \boxed{\frac{b}{ac}} \]

\[ \therefore \quad \frac{1}{a\alpha + b} + \frac{1}{a\beta + b} = \frac{b}{ac}. \]